Statistics and Probability PB Allied The Leading Mathematics-10 Statistics 301 => Development and demonstration of skills of collecting, arranging, presenting and analyzing the data. => Basic concept of probability on solving the problems of daily life and demonstration of the skills of addition and multiplication of the probability. Marks Weightage : 11 Estimated Working Hours : Competency Learning Outcomes 1. To find the central tendencies and quartiles from the data. 2. To establish the addition principle of probability and solve related problems. 3. To develop the concept of independent and dependent events. 4. To find the probability by using multiplication principle. 5. To find the probability by using tree diagram. 13. Statistics 13.1 Mean of Grouped Data 13.2 Median of Grouped Data 13.3 Mode of Grouped Data 13.4 Quartiles of Grouped Data 13.5 Median and Quartiles from Ogive 14. Probability 14.1 Probability on Mutually Exclusive Events 14.2 Probability on Independent and Dependent Events 14.3 Probability By Tree Diagram Chapters / Lessons STATISTICS AND PROBABILITY VI UNIT 22 Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of Items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks VI Statistics and Probability 24 2 2 2 3 2 4 2 2 8 2 11 FEE Concept: "Maths is Fun, Maths is Easy and Maths is Everywhere"
Statistics and Probability 302 Allied The Leading Mathematics-10 Statistics 303 WARM-UP Statistics 1) Statistics: A branch of mathematics dealing with the collection, analysis, interpretation, and presentation of numerical data is called statistics. It is a crucial process which helps to make the decision based on the data. 2) Data (Singular form; Datum): Data are measurements or observations that are collected as a source of information them. There are a variety of different types of data, and different ways to represent. 3) Observation: An observation is a fact or figure we collect about a given variable. It can be expressed as a number or as a quality. 4) Array: An arrangement of numbers or symbols in rows and columns is called an array. In statistics, it is a group of numbers in rows and columns with the smallest at the beginning and the rest in order of size up to the largest at the end. 5) Tally Marks: Tally mark is a graphical representation of counting the observation of a datum using small vertical line. Graph table has one vertical line which is made for each of the first four numbers and the fifth number is represented by a diagonal line across the previous four. 6) Frequency: A frequency in statistics is the number of times an event or observation happened in an experiment or a data. It can also be defined simply as a count of a certain event. 7) Frequency Table: A frequency table is a way to present data by counting and ordering to summarize larger sets into the simplest form. 8) Classification of Data: (i) Raw or Primary Data and (ii) Secondary Data. 9) Types of Data: (i) Individual or personal Data: A collected data which can be associated with a single observation in a sample is called an individual data. For example, marks of 10 students: 39, 42, 45, 55, 56, 65, 75, 86, 89, 90. (ii) Discrete Data: A numerical or quantitative data that includes fixed numbers with specific values determined by counting observations, is called a discrete data orseries. For example, Observation (x) 35 45 55 65 75 85 Frequency (f) 5 6 8 9 12 7 (iii) Continuous Data: Continuous data includes complex numbers and varying data values measured over a particulartime interval, is called a continuous data orseries. For example, Observation (x) 30-40 40-50 50-60 60-70 70-80 80-90 Frequency (f) 4 8 9 7 5 8 10) Cumulative Frequency Table or Distribution: A cumulative frequency distribution is the sum of the observation/class and all observations/classes below it in a frequency distribution. e.g., Observation (x) 30-40 40-50 50-60 60-70 70-80 80-90 Frequency (f) 4 8 9 7 5 8 Cumulative Frequency (cf) 4 4+8=12 9+12=21 7+21=28 5+28=33 8+33=41 R E V I S I T I N G S T A T I S T I C S
Statistics and Probability 302 Allied The Leading Mathematics-10 Statistics 303 11) Types of Cumulative Frequency Table: (i) Less Than Cumulative Frequency Table: A table in which the frequencies are obtained by adding successively the frequencies from the lowest to the highest size. (ii) More or Greater Than Cumulative Frequency Table: A table in which the frequencies are obtained by adding successively the frequencies from the highest to the lowest size. 12) Ogive or Cumulative Frequency Curve: A freely drawn graph obtained by cumulative frequency table is called Ogive or cumulative frequency curve. 14) Representation of Data: Bar Graph Histogram Frequency Polygon Less Than Ogive More Than Ogive Double Ogive 15) Central Tendency: The statistical measure that identifies a single or central value as representative of an entire distribution is known as central tendency (CT). The most common measures of central tendency are the arithmetic mean, the median, and the mode. CT \ Data Definition Individual Data Descrite Data Arithmetic Mean The average value of the data is called mean (x). x = ∑x N x = ∑fx N Median The middle term of the data is called median (Md). It divides the data into equal part (50%). Md = ( N + 1 2 )th term Md = Value of ( N + 1 2 )th term in cf Mode The most frequent value in the data is called mode. Mo = Most repeated term Mo = Term with the highest frequency 16) Quartiles: The partition terms that divide a data into four equal parts, are called quartiles. There are three quartiles; first quartile (Q1), second quartile (Q2) and third quartile (Q3). Quartiles Definition Individual Data Descrite Data Q1 First or lower quartile (Q1) divides the data at 25%. Q1 = ( N + 1 4 )th term Q1 = ( N + 1 4 )th term in cf Q2 Second or middle quartile (Q2) divides the data at 50%. Q2 = ( 2(N + 1) 4 )th term Q2 = Value of ( 2(N + 1) 4 )th term in cf = Md Q3 Third or upper quartile (Q3) divides the data at 50%. Q3 = ( 3(N + 1) 4 )th term Q3 = Value of ( 3(N + 1) 4 )th term in cf R E V I S I T I N G S T A T I S T I C S
Statistics and Probability 304 Allied The Leading Mathematics-10 Statistics 305 13.1 Mean of Grouped Data At the end of this topic, the students will be able to: ¾ find the mean of grouped data. ¾ solve the problems on the mean of grouped data. Learning Objectives I Introduction The most common measures of central tendency are the arithmetic mean, the median, and the mode. The central tendencies are frequently used in insurance company, real-state and other business, human resource field, engineering, physics, report card, GDP, family budget, etc. In statistical analysis or a research work it is always important to find a single value or impression that convey fairly adequate idea of the whole group. In statistics, the single expression is the average of the given data. The averages that represent the whole mass of data may be; • Arithmetic Mean or Arithmetic Average (Mean) • Median and quartiles • Mode Discussthemedian andmode in graphs alongside. In this class, we study the averages under the grouped data of the continuous series. Activity 1 Consider an ungrouped data of obtained marks of some students: 22, 41, 22, 21, 17, 18, 27, 37, 32, 47, 48, 50, 40, 42, 37, 28, 31, 51, 57, 18, 42, 20, 49 Let us group the above marks as 10 by 10 of the interval 10 starting from 10 because the last mark is 17, between 10 and 20. In the interval 10 - 20, we take only the marks from 10 to 19 and so on. Intervals 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 No. of students 3 5 4 7 3 This type of the arranging data is called a grouped data. A data arranging in the class intervals as groups is called a grouped data of continuous series. The grouped data are as the following form: CHAPTER 13 STATISTICS Mean Median Mode Mean Median Mode Non-symmetric curve
Statistics and Probability 304 Allied The Leading Mathematics-10 Statistics 305 Class Interval x1 - x2 x2 - x3 x3 - x4 …….. xn–1 - xn Frequency (f) f1 f2 f3 …….. fn The average value that represents whole grouped data is called the arithmetic mean or average. It is computed by average of the product of frequency and mid-value of the class interval. II Computation of Mean by Direct Method For the grouped data or the continuous series where data are grouped into frequency class, the arithmetic mean is obtained by using the formula; x = ∑fm N , where m = mid–value of the class interval. Look at the following example: Example 1: Table below gives the expenditure in rupees on water consumption in 50 houses in certain locality. Find the average expenditure in rupees. Expenditure on water (Rs.) 50 - 60 60 - 70 70 - 80 80 - 90 90 – 100 Number of houses (f) 6 15 20 5 4 Solution: Making fm–table, Expenditure on water (Rs.) 50-60 60-70 70-80 80-90 90-100 Total Number of houses (f) 6 15 20 5 4 f = N = 50 Mid-value (m) 55 65 75 85 95 –––––––– fm 330 975 1500 425 380 fm = 3610 Now, we know that Arithmetic mean ( x ) = ∑fm N = 3610 50 = 72.20. ∴ The average expenditure on water is Rs. 72.20. Computation of Mean by Assume Mean (Short-cut) Method Handling the huge mass of data and working with larger sum is somehow hard and tedious job in statistics. In such situation, we use the short cut method. For this we use the assumed mean; work out the deviation from the mean and use the following formula; x = A + ∑fd N , where A = Assumed mean, f = frequency, and d = m – A, deviation of the assumed mean from the mid-value. Example 2: Table below gives the wage distribution of the workers in a factory. Find the average wage paid by the factory: Wage (Rs.) 100 - 200 200 - 300 300 - 400 400 - 500 500 - 600 No. of workers (f) 55 80 95 40 30 Solution: Computation of average wage, suppose A = 350,
Statistics and Probability 306 Allied The Leading Mathematics-10 Statistics 307 Wage (x) Mid–value (m) f d = m – 350 fd 100 - 200 200 - 300 300 - 400 400 - 500 500 - 600 150 250 350 450 550 55 80 95 40 30 – 200 –100 0 100 200 – 11000 – 8000 0 4000 6000 N = 300 ∑fd = – 9000 Now, we have Arithmetic mean ( x ) = A + ∑fd N = 350 + – 9000 300 = 350 – 30 = 320. Hence, the average wage is Rs. 320. Computation of Mean by Step Deviation Method If we multiply and divide a number by some quantity, the number does not change. We can apply this technique to make the calculation short cut by a slight modification in the above formula as, x = A + Σfd' N × i, where; d' = m – A i and i = class interval. This method is used for large data. Let us workout the average wage of the worker from the data given above using this short cut method. Suppose A = 350 and i = 100. Wage (Rs.) (x) 100 - 200 200 - 300 300 - 400 400 - 500 500 - 600 Total No. of workers (f) 55 80 95 40 30 N = 300 Mid–value (m) 150 250 350 450 550 –– d' = m – A i –2 –1 0 1 2 – fd' –110 –80 0 40 60 Σfd' = – 90 Now, we have x = A + Σfd' N × i = 350 + – 90 300 × 100 = 350 – 30 = 320 ∴ Average wage of the workers is Rs. 320. Alert to Arithmetic Mean/s Don't confuse ∑fx and ∑fm. ∑fx is used for discrete data and ∑fm is used for continuous data.
