NCERT solutions for class 10

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STATISTICS

1. Arithmetic Mean
The mean of n observations x1, x2, x3, ...... xn is

x= x1 + x2 + x3 + x4 + ...... + xn = 1 n
n n
 xi

i =1

2. Let x be the mean of n numbers, if k is added to each number, new mean = x + k.

If k is subtracted from each number, new mean = x − k.

If each number is multiplied by k, new mean = kx. If each number is divided by k, new
mean = x .

k

3. Sum of observations = Number of observations × Mean.

4. When the raw data are presented in the form of a frequency distribution with equal or
unequal, exclusive or inclusive class-intervals, we assume that the frequency in any
class is centred at its class marks (or mid-point).

Class marks of any class-interval = Upper limit + Lower limit
2

5. Methods of finding arithmetic mean of grouped data

(i) Direct Method

If n observations in the raw data consist of k distinct value x1, x2, x3, ...... xn
of the variable x occurring with frequencies f1, f2, f3, ...... fn respectively, then

n

 x fi xi k
= f1x1 + f2x2 + ......+ fn xn = = 1 xi
f1 + f2 + ......+ fn i =1 n fi
n
i =1
fi

i =1

k

Where n = fi

i =1

denotes total frequency.
(ii) Assumed mean method

In this method, an arbitrary constant A is chosen as assumed mean some
where in the middle of all values of xi and find di = xi − A.

x = A + fi di
fi

(iii) Step deviation method

In this method, an arbitrary constant A is chosen which is called as origin or

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assumed mean somewhere in the middle of all values of xi. If h is the difference

of any two consecutive values of xi, then ui = xi − a .
h

Mean h k
(x) = A+ n = A + hu.
fi ui

i =1

For example:
Find the mean of the following data

Class interval 10 – 25 25 – 40 40 – 55 55 – 70 70 – 85 85 – 100

Number of students 2 3 766 6

(A) Direct method

Class interval Number of students (fi) Class mark (xi) (fi xi)
10 – 25 2 17.5 35.0
97.5
25 – 40 3 32.5 332.5
375.0
40 – 55 7 47.5 465.0
555.0
55 – 70 6 62.5 (fi xi) = 1860

70 – 85 6 77.5

85 – 100 6 92.5

Total fi = 30

The mean x of the given data is given by
x =  fi xi = 1860 = 62
 fi 30

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(B) Assumed mean method

Class Number of Class mark (xi) di = xi − 47.5 fidi
interval students (fi)
10 – 25 17.5 −30 −60
25 – 40 2 32.5 −15 −45
40 – 55 3 47.5
55 – 70 7 62.5 00
70 – 85 6 77.5 15 90
85 – 100 6 92.5 30 180
6 45 270
Total fi = 30
 (fidi) = 435

Here A (assumed mean) = 47.5
The mean x of the given data is given by

x = A + fidi
 fi

x = 47.5 + 435 = 47.5 + 14.5 = 62
30

(C) Step deviation method

Class Number of Class di = xi − 47.5 ui = xi − A fiui
interval students (fi) mark (xi) h
−30 −4
10 – 25 2 17.5 −15 −2 −3
25 – 40 3 32.5 0
40 – 55 7 47.5 0 −1 6
55 – 70 6 62.5 15 12
70 – 85 6 77.5 30 0 18
85 – 100 6 92.5 45 (fiui) = 29
fi = 30 1
Total
2

3

The mean x of the given data is given by

Mean = h k
(x) A+ n
fi ui

i =1

x = 47.5 + 15   29 
 30 

= 47.5 + 14.5 = 62

6. The algebraic sum of the deviations of a set of values from their arithmetic mean is

zero.

k

i.e.,  fi (xi − x ) = 0.
i =1

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7. Mode of Grouped Data

Mode of grouped frequency distribution:

(i) Locate a class with the maximum frequency. This class is known as modal class.

(ii) Mode = l +  f1 − f0 f2   h ,
2f1 − f0 −

where l = lower limit of the modal class.

f1 = frequency of the modal class.
f0 = frequency of the class preceding the modal class.
f2 = frequency of the class succeeding the modal class.
h = size of the class interval

For example:

Calculate the mode for the following frequency distribution:

Class 0–10 10– 20–30 30–40 40–50 50–60 60–70 70–
20 80

Frequency 5 8 7 12 28 20 10 10

As the class 40–50 has maximum frequency, so it is the modal class.

 l = 40, h = 10, f1 = 28, f 0 = 12 and f 2 = 20 .