Statistics and Probability 306 Allied The Leading Mathematics-10 Statistics 307 Points to be Remembered 1. The average value of the data is called mean (x). 2. The formula to calculate the mean of continuous data by direct method is x = ∑fm N , where m = mid-value of each class. 3. The formula to calculate the mean of continuous data by short-cut or assumed mean method is, x = A + ∑fd N , where A = Assumed mean, f = frequency, and d = m – A, deviation of the assumed mean from the mid-value (m). 4. The formula to calculate the mean of continuous data by step deviation method is, x = A + Σfd' N × i, where; d' = m – A i and i = class interval, A = Assumed mean, f = frequency. Example-1 (a) Write the formula to calculate the mean of grouped data by short cut method. (b) If the total number of frequencies (N) = 75 + a, ∑fm = 230 + 12a and mean (x) = 7, find the value of a. (c) Check the value of a. Solution: (a) The formula to calculate the mean of grouped data is x = A + ∑fd N , A = assumed mean, d = m – A and m = Mid-value of class interval. (b) Here, N = 75 + a, ∑fm = 230 + 12a and x = 7 Now, we have x = ∑fm N or, 7 = 230 + 12a 75 + a or, 230 + 12a = 525 + 7a or, 12a – 7a = 525 – 230 or, 5a = 295 or, a = 59. (c) Now, putting the value of a in N and ∑fm, we get N = 75 + 59 = 134, ∑fm = 230 + 12 × 59 = 938 ∴ x = ∑fm N = 938 134 = 7, which is true. Hence, the value of a is correct. Example-2 The mean of the given data is 390. Wage (Rs.) 150-250 250-350 350-450 450-550 550-650 No. of workers 4 6 12 p 3 (a) In which series does the given data represent ? (b) Construct fm-table on the basis of the given data.
Statistics and Probability 308 Allied The Leading Mathematics-10 Statistics 309 (c) What is the value of p in the given data? Find it. (d) Find the average wages of the workers of the two maximum frequencies. Solution: (a) The given data represents in continuous series. (b) Making fm-table; Wage (Rs.) (x) 150- 250 250- 350 350- 450 450- 550 550- 650 Total No. of workers (f) 4 6 12 p 3 N = 25 + p Mid-value (m) 200 300 400 500 600 fm 800 1800 4800 500p 1800 fm = 9200 + 500p (c) Now, we know that. x = ∑fm N or, 390 = 9200 + 500p 25 + p or, 9200 + 500p = 9750 + 390p or, 110p = 550 ∴ p = 5 (d) The maximum two frequencies are 12 and 6. Now, from the above table, The required average wages of the workers that have two maximum frequencies = 12 × 400 + 6 × 300 6 + 12 = 6600 18 = Rs. 366.67 Example-3 Observe the obtained marks of the full marks 50 in the table below. Obtained Marks (X) 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50 No. of std. (f) 5 7 16 18 20 (a) Construct fm-table by arranging class intervals. (b) Find the average mark of the students. (c) If the pass mark is 35%, how much percent is more or less the average mark than pass mark? Solution: (a) Making fm-table by arranging class intervals; Marks (X) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Total No. of std. (f) 5 2 9 2 2 N = 20 Mid-value (m) 5 15 25 35 45 ------ fm 25 30 225 70 90 fm = 440 (b) Now, we have Average mark (x) = ∑fm N = 440 20 = 22.
Statistics and Probability 308 Allied The Leading Mathematics-10 Statistics 309 (c) The pass mark is 35% of 50 = 35 100 × 50 = 17.5 marks. Here, 22 > 17.5. So, the average mark is more than the pass mark. ∴ More percent in average mark than pass mark = 22 – 17.5 17.5 × 100% = 4.5 17.5 × 100% = 25.71% Example-4 Observe the obtained marks of the full marks 50 in the table below. Class Interval (X) 0-100 200-300 400-500 600-700 800-900 Frequency (f) 7 9 8 6 5 (a) Construct fm-table. (b) Find the arithmetic mean of the students. Solution: (a) Making fm-table by arranging class intervals; Class Interval (X) 0-10 20-30 40-50 60-70 80-90 Total Frequency (f) 7 9 8 6 5 N = 35 Mid-value (m) 5 25 45 65 85 ------ fm 35 225 360 390 425 fm = 1435 (b) Now, we have Arithmetic mean (x) = ∑fm N = 1435 35 = 41. PRACTICE 13.1 Read Think Understand Do Keeping Skill Sharp 1. (a) Write the definition of the mean of a grouped data. (b) Write the formula to calculate the arithmetic mean of a continuous data when the sum of frequency Σf and the sum of product of frequency and the mid value m of the classinterval (c) If Σfm = 5k + 25 and Σf = k + 5 in a data, what is its mean? (d) In a grouped data if x = 35.5 and N = 14, what is the value of Σfm? (e) What is the meaning of Σfm in a grouped data ? 2. Circle ( ) the correct answer. (a) Which formula is used for finding the mean of the grouped data? (i) x = ∑fm N (ii) x = A + ∑fd N (iii) x = A + Σfd' N × i (iv) All of the above Not necessary to arrange the data in continuous series for Mean
Statistics and Probability 310 Allied The Leading Mathematics-10 Statistics 311 (b) If Σfm = 1794 and N = 23 in a continuous data, what is its mean ? (i) 1817 (ii) 1771 (iii) 78 (iv) 23 (c) If x = 25 and N = 5 in a continuous data, what is the value of Σfm? (i) 125 (ii) 100 (iii) 25 (iv) 5 (d) If x = 18 and Σfm = 180 in a continuous data, what is the value of Σf ? (i) 15 (ii) 10 (iii) 5 (iv) 32400 (e) If Σfm = 20 + 4a and N = 5 + a in a grouped data, what is its mean? (i) 24 (ii) 4 (iii) 25 (iv) 5 (f) In a grouped data, the sum of the ages of 50 students is 2450 years. Find the average age. (i) 49 (ii) 2450 (iii) 2500 (iv) 2400 Check Your Performance 3. (i) (a) Write the formula to calculate the mean of grouped data by direct method. (b) If the total number of frequencies (N) = 17 + p, ∑fm = 240 + 15p and mean (x) = 14.25, find the value of p. (c) Check the value of p. (ii) (a) Write the formula to calculate the mean of grouped data by step-deviation method. (b) In the continuous data, Σfx = 538 + 53n, N = 50 + 3n and x = 15.2. Find the value of n. (c) Check the value of n. Answer the following questions by observing the given data. 4. (i) Marks obtained by students in a monthly test of 50 marks are given below: 15, 25, 10, 12, 35, 20, 5, 7, 29, 42, 44, 18, 17, 22, 25, 33, 41, 34, 36, 37, 17, 19, 22, 32, 16, 18, 44, 33, 32, 34, 22, 37, 28, 24, 36, 25, 23, 34, 25, 28. (ii) A survey of mobile user of 50 students in a week is given below (in hours): 12, 13, 15, 35, 50, 40, 33, 32, 22, 14, 16, 17, 35, 32, 44, 22, 20, 18, 20, 12, 25, 16, 20, 19, 22, 21, 18, 15, 25, 18, 32, 35, 32, 34, 17, 19, 21, 23, 24, 13, 42, 27, 29, 40, 42, 15, 18, 21, 22, 17. (a) Construct the frequency table at the class interval of 10 using tally bars. (b) Find the arithmetic mean of the data. (c) How many students are below mean value? Write it by counting the raw data. 5. (i) Age in years 3-5 5-7 7-9 9-11 11-13 13-15 No. of students 50 65 80 95 70 40 (ii) Marks 10-20 20-30 30-40 40-50 50-60 No. of students 35 62 77 88 38 (iii) Expenditure (in Rs.) 40 ≤ x < 50 50 ≤ x < 60 60 ≤ x < 70 70 ≤ x < 80 80 ≤ x < 90 No. of students 135 150 200 160 155
Statistics and Probability 310 Allied The Leading Mathematics-10 Statistics 311 (a) What is the name of the given data? (b) Construct fm-table. (c) Find its mean. (d) How many students are below and up the class containing the mean value? 6. (i) The arithmetic mean of the following data is 41. Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 No. of students 1 4 6 4 14 p 3 1 (ii) The mean of the following data is 145. Height (cm) 0-50 50-100 100-150 150-200 200-250 250-300 No. of Students 8 15 32 p 12 7 (a) Construct fm-table. (b) Find the value of p. 7. (i) Height (in Inch) 0-10 0-20 0-30 0-40 0-50 0-60 No. of Children 8 11 15 22 25 30 (ii) Expenditure 20-70 30-70 40-70 50-70 60-70 No. of Students 25 21 15 9 4 (iii) Marks Obtained Below 20 Below 30 Below 40 Below 50 Below 60 No. of Students 8 12 19 25 30 (iv) Marks 10 More 20 More 30 More 40 More 50 More No. of Students 50 42 35 20 8 (a) Construct fm-table by arranging class intervals. (b) Find the average mark of the students. (c) If the pass mark is 35%, what percent is more or less the average mark than pass mark? 8. (i) Ages (in Month) 10-19 20-29 30-39 40-49 50-59 60-69 No. of Children 5 3 8 8 9 7 (ii) Height (cm) 0-10 20-30 40-50 60-70 80-90 No. of Students 4 8 12 9 7 (a) Construct fm-table. (b) Find the mean of the data. 3. (i) (a) x = ∑fm N (b) 3 (ii) (a) x = A + ∑fd' N × i (b) 30 4. (i) (b) 26.75 (c) 22 (ii) (b) 25.6 (c) 33 5. (i) (a) Grouped data (c) 8.95 (d) 115, 205 (ii) (a) Continuous data (c) 36 (d) 97, 126 6. (i) (b) 7 (ii) (b) 26 7. (i) (b) 28 (c) 33.33% (ii) (b) 44.6 (c) 82.04% (iii) (b) 38.78 (c) 84.6% (iv) (b) 34 (c) 61.9% 8. (i) (b) 43 (ii) (b) 48.5% Answers
Statistics and Probability 312 Allied The Leading Mathematics-10 Statistics 313 13.2 Median of Grouped Data At the end of this topic, the students will be able to: ¾ find the median of grouped data. ¾ solve the problems on the median of grouped data. Learning Objectives I Introduction Activity 1 Observe the line segments in the alongside. Which is the middle line segment in length among them? At first, measure all the line segments and arrange the line segments from shorter to longer as shown in the figure. Oh! The line segment AB is in the middle. So, it is the middle line segment in length. Now, what is median ? A median is a value or item, which divides the total number of values or items, or observations into two equal halves. This situation generally arises when data or values are arranged in a series. So, median is a kind of positional average. To find the median of the grouped data, it should be arranged in ascending or descending order and then computing the cumulative frequency of the data: Class Interval x1 - x2 x2 - x3 x3 - x4 ….. xn–1 - xn Frequency (f) f1 f2 f3 ….. fn Cumulative frequency(cf) f1 f1 + f2 f1 + f2 + f3 f1 + f2 + f3 + .... + fn From the cumulative frequency table, we identify the median class and find out the value of median. II Computation of Median If the values are continuous or the data are presented in the form of frequency distribution with class intervals, then median lies in the class corresponding to the N 2 th item and the formula for calculating the median is given below: A B C E G I K M N L H J F D G H 2 cm I J 1.7 cm A B 1.5 cm M N 1.4 cm K L 0.8 cm 1.2 cm E F 1.8 cm C D
Statistics and Probability 312 Allied The Leading Mathematics-10 Statistics 313 Median (Md) = L + = N 2 – cf f × h, Where, L = lower limit of the median class, N = total frequency cf = cumulative frequency preceding the median class, f = frequency of the median class, h = size or class interval of the median class Look at the following example; Find the median for the following data: Marks (x) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency (f) 5 18 32 24 11 Representing the data in cumulative frequency table, we get, Marks (x) 0 - 10 0 - 10 20 - 30 30 - 40 40 - 50 Total Frequency (f) 5 18 32 24 11 N = 90 cf 5 23 55 79 90 same Now; N 2 = 90 2 = 45 ∴ Median lies in the class where the cumulative frequency is at most ≥ 45. Here, 45th item lies in the class where cumulative frequency is 55. Hence median class = 20 – 30. This means, the median lies in between 20 – 30. In such situation we use the following formula to estimate the median. Median (Md) = L + N 2 – cf f × i. Where, L = Lower limit of the class where median lies. cf = cumulative frequency of the class preceding the median class. f = frequency of the median class; and i = class interval For the data given above, L = 20 cf = 23 f = 32 i = 10 N 2 = 45 ∴ Median (Md) = L + N 2 – cf f × i = 20 + 45 – 23 32 = 20 + 22 32 × 10 = 20 + 6.88 = 26.88.
Statistics and Probability 314 Allied The Leading Mathematics-10 Statistics 315 III Median From Ogive We can easily compute the median class and the value of median from the Ogive or cumulative frequency curve. Observe the less than Ogive of the given data alongside. x 5-15 15-25 25-35 35-45 45-55 f 15 8 10 20 17 cf 15 23 33 53 70 Here, the total number of students (N) = 70 The position of median is N 2 = 70 2 = 35 Draw a line parallel to x-axis from the point 35 on the vertical scale that intersects the Ogive at a point. From the intersecting point, draw a perpendicular line to x-axis that falls between (25 - 35). Thus, the required median class is (25 - 35). The value of median is approximately 32. We can easily find the median class and the value of median from the more than Ogive as mentioned above. But, in the single graph of the less than and greater than Ogive, the less than Ogive and greater than Ogive intersect at a point. From that point, draw a perpendicular line to x-axis that falls at the point 108 between (100 - 120). Thus, the required median class is (100 - 120) and the value of the median is 108. Points to be Remembered 1. The middle item of the data is called median (Md). It is positional average item of the data and divides the total number of values or items, or observations into two equal halves. 2. Median lies in the class corresponding to the N 2 th item/class and this value is looked with cumulative frequency of the data. This class is called median class. 3. The value of median is calculated by the formula; Median (Md) = L + = N 2 – cf f × h, where, L = lower limit of the median class, N = total frequency, cf = cumulative frequency preceding the median class, f = frequency of the median class, h = size or class interval of the median class 0 60 80 100 120 140 Class Intervals Cumulative Frequency 160 180 20 40 60 80 100 120 140 160 180 Less than ogive Greater than ogive 60 30 70 40 50 20 10 0 5 15 25 35 45 55 65
Statistics and Probability 314 Allied The Leading Mathematics-10 Statistics 315 Example-1 The given data shows the wages of some workers worked in a factory. Wage in Rs/hr 15-25 25-35 35-45 45-55 55-65 No. of workers 4 6 12 8 6 (a) In which series does the given data represent ? (b) Construct cumulative frequency (cf) table on the basis of the given data. (c) Find the median wages of the workers. (d) Where the workers are more above or below the median class ? Solution: (a) The given data represents in continuous series. (b) Constructing cumulative frequency table, Wages (x) Numbers of workers (f) Cumulative frequency (cf) 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65 4 6 12 8 6 4 10 22 30 36 N = 36 (c) Now, we have Median class = N 2 th class = 36 2 th class = 18th class < 22 = (35 – 45). Here, L = 35, cf = 10, f = 12, i = 10 We know, Median (Md) = L + N 2 – cf f × i = 35 + 18 – 10 12 × 10 = 35 + 8 12 × 10 = 35 + 6.67 = 41.67. Thus, the medium wages of the workers is Rs. 41.67 per hour. (d) The number of the workers above the median class = 8 + 6 = 14. The number of the workers above the median class = 6 + 4 = 10. Hence, the workers are more above the median class.
Statistics and Probability 316 Allied The Leading Mathematics-10 Statistics 317 Example-2 The median mark of the given data is 24. Marks 0 - 10 10 - 20 30 - 40 40 - 50 20 - 30 No. of Students 4 12 9 5 x (a) Construct cumulative frequency (cf) table by arranging class intervals. (b) Find the value of the x. (c) Find the total number of students who appeared in the exam. Solution: (a) Constructing the cf-table by arranging the given data in ascending order; Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Total No. of Frequency 4 12 x 9 5 30 + x cf 4 16 16 + x 25 + x 30 + x ---- (b) By given, Median (Md) = 24. So, Median Class = (20 - 30) where, L = 20, cf = 16, f = x, i = 10 We know, Median (Md) = L + N 2 – cf f × i or, 24 = 20 + 30 + x 2 – 16 x × 10 or, 24 – 20 = 30 + x – 32 2x × 10 or, 4 = 30 + x – 32 2x × 10 or, 8x = (x – 2) × 10 or, 8x = 10x – 20 or, 2x = 20 ∴ x = 10 (c) The total number of students who appeared in the exam, is 30 + 10 = 40. Example-3 The given table shows the information about the height of the plants. Height (in cm) 10 -19 20 - 29 30 - 39 40 - 49 50 - 59 No. of Plants 5 7 8 6 4 (a) In which series does the given grouped data represent ? (b) Construct cumulative frequency (cf) table by expressing in continuous series. (c) Find the median height of the plants. Solution: (a) The given grouped data represents the discontinuous series. (b) Since the given grouped data is discontinuous, so each class makes continuity with other classes by using correction factor.
Statistics and Probability 316 Allied The Leading Mathematics-10 Statistics 317 Correction factor = Lower limit of second interval – Upper limit of first interval 2 = 20 – 19 2 = 1 2 = 0.5 The class intervals are made continuous by adding 0.5 in upper limit and subtracting 0.5 in lower limit of each class interval. Constructing cumulative frequency (cf) table by expressing in continuous series, Marks 10 - 19.5 19.5 - 29.5 29.5 - 39.5 39.5 - 49.5 49.5 - 59.5 Total No. of Frequency 5 7 8 6 4 30 cf 5 12 20 26 30 ---- (c) Now, Median class = N 2 th class = 30 2 th class = 15th class < 20 = (29.5 - 39.5) Here, L = 29.5, f = 8, cf = 12, i = 10 We know, Median (Md) = L + N 2 – cf f × i = 29.5 + 15 – 12 8 × 10 = 29.5 + 3 8 × 10 = 29.5 + 3.75 = 33.25. Thus, the medium height of the plants is Rs. 33.25 cm. PRACTICE 13.2 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the median of a continuous data? (b) What is the median class in the continuous data having N terms? (c) Write the formula to calculate the media of a continuous data. (d) What is the position of the median class in the grouped data having 38 terms? (e) Which is the lower limit in the median class (30 – 40)?
Statistics and Probability 318 Allied The Leading Mathematics-10 Statistics 319 (f) What are the median classes in the following Ogives ? 5 15 25 35 10 20 30 40 O 10 20 30 40 50 60 70 80 5 15 25 35 10 20 30 40 O 10 20 30 40 50 60 70 80 2 6 10 14 4 8 12 16 O 10 20 30 40 50 60 70 80 (i) (ii) (iii) 2. Circle ( ) the correct answer. (a) Which formula is used to calculate the median of the grouped data? (i) Median (Md) = L + N 2 + cf f × i (ii) Median (Md) = L + ( N 2 – cf) × i f (iii) Median (Md) = L – N 2 – cf f × i (iv) Median (Md) = L – N 2 + cf f × i (b) In the median formula Md = L + ( N 2 – cf) × i f , what is the meaning of L ? (i) Low value (ii) Lower limit (iii) Upper limit (iv) Lower median (c) In the median formula Md = L + N 2 – cf f × i, what is the meaning of f ? (i) Frequency of upper class (ii) Frequency of lower class (iii) Frequency of median class (iv) Frequency of preceding of median class (d) Which is the median class in the given data ? Marks 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Cumulative Frequency 4 5 3 5 3 (i) 20 - 30 (ii) 30 - 40 (iii) 40 - 50 (iv) 10 - 20 (e) What is the frequency of the median class in the given data ? Height (in inch) 24-28 28-30 30-32 32-34 34-36 36-40 No. of Students 8 6 2 4 7 3 (i) 6 (ii) 7 (iii) 4 (iv) 2
Statistics and Probability 318 Allied The Leading Mathematics-10 Statistics 319 Check Your Performance Answer the following questions by observing the given data. 3. (i) The expenditure in rupees on Tiffin of 45 students in a school is given below. 25, 26, 20, 28, 24, 30, 32, 50, 52, 51, 60, 28, 15, 45, 35, 32, 31, 17, 26, 30, 36, 25, 46, 31, 42, 25, 24, 20, 26, 32, 30, 20, 25, 35, 40, 52, 62, 65, 25, 35, 45, 25, 38, 29, 42. (ii) The electricity consumption in 30 families for the month of Shrawan-2080 is given below in units. 75, 56, 54, 44, 67, 65, 61, 52, 50, 62, 45, 42, 54, 62, 52, 68, 79, 60, 72, 53, 59, 49, 41, 45, 72, 76, 53, 46, 41, 58. (a) In which data does the given information represent ? (b) Construct the frequency table at the class interval of 10 for (a) and 5 for (b) using tally bars. (c) Find the median units of the data. (d) How many students are below median value? Write it by counting the raw data. 4. (i) The heights of 30 pea plants were measured, correct to the nearest cm, six weeks after planting. The frequency distribution table is given below. Find the median. Height (cm) 4 - 6 7 - 9 10 - 12 13 - 15 16 - 18 19 - 21 22 - 24 Frequency (f) 1 2 11 8 5 2 1 (ii) Find out median from the data given below: Marks 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Cumulative frequency 10 35 77 100 108 (a) In which series does the given data represent ? (b) Construct frequency distribution table with cumulative frequencies. (c) Find the median value of the given data. (d) Where the frequency is more above or below the median class ? 5. (i) The given data shows the marks obtained by 30 students of a class. Marks 0–10 0–20 0–30 0–40 0–50 No. of students 1 10 21 28 30 (ii) The given data shows the height of some students of a class. Height (in cm) 40–50 30–40 50–60 80–90 70–80 60–70 No. of students 4 6 7 8 7 5 (a) Construct cumulative frequency (cf) table by arranging classes. (b) Find the median value of the given data. (c) Find the average value of the students above the median class.