M0 =l + h .(2f1(f1− −f 0) ) 
f0 −f 
2 

= 40 + 10  (2  (28 − 12) 20) 
 28 − 12 − 


= 40 + 10  16  =  40 + 20  = (40 + 6.67) = 46.67 .
24   3 

Hence, mode = 46.67.

8. Median of Grouped Data
Calculation of median for grouped data:
(a) When the data is discrete.
(i) Arrange the data in ascending or descending order of magnitude alongwith their
frequencies.
(ii) Make a column of cumulative frequencies.

(iii) • If the total frequency n is odd, then  n + 1 th observation is the median.
2

• If the total frequency n is even, then the mean of n th and  n + 2  th
2 2

observation is median.
(b) When the data is continuous and in the form of frequency distribution.

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 n − c 
Median = l +  2   h

 f 
Where l = lower limit of median class

f = frequency of median class
c = cumulative frequency of class preceding the median class
h = class size,
n = number of observations.

For example:

Find the median for the following frequency distribution:

Class interval 0–8 8 – 16 16 – 24 24 – 32 32 – 40 40 – 48
15 7
Frequency 8 10 16 24

We prepare the cumulative frequency table, as given below:

Class Frequency (fi) Cumulative frequency
0–8 8 8
8 – 16 10 18

16 – 24 16 34
24 – 32 24 58
32 – 40 15 73
40 – 48 7 80

N = fi = 80

Now, N = 80  (N / 2) = 40.

The cumulative frequency just greater than 40 is 58 and the corresponding class is
24– 32.

Thus, the median class is 24 – 32.

 l = 24, h = 8, f = 24, c = 34, and

  N −c  
 2  
Median, Me =l + h  
 f
 

= 24 + 8  (40 − 34) = (24 + 2) = 26 .

24 

Hence, median = 26.
9. Graphical representation of cumulative frequency distribution.

(i) Cumulative frequency curve or an ogive (of the ‘less than’ type)

Represent the data in the table graphically.

(a) Mark the upper limits of the class intervals on the horizontal axis and their
corresponding cumulative frequencies on the vertical axis.

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(b) Plot the points corresponding to the ordered pairs given by (upper limit,
corresponding cumulative frequency) and join them by a free hand smooth
curve.

(ii) Cumulative frequency curve or an ogive (of the ‘more than’ type)

(a) Mark the lower limit of the class intervals on the horizontal axis and their corresponding cumulative
frequencies on vertical axis.

(b) Plot the points corresponding to the ordered pairs given by (lower limit,
corresponding cumulative frequency) and join them by a free hand smooth
curve.
Median of a grouped data can be obtained graphically as the x-coordinate of
the point of intersection of the two ogives (more than type and less than type)
for this data.

For example:
The following table gives production yield per hectare of wheat of 100 farms of a
village.

Production yield 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 75 – 80
(in kg/ha)

Number of farms 2 8 12 24 38 16

Change the distribution to a "more than" type distribution and draw its Ogive.

"More than" type distribution Cumulative frequency

Production yield (kg/ha) 100
98
More than or equal to 50 90
More than or equal to 55 78
More than or equal to 60 54
More than or equal to 65 16
More than or equal to 70
More than or equal to 75

Construction of Ogive.

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Y
(50, 100)

100 (55, 98)

90 (60, 90)

Production yield (in kg/ha) 80 (65, 78)
70

60

(70, 54)
50

40

30

20
(75, 16)

10

0 X
50 55 60 65 70 75 80
(No. of farms)

10. Empirical relationship among Mean, Median and Mode is as below:
3 Median = Mode + 2 Mean.

For example:
If median of 20 observations is 50 and mode is also 50, find the mean.
Median of 20 observations = 50
Mode of 20 observations = 50

3 Median = Mode + 2 Mean
or Mean = Mode + 3 (Median − Mode)

2
= 50 + 3 (50 − 50)

2
= 50 + 3  0 Mean = 50

2

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