Statistics and Probability 320 Allied The Leading Mathematics-10 Statistics 321 6. (i) The median of the given table is 40. Marks obtained 10–20 20–30 40–50 30–40 50–60 No. of students 4 6 k 15 5 (ii) The median of the given table is 45. Weight (in kg) 10–20 20–30 30–40 40–50 50–60 60–70 70–80 No. of students 1 4 5 4 11 k 3 (a) Construct cumulative frequency (cf) table by arranging classes. (b) Find the value of k. (c) Find the total number of students. 7. (i) Observe the given data below. Age ( in years) 0-9 10-19 20-29 30-39 40-49 No. of people 6 4 5 3 2 (ii) The median of the given table is 45. Weight (in kg) 50-99 150-199 250-299 350-399 450-499 550-599 No. of students 5 8 9 12 7 9 (a) In which series does the given grouped data represent ? (b) Construct cumulative frequency (cf) table by expressing in continuous series. (c) Find the median value of the data. 3. (i) (a) Raw data (c) 32.69 (d) 27 (ii) (a) Raw data (c) 55 (d) 15 4. (i) (a) Discontinuous series (c) 12.875 (d) below (ii) (a) Discontinuous series (c) 49.04 (d) below 5. (i) (b) 24.54 (c) 37.22 (ii) (b) 63 (c) 80.33 6. (i) (b) 20 (c) 50 (ii) (b) 1 (c) 24 7. (i) (a) Discontinuous series (c) 19.5 (ii) (a) Discontinuous series (c) 349.5 Answers
Statistics and Probability 320 Allied The Leading Mathematics-10 Statistics 321 13.3 Mode of Grouped Data At the end of this topic, the students will be able to: ¾ find the mode and range of grouped data. Learning Objectives I Introduction In statistics, we have learned that the mode is the most repeated value among the given data. Among the given set of observations, mode is the value which has the maximum frequency. Sometimes, it is possible to have more than one value which has the same maximum frequency. In such a case, we can say that the data is multi-modal. Mode is used to find the ideal size or item from a collection of items. For example: in business forecasting, ready-made manufacturing garments, manufacturing shoes, etc. Suppose a watch manufacturer is designing a series of watches. To know which of those is sold most and in high demand, we can use mode. The same way when a manufacturer designs garments. He can know the high demand size or colour using mode. Here, the mode is the most appropriate measure because it shows the highest number of repetitions in the series. Now, we are going to learn about the mode of the grouped data with its formula. In the case of grouped data, it is not possible to identify the mode of the data, by looking at the frequency of data. In this scenario, we can determine the mode value by locating the class with the maximum frequency called modal class. Inside a modal class, we can locate the mode value of the data by using the formula, Mode (Mo) = l + f1 – f0 2f1 – f0 – f2 × i OR Mode (Mo) = l + ∆1 – f0 ∆1 + ∆2 × i where, ∆1 = f1 – f0, ∆2 = f1 – f2 f1 = the frequency of the modal class f0 = the frequency of the class preceding the modal class f2 = the frequency of the class succeeding the modal class i = the size of the class intervals l = the lower limit of the modal class A sample of grouped data is as below; Mode Median Mean
Statistics and Probability 322 Allied The Leading Mathematics-10 Statistics 323 Class (x) a - b c - d e - f j - k l - m l - m ...... y - z Frequency (f) fa fc f1 f0 f1 Modal Class f2 ..... fy Now, let us understand how to find the mode of grouped data using this formula with the help of an example. Example-1: Compute the mode for the given frequency distribution. Size (x) 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24 24-28 28-32 32-36 36 - 40 Frequency (f) 5 7 9 17 12 10 6 3 1 0 Solution: In the given data, 17 is the maximum frequency and the corresponding class is 12 - 16. So, the modal class is 12 - 16 such that, f1 = 17, l = 12, i = 4, f0 = 9 and f2 = 12. Now, we know that mode (Mo) = l + f1 – f0 2f1 – f0 – f2 × i = 12 + 17 – 9 34 – 9 – 12 × 4 = 12 + 8 13 × 4 = 12 + 32 13 = 12 + 2.46 = 14.46 Therefore, the mode of the given distribution is 14.46. Relation between Mean, Median and Mode Mean is the average of the data set which is calculated by adding all the data values together and dividing it by the total number of data sets. Median is the middle value among the observed set of values and is calculated by arranging the values in ascending order or in descending order and then choosing the middle value. Mode is the number from a data set which has the highest frequency and is calculated by counting the number of times each data value occurs. In the skewed (neither symmetric nor normal) distribution, the difference between mean and mode is almost equal to three times the difference between the mean and median. i.e., Mean – Mode = 3(Mean – Median) OR Mode = 3Median – 2Mean This relationship among mean, median and mode is known as the “empirical relationship” which is also defined as Mode is equal to the difference between 3 times the median and 2 times the mean.
Statistics and Probability 322 Allied The Leading Mathematics-10 Statistics 323 Mean > Median > Mode Positively skewed frequency distribution Negatively skewed frequency distribution x variable Mean = Median = Mode Mode Mode Median Median Mean Mean Frequency Mean < Median < Mode Mean = Median = Mode Example-1 It is given that in a frequency distribution, median = 10 and mean = 12. (a) Differentiate between mean, median and mode. (b) Write the relation between mean, median and mode. (c) Find the approximate value of the mode. (d) Compare the mean, median and mode. Solution: (a) Mean, median and mode are three measures of central tendency. Mean is the average value of the data that is obtained by dividing the sum of the observations by the number of observations. Median is the middlemost item in the ordered list of observations, whereas the mode is the most frequently occurring item. (b) The relation among mean, median and mode is, Mode = 3 media – 2 mean (c) Here, median = 10 and mean = 12, value of the mode = ? Now, we know that the relation among mean, median, and mode, Mode = 3 media – 2 mean = 3 × 10 – 2 × 12 = 30 – 24 = 6. Therefore, the value of mode is 6. (d) Mode < Median < Mean Example-2 Observe the given data and answer the following questions: Weight (x) 60-62 63-65 66-68 69-71 72-74 No. of Students (f) 5 18 42 27 8 (a) In which series does the given grouped data represent ? (b) Arrange the given data in continuous series. (c) Find the mode. Solution: (a) The given grouped data represents the discontinuous series. (b) Since the given grouped data is discontinuous, so each class makes continuity with other classes by using correction factor.
Statistics and Probability 324 Allied The Leading Mathematics-10 Statistics 325 Correction factor = Lower limit of second interval – Upper limit of first interval 2 = 63 – 62 2 = 1 2 = 0.5 The class intervals are made continuous by adding 0.5 in upper limit and subtracting 0.5 in lower limit of each class interval. Arranging the given data in continuous series, Weight (x) 59.5-62.5 62.5-65.5 65.5-68.5 68.5-71.5 71.5-74.5 No. of Students (f) 5 18 42 27 8 (c) In the given data, 42 is the maximum frequency and the corresponding class is 65.5-68.5. So, the modal class is 65.5-68.5 such that, f1 = 42, l = 65.5 i = 3, f0 = 18 and f2 = 27. Now, we know that mode (Mo) = l + f1 – f0 2f1 – f0 – f2 × i = 65.5 + 27 – 18 84 – 18 – 27 × 3 = 65.5 + 9 39 × 3 = 65.5 + 9 13 = 65.5 + 0.69 = 66.19 Therefore, the mode of the given distribution is 66.19. PRACTICE 13.3 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the mode of grouped data? (b) Write the modal class in the given data. x 5-10 10-15 15-20 20-25 25-30 f 5 7 12 6 7 2. Circle ( ) the correct answer. (a) Which formula is used to calculate the mode of the grouped data? (i) Mode (Mo) = l + f1 + f0 2f1 + f0 + f2 × i (ii) Mode (Mo) = l + f1 – f0 2f1 + f0 + f2 × i (iii) Mode (Mo) = l + f1 – f0 2f1 – f0 – f2 × i (iv) Mode (Mo) = l + f1 + f0 2f1 – f0 – f2 × i
Statistics and Probability 324 Allied The Leading Mathematics-10 Statistics 325 (b) Which formula is used to calculate the mode of the grouped data? (i) Mode (Mo) = l + ∆1 + f0 ∆1 + ∆2 × i (ii) Mode (Mo) = l + ∆1 – f0 ∆1 + ∆2 × i (iii) Mode (Mo) = l + ∆1 + f0 ∆1 – ∆2 × i (iv) Mode (Mo) = l – ∆1 – f0 ∆1 + ∆2 × i (c) In the mode formula Mo = l + f1 – f0 2f1 – f0 – f2 × i, what is the meaning of f1 ? (i) Frequency of the median class (ii) Frequency of the modal class (iii) Frequency of the middle class (iv) Lower limit of modal class (d) In the mode formula Mo = l + f1 – f0 2f1 – f0 – f2 × i, what is the meaning of f0 ? (i) Frequency of the class preceding the modal class (ii) Frequency of the class successive the modal class (iii) Frequency of the modal class (iv) Frequency of the median class (e) Which is the modal class in the given data below ? x 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 f 5 9 7 6 5 (i) 10 - 20 (ii) 30 - 40 (iii) 40 - 50 (iv) 20 - 30 Check Your Performance 3. (i) It is given that in a frequency distribution, median = 21 and mean = 16. (a) Why do mean, median and mode of a data differentiate ? (b) Write the relation among mean, median and mode. (c) Find the value of the mode. (d) Compare the mean, median and mode. (ii) It is given that in a frequency distribution, mean = 25 and mode = 20. (a) Find the value of the median. (b) Compare the mean, median and mode. (iii) It is given that in a frequency distribution, mean = 40 and mode = 32. (a) Find the value of the median. (b) Compare the mean, median and mode. Answer the following questions by observing the given data. 4. (i) x 0-20 20-40 40-60 60-80 80-100 f 7 10 12 6 4 (ii) Height (in inch) 20-25 25-30 30-35 35-40 40-45 45-50 No. of Students 7 3 11 5 2 9
Statistics and Probability 326 Allied The Leading Mathematics-10 Statistics 327 (iii) x 5-10 10-15 15-20 20-25 25-30 f 3 4 8 9 5 (iv) Height (in inch) 10-20 20-30 30-40 40-50 50-60 60-70 No. of Students 10 15 25 30 11 6 (a) In which series does the given grouped data represent ? (b) Find the mode. 5. (i) x 40-49 50-59 60-69 70-79 80-89 90-99 f 6 8 12 14 7 3 (ii) Height (in inch) 20-29 30-39 40-49 50-59 60-69 No. of Students 8 12 20 7 3 (iii) Height (in inch) 51-60 41-50 31-40 21-30 11-20 1-10 No. of Students 4 7 10 8 6 5 (a) In which series does the given grouped data represent ? (b) Arrange the given data in continuous series. (c) Find the mode. 6. (i) x 10-20 30-40 40-50 50-60 60-70 f 9 8 6 4 3 (ii) x 5-15 15-25 25-35 35-45 45-55 55-65 y 12 15 18 12 10 6 (a) Find the mean, median and mode. (b) Compare the mean, median and mode. 3. (i) (c) 31 (d) Mean < Median < Mode (ii) (a) 23.33 (b) Mode < Median < Mean (iii) (a) 37.33 (b) Mean < Median < Mode 4. (i) (a) Continuous series (b) 45 (ii) (a) Continuous series (b) 32.857 (iii) (a) Continuous series (b) 21 (iv) (a) Continuous series (b) 42.08 5. (i) (a) Discontinuous series (c) 71.72 (ii) (a) Discontinuous series (b) 43.30 (iii) (a) Discontinuous series (c) 34.5 6. (i) (a) 39.6, 37.5, 25; Mode < Median < Mean (ii) (a) 31.5, 30.27, 28.34; Mode < Mean < Median Answers
Statistics and Probability 326 Allied The Leading Mathematics-10 Statistics 327 13.4 Quartiles of Grouped Data At the end of this topic, the students will be able to: ¾ find the quartiles of grouped data. ¾ solve the problems on the quartiles of grouped data. Learning Objectives I Introduction Median, as described above, clearly divides the total number of values of the variable or variates into two equal halves or portions. If we divide the total number of variate values into four equal parts, then there will be three such variate values. These values also act as the partitions of the total number of values. In this case, they are known as quartile values and not median value. The three partition values are successively called the first, second and third quartile values. Since median lies at the middle, it is the second quartile and is denoted by Q2. The first quartile lies at the centre of the first half or lower half of the total variate values. It is denoted by Q1. The variate value at the centre of the second half or lower half is the third quartile and is denoted by Q3. They are also cleared by the following charts: First half (or lower half) First quartile (Q1) Third quartile (Q3) Median (Q2) Second half (or upper half) Max. no. obtained by lower 75% Min. no. obtained by upper 25% Max. no. obtained by lower 50% Min. no. obtained by upper 50% Max. no. obtained by lower 25% Min. no. obtained by upper 75% 75% 100% 50% 25% 0% 25% 0% Q3 Q2=M 50% 75% 100% Q1 Lower Upper Besides the median and quartile values, we have other measures of central tendency. They are deciles dividing the total number of data into ten equal parts, percentiles dividing the total number of values into one hundred equal parts. Q1 = First quartile; below and above which lie the 25% and 75% of the data respectively. It is also called the lower quartile. Q2 = Second quartile or the median; below or above which lie the 50% of the data. It is also called the middle quartile. Q3 = Third quartile, below or above which lie the 75% and 25% of the data respectively. It is also called the upper quartile.
Statistics and Probability 328 Allied The Leading Mathematics-10 Statistics 329 II Computation of Quartiles To find the quartiles for the grouped data; we use the following formulae. Q1 = L + N 4 – cf f × i, Q2 = L + N 2 – cf f × i = Md and Q3 = L + 3N 4 – cf f × i Where, L = lower limit of the quartile class. cf = frequency of the class preceding the quartile class f = frequency of the quartile class i = class interval. Look at the following example; For the following data calculate the two quartiles: Marks 10–20 20–30 30–40 40–50 50–60 60–70 No. of students 4 6 10 15 1 7 Solution: We begin with preparation of the following cumulative frequency table: Marks Frequency Cumulative frequency 10 – 20 4 4 20 – 30 6 10 30 – 40 10 20 40 – 50 15 35 50 – 60 1 36 60 – 70 7 43 For first quartile, Clearly, N 4 = 43 4 = 10.75, and, so the 1st quartile lies in the class 30 – 40. Then, L = 30, cf = 10, f = 10, i = 40 – 30 = 10. Using the formula, Q1 = L + N 4 – cf f × i = 30 + 10.75 – 10 10 × 10 = 30 + 0.75 = 30.75. For second quartile, Clearly, 2N 4 = 43 4 = 21.5, and, so the 2nd quartile lies in the class 40 – 50. Then, L = 40, cf = 20, f = 15, i = 50 – 40 = 10. Using the formula, Q3 = L + 3N 4 – cf f × i
Statistics and Probability 328 Allied The Leading Mathematics-10 Statistics 329 = 40 + 21.5 – 20 4 × 10 = 40 + 1 = 41, which is also median. For third quartile, Clearly, 3N 4 = 129 4 = 32.25, and, so the 3rd quartile lies in the class 40 – 50. Then, L = 40, cf = 20, f = 15, h = 50 – 40 = 10. Using the formula, Q3 = L + 3N 4 – cf f × i = 40 + 32.25 – 20 15 × 10 = 40 + 8.17 = 48.17. II Quartiles From Ogive We can easily compute the quartile classes and the values of quartilesfrom the Ogive or cumulative frequency curve. Observe the less than Ogive of the given data alongside. x 5-15 15-25 25-35 35-45 45-55 f 15 8 10 20 17 cf 15 23 33 53 70 Here, the total number of students (N) = 70 The position of the first quartile is N 4 = 70 4 = 17.5 Draw a line parallel to x-axis from the point 17.5 on the vertical scale that intersects the Ogive at a point. From the intersecting point, draw a perpendicular line to x-axis that falls between (5 - 15). Thus, the required the first quartile class is (5 - 15). The value of the first quartile is approximately 12. We can easily find the third quartile class and the value of the third quartile from the less than Ogive as mentioned above. Example-1 Find Q1 and Q3 for the data given below: x 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 f 3 7 12 18 16 11 5 (a) Write the formulae to calculate the first and the third quartiles of the continuous series. (b) Construct the cumulative frequency table from the given data. (c) Find the classes where the first and third quartiles lie. (d) Calculate the values of the first and third quartiles. 60 30 70 40 50 20 10 0 5 15 25 35 45 55 65
Statistics and Probability 330 Allied The Leading Mathematics-10 Statistics 331 Solution: (a) The formulae to calculate the first and the third quartiles of the continuous series are, Q1 = L + N 4 – cf f × i and Q3 = L + N 4 – cf f × i, where, L = lower limit of the quartile class, f = frequency of the quartile class cf = frequency of the class preceding the quartile class, i = class interval (b) Constructing the cumulative frequency table, x 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 f 3 7 12 18 16 11 5 cf 3 10 22 40 56 67 72 (c) The Q1 class = N 4 th class = 72 4 th class = 18th class = 40 – 50. The Q3 class = 3N 4 th class = 3 × 72 4 th class = 54th class = 60 – 70. (d) Now, for Q1, L = 40, cf = 10, f = 12, i = 10 Q1 = L + N 4 – cf f × i = 40 + 18 – 10 12 × 10 = 40 + 80 12 = 40 + 6.67 = 46.67. Again, for Q3, L = 60, cf = 40, f = 16, i = 10 Q3 = L + 3N 4 – cf f × i = 60 + 54 – 40 16 = 60 + 8.75 = 68.75. Example-2 If the third quartile (Q3) of the given data is 44, find the value of m: x 0-10 10-20 20-30 30-40 40-50 50-60 f 6 3 9 7 m 6 (a) Construct the cumulative frequency table from the given data. (b) Find the class where the first quartile lies. (c) Calculate the value of m. Solution: (a) Constructing the cumulative frequency table; x 0-10 10-20 20-30 30-40 40-50 50-60 f 6 3 9 7 m 6 cf 6 9 18 25 25 + m 31 + m (b) By given, the third quartile (Q3) = 44. So, the Q3 class = 40 – 50. (c) Here, Q3 class = 40 – 50. So, L = 40, cf = 25, f = m, i = 10
Statistics and Probability 330 Allied The Leading Mathematics-10 Statistics 331 Now, Q1 = L + 3N 4 – cf f × i or, 44 = 40 + 3(31 + m) 4 – 25 m × 10 or, 4 = 93 + 3m – 100 4m × 10 or, 16m = (3m – 7) × 10 or, 16m = 30m – 70 or, 70 = 30m – 16m or, 14m = 70 or, m = 70 14 or, m = 5. PRACTICE 13.4 Read Think Understand Do Keeping Skill Sharp 1. (a) Define quartiles of continuous data. (b) What is the first quartile class in the grouped data having N observations? (c) Which quartile is equal to the median? (d) What is the first quartile? (e) What is the third quartile? (f) Write the formula to calculate the value of Q1 in the grouped data. (g) What are the first quartile classes and the values of the first quartiles in the following Ogive? O 10 30 50 70 20 40 60 80 5 10 15 20 25 30 35 40 2 6 10 14 4 8 12 16 O 10 20 30 40 50 60 70 80 10 30 50 70 20 40 60 80 O 20 40 60 80 100 120 140 180 (i) (ii) (iii)
Statistics and Probability 332 Allied The Leading Mathematics-10 Statistics 333 (h) What are the third quartile classes and the values of the third quartiles in the following Ogive? 5 15 25 35 10 20 30 40 O 3 6 9 12 15 18 21 24 2 6 10 14 4 8 12 16 O 20 40 60 80 100 120 140 160 5 15 25 35 10 20 30 40 O 10 20 30 40 50 60 70 80 (i) (ii) (iii) 2. (a) What is the position of the first quartile class of a continuous series having 34 terms? (b) What is the position of the third quartile class of a continuous series having 57 terms? (c) Write the first quartile class in the given data. x 5-10 10-15 15-20 20-25 25-30 f 4 6 11 9 10 (d) Write the third quartile class in the given data. x 20-30 30-40 40-50 50-60 60-70 f 6 9 4 5 12 3. Circle ( ) the correct answer. (a) Which formula is used to calculate the first quartile of the grouped data? (i) Q1 = L – N 4 – cf f × i (ii) Q1 = L + N 4 – cf f × i (iii) Q3 = L + 3N 4 – cf f × i (iv) Q3 = L – 3N 4 – cf f × i (b) Which formula is used to calculate the first quartile of the grouped data? (i) Q1 = L – N 4 – cf f × i (ii) Q1 = L + N 4 – cf f × i (iii) Q3 = L + 3N 4 – cf f × i (iv) Q3 = L – 3N 4 – cf f × i (c) In the quartile formula Q1 = L + N 4 – cf f × i, what is the meaning of i ? (i) Q1 class interval (ii) Q3 class interval (iii) Frequency of the Q1 class (iv) Lower limit of Q1 class
Statistics and Probability 332 Allied The Leading Mathematics-10 Statistics 333 (d) Write the class interval in the data given below where the third quartile lies. x 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 f 8 7 8 4 3 (i) 40 - 50 (ii) 30 - 40 (iii) 10 - 20 (iv) 50 - 60 (e) Which is the cf taken to calculate the value of the third quartile for the given data? x 5 – 8 8 – 11 11 – 14 14 – 17 17 – 20 20 – 23 23 – 26 cf 7 10 14 19 22 28 30 (i) 17 (ii) 31 (iii) 72 (iv) 22 (f) Which is the frequency (f) taken to compute the value of the first quartile for the given data? x 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 cf 4 9 12 17 20 (i) 4 (ii) 9 (iii) 17 (iv) 12 Check Your Performance Answer the following questions by observing the given data. 4. (i) Marks obtained by students in an examination of 100 marks are given below: 95, 85, 90, 52, 75, 68, 75, 57, 99, 42, 44, 48, 37, 52, 75, 33, 41, 34, 36, 37, 97, 79, 82, 32, 56, 78, 64, 83, 32, 34, 82, 37, 98, 74, 86, 55, 63, 34, 45, 78. (ii) The monthly expenditure of certain families is given below in '000 rupees: 12, 13, 15, 35, 50, 40, 33, 32, 22, 14, 16, 17, 35, 32, 44, 22, 20, 18, 20, 12, 25, 16, 20, 19, 22, 21, 18, 15, 25, 18, 32, 35, 32, 34, 17, 19, 21, 23, 24, 13, 42, 27, 29, 40, 42. (a) Construct a frequency distribution table by taking class interval 10. (b) Write the cumulative frequency in the same table. (c) Find the classes where the first and the third quartiles lie. (d) Calculate the values of Q1 and Q3. 5. (i) Table below gives the wages of 264 workers in a factory: Wages (Rs.) 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 No. of workers 50 54 85 45 30 (ii) Observe the table below. Marks obtained 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 No. of students 3 12 20 17 23 5 (a) Write the cumulative frequency in the given data. (b) Find the classes where the first and the third quartiles lie. (c) Find the values of the first and the third quartiles.
Statistics and Probability 334 Allied The Leading Mathematics-10 Statistics PB 6. (i) Find the maximum marks obtained by lower 25% from the data given below: Marks obtained 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 No. of students 4 5 10 8 7 6 (ii) Find the maximum marks obtained by upper 25% from the given data: Marks obtained 50 - 60 50 - 70 50 - 80 50 - 90 50 - 100 No. of students 8 14 26 34 40 7. (i) Table below gives the shoe size for the students in a school: Shoe size 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 9 11 p 20 30 16 (a) Write the cumulative frequency in the given data. (b) If the first quartile is 25, find the class where the first quartile lie. (c) Find the value of p. (ii) The typing speed of some students is given below. Words per minute 0 - 20 20 - 40 80 - 100 60 - 80 40 - 60 No. of students 12 k 6 8 11 (a) Write the cumulative frequency in the given data. (b) If Q3 = 70, find the class where the third quartile lies. (c) Find the value of k. 4. (i) (c) 10 - 20, 70 - 80 (d) 40, 80 (ii) (c) 10 - 20, 30 - 40 (d) 17.03, 31.38 5. (i) (c) 40 - 50, 60 - 70 (d) 42.59, 64 (ii) (c) 30 - 40, 50 - 60 (d) 32.5, 53.47 6. (i) 31 (ii) 85 7. (i) (b) 20 - 30 (c) 6 (ii) 60 - 80 (c) 3 Answers Project Work Collect the marks obtained of a unit test of your friends in your class in Mathematics. Make a continuous data taking appropriate class intervals from it. Calculate the following: (i) Arithmetic Mean (ii) Median (iii) Quartiles
PB Allied The Leading Mathematics-10 Probability 335 Statistics and Probability CHAPTER 14 PROBABILITY WARM-UP Discuss the following technical terms on Probability. Technical word Definition Example Probability Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur or chance. Occurrence of head in tossing a coin Experiment The process by which an observation or measurement is made, is called an “Experiment”. Tossing a coin Outcomes The results of the experiment are called “Outcomes” ‘Head’ (H) or ‘Tail’ (T) Sample Space The set of all possible outcomes of an experiment is called a “Sample Space” S = {H, T} Elementary Events When and experiment is performed, it can result in one or more outcomes, which are know as “Elementary or simple Events” H and T are events Random Experiment An experiment which is performed under the similar (identical) conditions such that every result comes fairly is called “Random or Statical Experiment”. Throwing a dice with faces 1 to 6. Event An event of an experiment is a subset of simple events of the sample space. It is the occurrence of result. Had of a coin Sample Point An event of a sample space is called sample point. It is the outcome of a random experiment. 1 of dice Trial The performance of the random experiment is called trial. Tossing a coin Sure Event If the probability of occurrence of an event is 100% sure, then it is called sure or certain event whose probability is 1. Outcomes of a red or black card from a deck of 52 cards Impossible Event The non-occurence of the event in any trial is called impossible event. Turning 0 on throwing a dice Probability Scale Probabilities are given on a scale of 0 to 1 (0 < P(E) < 1), as decimal or as fractions. Sometimes probabilities are expressed as percentage using a scale of 0% to 100%. Probability of turning head is 0.5 or 1 2 R E V I S I T I N G P R O B A B I L I T Y
336 Allied The Leading Mathematics-10 Probability 337 Statistics and Probability Favourable cases A set of events of a trial in which a given event may happen, are called favourable cases for that event. The favourable case of getting H is 1. Equally likely events Two or more events which have an equally likely chance or equal probability of occurrence are said to be equally likely. Chance of head or coin in tossing a unbiased/fair coin Complementary events Two events that exist such that one event will occur if and only if the other does not take place, are called Complementary events. Complement of E is E. Tail (T) and head (H), where P(H) = 1 - P(H) or 1 - P(T) Exclusive event The events that do not occur at the same time are called mutually exclusive events. Getting either Head or tail on a coin. Exhaustive event A set of events are called exhaustive events if at least one of them necessarily occurs whenever the experiment is performed. Head or tail occurs when flipping a coin. Probability of occurrence of event Probability of occurrence of event is a positive number which represents the chances of occurrence of an event. Prob. of getting head, P(H) = 1 2 , H ⊂ S. Classical probability Classical probability is the statistical concept that measures the likelihood (probability) of something happening. The probability of an event A is denoted by P(A) and, P (E) = No. of favourable outcomes No. of possible outcomes . Prob. of getting even numbers on dice = 3 6 = 1 2 Empirical probability The empirical or experimental probability of an event (E) is defined by the ratio m:n of the total number (m) of successes to the total number (n) of the experiments, and is denoted by P(E) and, P (E) = m n . Prob. of getting head on a coin of 100 tossing = 57 100 Some Experiments and Their Events: Tossing Coin Birthing Baby Throwing Dice Drawing Card Turing Spinner Club Playing Cards on A Deck of 52 Cards Spade Heart Diamond R E V I S I T I N G P R O B A B I L I T Y
336 Allied The Leading Mathematics-10 Probability 337 Statistics and Probability 14.1 Probability on Mutually Exclusive Events At the end of this topic, the students will be able to: ¾ find the probability by using additive law for mutually exclusive events. Learning Objectives I Introduction The concept of probability with definition, and its application in simple situations have been dealt in the previous class. Probability is a mathematical term used to talk about the likelihood of something happening. The probability is the basic concept of games. It is the chance of winning or losing a game. As the same situation, a gambler's dispute in 1654 led to the creation of a mathematical theory of probability by two famous French mathematicians, Pierre de Fermat (1601-1665) and Blaise Pascal (1623-1662). The probability is used in many fields such as forecasting the weather, sports outcomes, card games and other games of chance, batting average in cricket, insurance, traffic signals, medical diagnosis, election results, lottery probability, stock market predictions etc. Based on the previous concepts, the application of the laws of addition and multiplication in calculation of the probabilities of mutually exclusive events, method of calculating probability of two dependent and independent events, are discussed in this chapter. Activity 1 Observe a deck of the playing cards. Fermat Blaise Pascal
338 Allied The Leading Mathematics-10 Probability 339 Statistics and Probability Consider the rell card R, Ace card A, and King card K. How many cards are there in the deck of the playing cards ? How many red cards (R) are there in the deck of the playing cards ? How many Ace (A) and King (K) cards are there in the deck of the playing cards ? How many red Ace (RA) cards are there in the deck of the playing cards? How many red King (RK) cards are there in the deck of the playing cards? Now, how many chances of getting each type of card separately? Similarly, how many chances of getting a red card? How many chances of getting an Ace card? How many chances of getting a King card? How many chances of getting a red Ace card? How many chances of getting a red King card? Is it possible to get an Ace card and a King card at the same time ? Is it possible to get an Ace card and a red card at the same time ? How many cards ma be repeated? Is it possible to get a King card and a red card at the same time? How many cards ma be repeated? Now, what is the probability of getting an Ace card ? ⇒ P(A) = 4 52 = 1 13 What is the probability of getting a King card ? ⇒ P(K) = 4 52 = 1 13 What is the probability of getting a red card ? ⇒ P(R) = 26 52 = 1 2 What is the probability of getting a red Ace card ? ⇒ P(RA) = 2 52 = 1 26 What is the probability of getting a red King card ? ⇒ P(RK) = 2 52 = 1 26 How many total cards of Ace or King or both are there in the deck of cards ? What is the probability of getting either an Ace or a King card ? ⇒ P(A or K) = 8 52 = 2 13 Here, we know, 2 13 = 1 13 + 1 13 ⇒ P(A or K) = P(A) + P(K). Why ? But, P(A or K)is not always equal to P(A) + P(B). In other hand, how many total cards of Ace or red or both are there in the deck of cards ? What is the probability of getting either an Ace or a red card? ⇒ P(A or R) = 28 52 Here, we know, 28 52 = 4 52 + 26 52 – 2 52 ⇒ P(A or R) = P(A) + P(R) – P(RA). Why ? The total probability of either one or other event is the sum of their separate probabilities if the events are mutually exclusive events. 44 U 4 4 A K 26 R 24 RA 2 2 A U
338 Allied The Leading Mathematics-10 Probability 339 Statistics and Probability Addition law for mutually exclusive events A and B; P(A ∪ B) = P(A) + P(B) if A ∩ B = φ So, P (A or B) = P(A) + P(B) Note: To find the total probability of either one or other events whether they are mutually exclusive or not exclusive. P(A∪B) = P(A) + P(B) – P(A ∩ B) Proof: (Optional): Let S be the sample space and A and B any two events of the sample space S. Again, let n(S) = n, n(A) = s, n(B) = r and n(A ∩ B) = t We have, P (A ∪ B) = n(A ∪ B) n(S) = (r – t) + t + (s – t) n = r + s – t n = r n + s n – t n = P(A) + P(B) – P(A∪B) ∴ P (A ∪ B) = P(A) + P(B) – P(A ∩ B) For example, Use formula P(A ∪ B) = P(A) + P(B) – P(A ∩ B) to verify the results of (i) P(N ∪ E) and (ii) P(T ∪ H) discussed in activity 2. Solution: (i) P(N ∪ E) = P(N) + P(E) – P(N ∩ E) = 7 16 + 4 16 – 0 = 11 16 [∴ N ∩ E = φ] (ii) P(T ∪ H) = P(T) + P(H) – P(T ∩ H) Now, T = {A, B, N, E, F, T, R, S}, H = {C, F, K, T, N} ∴ T ∩ N = {F, N} P(T ∪ H) = P(T) + P(H) – P(T ∩ H) = 8 16 + 5 16 – 2 16 = 11 16 Example-1 Suppose a sample space S and events A and B are shown graphically: (i) Are events A and B mutually exclusive? (ii) What is the probability of getting A or B? Solution: Here, n(A) = 3, n(B) = 2, n(S) = 7 (i) Since A ∩ B = φ, they are mutually exclusive. (ii) P(A or B) = P(A) + P(B) = n(A) n(S) + n(B) n(S) = 3 7 + 2 7 = 5 7 r s B A S r – t t s – t B A S O1 O3 O2 O5 O4 O6 O7 S A B
340 Allied The Leading Mathematics-10 Probability 341 Statistics and Probability Example-2 Event A and its complement A are shown graphically: (i) Are events A and A mutually exclusive? (ii) What is the probability of getting A or A ? Solution: Here, n(A) = 5, n(A) = 7, n(S) = 5 + 7 = 12 (i) Since A ∩ A = φ, and A and A are mutually exclusive. (ii) P(A or A) = P(A) + P(A) = n(A) n(S) + n(A) n(S) = 5 12 + 7 12 = 5 + 7 12 = 12 12 = 1. Example-3 A card is selected from a pack of 52 cards. What is the probability that the card is a 7 or a 10? Solution: Suppose E = one card is a 7 and T = a card is a 10. Then n(E) = 4, n(T) = 4, n(S) = 52 Since E and T are mutually exclusive events so, P(E or T) = P(E) + P(T) = n(E) n(S) + n(T) n(S) = 4 52 + 4 52 = 1 13 + 1 13 = 2 13 Example-4 Sarita throws a die numbered from 1 to 6. What is the probability that she gets a multiple of 5 or a multiple of 3? Solution: Let M5 be the event multiple of 5. i.e., M5 = {5} ∴ n(M5) = 1 and M3 be the event multiple of 3. i.e., M3 = {3,6} ∴ n(M3) = 2 Since they are mutually exclusive events so, P(M5 or M3) = P(M5) + P(M3) = n(M5) n(S) + n(M3) n(S) = 1 6 + 2 6 = 3 6 = 1 2 . Example-5 Spin the pointer; (a) Find the probability that the pointer stops in B section. (b) Find the probability that the pointer stops in B or G section. (c) Find the probability that the pointer stops in B or R. Solution: Here, (i) P(B) = 120° 360° = 1 3 (ii) P(G) = 60° 360° = 1 6 ∴ P(B or G) = P(B) + P(G) = 1 3 + 1 6 = 2 + 1 6 = 3 6 = 1 2 . (iii) P(R) = 90° 360° = 1 4 ∴ P(B or R) = P(B) + P(R) = 1 3 + 1 4 = 4 + 3 12 = 7 12. A U A G R Y B 120° 60°
340 Allied The Leading Mathematics-10 Probability 341 Statistics and Probability PRACTICE 14.1 Read Think Understand Do Keeping Skill Sharp 1. For each of these activities say whether or not the two given events are mutually exclusive: (a) Tossing a coin: showing a head, showing a tail. (b) Play a football match: ending in a win, ending in a draw. (c) Choosing a school child at random: getting a boy, getting a child 13 years old. (d) Choosing a person at random: getting a child, getting a person 85 years old. 2. (a) What is probability ? Define it. (b) Define mutually exclusive events. (c) What is the probability of mutually exclusive events? (d) If P and R are two mutually exclusive events in an experiment, what is the probability of getting P or R ? (e) A and M are two mutually exclusive events. If P(A) = 1 4 and P(M) = 1 5 , what is the probability of getting A or M ? 3. Circle ( ) the correct answer. (a) If H and T are two mutually exclusive events in tossing a coin, then ............... (i) P(H or T) = P(H) - P(T) (ii) P(H or T) = P(H) × P(T) (iii) P(H or T) = P(H) + P(T) (iv) All of the above (b) If C and D are two mutually exclusive events in tossing a coin, which is true ? (i) P(C ∩ D) = 0 (ii) P(C ∪ D) = 0 (iii) P(C ∩ D) = ϕ (iv) P(C ∩ D) = P(C) + P(D) (c) Which is the correct probability of getting an Ace from a well shuffled pack of 52 cards ? (i) 1 4 (ii) 1 52 (iii) 4 13 (iv) 1 13 (d) What is the probability of getting a spade card from a well shuffled pack of 52 cards ? (i) 1 52 (ii) 1 4 (iii) 1 13 (iv) 13 39 (e) Which is the correct probability of getting an even number when throwing a die ? (i) 1 2 (ii) 1 2 (iii) 1 3 (iv) 1 6 (f) The probability of getting a red ball from an urn containing 6 red and 3 white identical balls is ............. . (i) 1 6 (ii) 3 9 (iii) 2 3 (iv) 1 6
342 Allied The Leading Mathematics-10 Probability 343 Statistics and Probability Check Your Performance 4. A coin is tossed, calculate the probability of getting : (a) Head (b) Tail (c) Head or Tail (d) Both Head and Tail 5. (i) If Siman throws a die, calculate the probability of getting: (a) 1 (b) 6 (c) 1 or 6 (d) even or 5 (ii) A number card is drawn from the cards from 1 to 8. (a) odd or multiple of 4 (b) prime or 1 or 8 6. Ansari draws a card at random from the given cards. Calculate the probability of getting ; (a) Ace (b) King (c) Jack or Queen (d) Ace or King (e) Ace or faced card (f) Red or Black Queen (g) Red Jack or Spade King or black Queen (h) King or non-faced card (i) Diamond or black Ace or black King 7. A bag contains 3 black, 2 white and 5 green identical beads. A bead is chosen at random. Calculate the probability of getting; (a) Black or White (b) Green or White (c) Black or Green or White 8. A bag contains 20 balls in which there are 4 red, 10 blue, 5 yellow and 1 violet. If Hari chooses a ball at random, calculate the probability that he chooses, (a) a red ball (b) a blue ball (c) a yellow ball (d) a red or a blue ball (e) a yellow or a blue ball (f) a red or a violet ball 9. The given spinner has eight equal sectors. (i) Which of the following events are mutually exclusive ? Give reason. (a) Spinning 1 and Spinning 7 (b) Spinning 2 and Spinning 8 (c) Spinning 3 and Spinning 3 (d) Spinning 1, Spinning 2 and spinning 7 (ii) Find the probabilities of getting; (a) Spinning 2 or Spinning 6 (b) Spinning 5 or Spinning 8 (c) Spinning 1 or Spinning 4 or Spinning 7 B B B W W G G G G G 1 3 2 4 5 6 7 8
342 Allied The Leading Mathematics-10 Probability 343 Statistics and Probability 10. Prabina a weather forecaster says that the probability of raining tomorrow is 0.4 and of snowing is 0.2. Calculate the probability of getting; (a) rain or snow (b) no snow or rain 11. All denominations of US $ notes are of the same size, if an exchanger has the notes denominates. Denomination No. of Notes $ 1 5 $ 5 10 $ 10 20 $ 20 25 If the exchanger withdraws a note from the pack. What is the probability of getting a note of denominations. (a) P($ 1) (b) P($ 5) (c) P($ 10) (d) P($ 20) (e) P ($ 1 or $ 20) (f) P($ 10 or $ 5) 12. A couple plans to give birth to two babies. Find the probabilities of getting; (a) both sons (b) first daughter and second son (c) one son and one daughter (d) at least one son (e) both the same sex (f) no sons 13. A couple plans to give birth to three babies. Find the probabilities of getting; (a) all daughters (b) two sons and third daughter (c) at least one daughter (d) two daughters and one son 14. Two dice of the colours red and blue are thrown. (a) Write the sample space in ordered pair form. (b) Find the probability of getting; (i) 1 on blue die or 6 on red die (ii) 3 on blue die or 5 on red die (c) Show the sum of the elements of each ordered pair in addition table. (d) Calculate the probability of getting; (i) sum 10 or 7 (ii) sum 8 or 11 (iii) sum 4 or 10 or 12 (iv) one die shows a 1 or next die shows 6
344 Allied The Leading Mathematics-10 Probability 345 Statistics and Probability 15. Ravi picks up one of these cards at random. Find the probability of the following events: 9 15 12 11 8 24 13 18 2 14 25 32 28 5 1 16 10 3 27 30 7 (a) an even number (b) an odd number (c) a number greater than 6. (d) an odd number or an even number (e) a number greater than 6 or factors of 6 (f) a multiple of 10 or prime number (g) a multiple of 7 or prime number. 16. From the deck of number cards numbered from 5 to 30, a card is drawn at random. Calculate the probability of getting; (a) divisible by 5 or divisible by 7 (b) prime number or multiple of 3 17. From the deck of letter cards of the word the given word, a card is drawn at random. (i) PROBABILITIES (ii) MISSISSIPPI Calculate the probability of occurring the letter in each word; (a) P or I (b) S or I (c) P or S or I 18. (i) A die is rolled 100 times and the obtained results are shown in the table below: Result 1 2 3 4 5 6 No. of occurrence 23 25 26 32 21 27 Calculate the probability for the following results: (a) 1 or 6 (b) 2 or 4 (c) 3 or greater than 4 (ii) A spinner of the given colours in equal sectors is rotated and the obtained results are shown in the table below: Result Red (R) White (W) Green (G) Black (B) Yellow (Y) Frequency 12 14 20 18 11 (a) How many times did the spinner rotate? (b) What is the probability of occurring W or Y ? Find it. (c) Calculate the probability of occurring the frequency less than 18.
344 Allied The Leading Mathematics-10 Probability 345 Statistics and Probability 4. (a) 1 2 (b) 1 2 (c) 1 (d) 0 5. (i) (a) 1 6 (b) 1 6 (c) 1 3 (d) 2 3 (ii) (a) 3 4 (b) 3 4 6. (i) (a) 1 13 (b) 1 13 (c) 2 13 (d) 2 13 (e) 4 13 (f) 7 13 (g) 5 52 (h) 11 13 (i) 17 52 7. (a) 1 2 (b) 7 10 (c) 1 8. (a) 1 5 (b) 1 2 (c) 1 4 (d) 7 10 (e) 3 4 (f) 1 4 9. (a) 1 4 (b) 1 4 (c) 3 8 10. (a) 0.6 (b) 0.4 11. (a) 1 12 (b) 1 6 (c) 1 3 (d) 5 12 (e) 1 2 (f) 1 2 12. (a) 1 4 (b) 1 4 (c) 1 2 (d) 3 4 (e) 1 2 (f) 1 4 13. (a) 1 8 (b) 1 8 (c) 7 8 (d) 3 8 14. (b) (i) 1 3 (ii) 1 3 (d) (i) 1 4 (ii) 7 36 (iii) 7 36 15. (a) 11 21 (b) 10 21 (c) 17 21 (d) 1 (e) 20 21 (f) 2 4 (g) 3 7 16. (a) 1 4 (b) 1 4 17. (i) (a) 4 13 (b) 4 13 (c) 5 13 (iii) (a) 6 11 (b) 8 11 (c) 10 11 18. (i) (a) 1 2 (b) 57 100 (c) 37 50 (ii) (a) 75 (b) 1 3 (c) 37 75 Answers Project Work Roll a die 50 times and record each outcome in the table. Calculate the probabilities of getting prime numbers or composite numbers.
346 Allied The Leading Mathematics-10 Probability 347 Statistics and Probability 14.2 Probability on Independent and Dependent Events At the end of this topic, the students will be able to: ¾ find the probability using multiplicative law for independent and dependent events. Learning Objectives I Probability on Independent Events Activity 1 Consider the following experiment of rolling a blue die and a black die. The sample space is shown in the table below: Black die 1 2 3 4 5 6 1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) 2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) 3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) 4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) 5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Find the probability of the events; (i) G4: The blue die is 4. (ii) G3: The black die is 3. (iii) G: The blue die is 4 and black die is 3. Solution: (i) G4 = {(4 1), (4 2), (4 3), (4 4), (4 5), (4 6)} P(G4 ) = 6 36 = 1 6 (ii) G3= {(1 3), (2 3), (3 3), (4 3), (5 3), (6 3)} P(G3 ) = 6 36 = 1 6 (iii) G4 ∩ G3 = {(4 3)} = G P(G4 ∩ G3 ) = 6 36 = 1 6 × 1 6 = P(G4 ) × P(G3 ) Two events A and B of the same experiment are said to be independent if the occurrence of the event A does not affect to the occurrence of the next event B and therefore; P(A ∩ B) = P(A) × P(B). Otherwise, A and B are dependent events. In such a way, for independent events A, B, C, .........., we have P(A ∩ B ∩ C ∩ ......) = P(A) × P(B) × P(C) × ............... Blue die
346 Allied The Leading Mathematics-10 Probability 347 Statistics and Probability II Probability on Dependent Events Activity 2 Consider the following experiment of pulling two black cards from a deck of playing cards. One card is pulled from the deck of playing cards and is not replaced. Another card is pulled from the same deck. Then the sample space of the second event is changed. The sample space of the first event is 26 black cards, but the sample space of the second event is now 25 black cards and also the total number of cards reduced by one card. Thus, the events are dependent. In this case, the probability of second card is dependent on the pulled card of the same deck of the playing cards. What is the probability that the both cards are black without replacement? We have, P(first black card) = 26 52 and P(second black card) = 26 – 1 52 – 1 = 25 51. Therefore, P(both black cards) = 26 52 × 25 51 = 25 102. If the result of one event is affected by the result of another event, the events are said to be dependent. If A and B are dependent events, the probability of both events occurring is the product of the probability of the first event and the probability of the second event once the first event has occurred. If A and B are dependent events, and A occurs first, P(A and B) = P(A) × P(B, once A has occur) and is written as, P(A and B) = P(A) × P(B) (Read as probability of B when A is given) Example-1 A drawer contains 2 red paperclips, 3 green paperclips, and 4 blue paperclips. One paperclip is taken from the drawer and then replaced. Another paperclip is taken from the drawer. What is the probability that the first paperclip is red and the second paperclip is blue? Solution: Because the first paper clip is replaced, the sample space of 9 paperclips does not change from the first event to the second event. The events are independent. Then, P(red) = and P(blue) = and therefore; P(red then blue) = P(red) × P(blue) = 2 9 × 4 9 = 8 81.
348 Allied The Leading Mathematics-10 Probability 349 Statistics and Probability Example-2 The events of tossing a coin are {T}, {H}, {T, H}, φ. Determine whether {T} and {H} are independent. Solution: Since P({T} ∩ {H}) = P(φ) = 0, but P({T}) = 1 2 and P({H}) = 1 2 P({H}) × P({T}) = 1 2 × 1 2 = 1 4 . ∴ P({T}∩{H}) = 0 ≠ 1 4 = P{H} × P{T} Hence, {T} and {H} are not independent. Example-3 A bag contains 5 red TT balls and 3 green TT balls. One ball is taken from the bag and then replaced. Another ball is taken from the bag. What is the probability that (a) both balls are green and (b) none of them are green? Solution: Here, The number of red TT balls, n(R) = 5 The number of green TT balls, n(G) = 3 The total number of TT balls, n(S) = 5 + 3 = 8 ∴ The probability of green TT balls, P(G) = n(G) n(S) = 3 8 Now, we have (a) The probability of both green TT balls, P(G and G) = P(G) × P(G) = 3 8 × 3 8 = 9 64. (b) The probability of both non-green TT balls, P(G and G) = P(G) × P(G) = {1– P(G)}×{1– P(G)} = 1 – 3 8 × 1 – 3 8 = 5 8 × 5 8 = 25 64. Example-4 Two cards are thrown from a well-shuffled deck of 52 cards without replacement. Find the probability that one card is Ace and another is Heart. Solution: Here, the total number of cards, n(S) = 52 The number of Ace card, n(A) = 4 The number of Heart card, n(H) = 13 ∴ The probability of getting Ace, P(A) = n(A) n(S) = 4 52 = 1 13.
348 Allied The Leading Mathematics-10 Probability 349 Statistics and Probability The probability of getting Heart, P(H) = n(H) n(S) = 13 51 Now, we have The probability of getting Ace and Heart, P(A and H)= P(A) × P(H) = 1 13 × 13 51 = 1 51 PRACTICE 14.2 Read Think Understand Do Keeping Skill Sharp 1. (a) Define independent events. (b) If P and Q are two independent events, what is the probability of getting both P and Q? (c) What is the probability of getting both Heads on tossing a coin twice ? (d) What is the probability of occurring 1 and 6 on throwing a die ? (e) Write the probability of getting 1 and 4 on spinning the given spinner. (f) What is the probability of occurring the letter M and T on drawing two letter cards of the word " MATHEMATICS" with replacement? (g) What is the probability of getting 3 and 5 on drawing a card from a pack of the number cards numbered from 3 to 10 without replacement ? (h) Write the probability of getting King and Heart card on drawing two cards from a well shuffled pack of 52 cards. (i) Write the probability of getting both Kings on drawing two cards from a well shuffled pack of 52 cards without replacement. (j) What is the probability of getting a Queen and a Jack on drawing two cards from a well shuffled pack of 52 cards without replacement ? 2. Circle ( ) the correct answer. (a) If H and T are two independent events in a tossing a coin, then ............... (i) P(H or T) = P(H) - P(T) (ii) P(H and T) = P(H) × P(T) (iii) P(H or T) = P(H) + P(T) (iv) P(H and T) = P(H) + P(T)
350 Allied The Leading Mathematics-10 Probability 351 Statistics and Probability (b) If A, B and C are mutually exclusive events of an experiment, which is true ? (i) P(ABC) = P(A) × P(B) (ii) P(ABC) = P(A) × P(C) (iii) P(ABC) = P(A) × P(B) × P(C) (iv) P(ABC) = P(A) + P(B) + P(C) (c) What is the probability of getting an Ace and a King from a well shuffled pack of 52 cards? (i) 1 13 (ii) 4 13 (iii) 4 169 (iv) 1 169 (d) What is the probability of birthing two sons from a couple? (i) 1 2 (ii) 1 4 (iii) 1 8 (iv) 2 5 (e) The probability of getting a 1 and a 6 on throwing a die .... (i) 1 3 (ii) 1 2 (iii) 1 36 (iv) 1 6 Check Your Performance 3. (a) Show the possible outcomes when two coins are tossed one after other in the table. Calculate the probability of (HH) or (TT). (b) The coin is tossed and the spinner is spun. (i) Prepare the sample space using table. (ii) Calculate P(H3) and P(T2). 4. These three cards are placed face down. One is chosen at random noted and replaced. The cards are shuffled, and another card is chosen at random. Calculate the probability of picking; (a) a pair of cards which are the same (b) a King and Queen in either order 5. A card is drawn from a deck of 52 cards and then replaced. After shuffling, a second card is drawn. Find; (a) P(two hearts) (b) P(one heart) (c) P(no hearts) 6. Two bags each contain 3 yellow balls and 2 red balls. One ball is drawn at random from each bag with replacement. Calculate the probability of drawing; (a) two red balls. (b) one red and one yellow balls. (c) two balls of same colour. 2 1 3 Y Y Y R R Y R R Y Y