Class 9 O.Maths

296 PRIME Opt. Maths Book - IX v) A(4, 1) and B(1, –2) 5. Find the image of the following triangles under the given vector. Also plot the object and image in graph. i) A(1, –2), B(–3, 4) and C(3, –3) under a vector T= 2 3 < F ii) P(2, 5), Q(–2, 1) and R(5, 2) under a vector T = 2 3 – – < F iii) A(–2, –4), B(–4, –1) and C(3, 0) under a vector T = 1 3 < F iv) K(2, 3), L(4, 6) & M(6, 1) under a vector T = 2 4 – < F v) X(–2, 5), Y(3, 1) & Z(–4, –1) under a vector T = 3 –4 < F 6. i) If a translation T gives image of an object A(1, 3) → A’(3, 4). Find the value of translation vector T. ii) If a point P(3, –1) is translated to P’(5, 2) under a vector T, find the value of translation vector T. iii) If a vector T = 1 3 < F translate a point P to P’(3, 7), find the co-ordiante of the point P. iv) Find the co-ordiante of image of DABC having vertices P(1, 2), Q(3, 5) and R(6, –2) under a translation vector PQ . Also plot DP’Q’R’ and DP’Q’R’ on graph paper. v) Find the co-ordinate of the image of DABC having A(3, 1), B(7, 3), C(5, –2) under a translation vector AB . 7. PRIME more creative questions i) Find the image of ABCD having vertices A(1, 8), B(–3, 9), C(0, 13) & D(4, 12) under translation about T1 = AB . Also transform the image so obtained under T2 = 3 2 – – < F and plot the image and object in graph. ii) Transform a triangle having vertices A(–4, 6), B(3, –2) and C(1, 2) under translation about T1 = 1 –2 < F followed by translation about T2 = 2 3 – < F. Also plot the object and image in graph. iii) Transform a triangle having vertices A(–3, 2), B(1, 5) and C(2, 3) under reflection about x = 2 followed by translation about AB . Also plot the object and image in graph. iv) Reflect p(x, y) under x = 1 followed by x = 3 and compare its image with translation about T 4 0 . Write down the conclusion of it. v) What is translation? Write down its properties. Where is it applicable in our daily life? Discuss in group and prepare a report.

PRIME Opt. Maths Book - IX 297 Answer 1. Show to your teacher. 2. Show to your teacher. 3. i) A’(5, 3) ii) P’(0, 6) iii) M’(–1, 0) iv) N’(6, –5) v) O’(–1, –3) 4. i) P’(1, –4) ii) P’(5, 1) iii) P’(5, –6) iv) P’(6, –3) v) P’(0, –5) 5. i) A’(3, 1), B’(–1, 7) & C’(5, 0); graph ii) P’(0, 2), Q’(–4, –2) & R’(3, –1); graph iii) A’(–1, –1), B’(–3, 2), C’(4, 3); graph iv) K’(0, 7), L’(2, 10), M’(4, 5); graph v) X’(1, 1), Y’(6, –3), & Z’(–1, –5); graph 6. i) T = 2 1 < F ii) T = 2 3 < F iii) P(2, 4) iv) P’(3, 5), Q’(5, 8) & R(8, 1) v) AB 4 2 = d n; A’(7, 3), B’(11, 5) & C’(9, 0) 7. iii) A’(7, 2), B’(3, 5), C’(2, 3); A’’(3, 5), B’’(–1, 8), C’’(–2, 0); graph iv) P(x, y) → P’(2 – x, y) → P’’(x + 4, y); both are equal v) Discuss with your teacher.

298 PRIME Opt. Maths Book - IX 7.4 Enlargement Let us consider DABC is an object. It’s image DA’B’C’ is drawn taking o as the constant point and OA’ = 20A, OB’ = 20B and OC’ = 20C. The image so formed is 2 times bigger than the object DABC where the fixed point O is called the centre of enlargement and constant 2 (double) is called the Scale factor where k = ` ` OA OA OB OB = = ` OC OC = 2. The size of image depends upon the constant value (k) (which is used to draw the image by taking less or more than the length of OA, OB and OC from the centre) which is called the scale factor. The transformation where the image is enlarge or contract in a certain proportion according to the centre and scale factor is called the enlargement. • In the enlargement E [0, k] where, the centre of enlargement is origin and ‘k’ is the scale factor. • In the enlargement E[A, K] where A is the centre of enlargement other than origin and ‘k’ is the scale factor. If k = 1, the object and image are invariant. If k > 1, the image is enlarged than object. If k < 1, the image is reduced than object. If k < 0, the image is in opposite direction than the object. If k = –1, the image and object are equal in size but in opposite direction. Example: Enlarge the given arrow about the centre of enlargement at ‘O’ and scale factor –2. Here, Join OA and produced to OA’ where OA’ = 20A similarly OB’ = 20B OC’ = 20C OD’ = 20D Then, Arrow A’B’C’D’ is the image of arrow ABCD. C’ B’ A’ C’ B A O D’ O C B A A’ B’ C’ D’

1 1 1 1 – – × – + + + + a k y` (k 1 1 1 + 1 – × – + 1 2 m x 1 2 + a k PRIME Opt. Maths Book - IX 299 7.4.1 Enlargement using Co-ordinates 1. Having centre origin and scale factor ‘k’. i.e. E[o, k]. Let p1 (x1 , y1 ) be the image of an object p(x, y) under an enlargement having, origin 0(0, 0) is the centre of enlargement and ‘k’ is the scale factor where, scale factor (k) = ' OP OP = k i.e. OP’ = k . OP OP = 1 ∴ PP’ = k – 1. Using internal section formula for OP’. (x , y) = , m m m x m m 1 2 2 1 m1 y2 m2 y1 + + + P(x , y) = ` ( ) , ) k x k 1 1 k 1 1 0 0 – × + – + a k P(x , y) = ' , ' k x k y a k Equating the corresponding elements, ' k x = x and ' k y = y ∴ x’ = kx and y’ = ky ∴ p(x , y) → p’(kx , ky) ii. Having center origin (a, b) and scale factor ‘k’. Let, A’ (x’, y’) be the image of a point A(x, y) under, Center of enlargement p(a,b) and scale factor ' PA PA = K where, PA’ = K. PA PA = 1, AA’ = K–1 Then, Using section formula internally for PA’. (x,y) = , m m m x m x m m m y m y 1 2 1 2 2 1 1 2 1 2 2 1 + + + + a k A(x.y) = ` ( ) , ` ( ) k x k a k y k b 1 1 1× –1 or, (x,y) = ` , ` k x ka a k + – y + kb – b a k By equating corresponding elements, or, x = ' k x + ka–a , y = ' k y + kb–b or, x’ = kx – ka + a , y’ = ky – kb + b , or, x’ = k(x–ka)+a , y’ = k(y–b)+b ∴ p(x , y) → p’[k(x–a)+a , k(y–b)+b] Y Y’ X P(x, y) k P’(x’, y’) O (0, 0) X’ A(x, y) k – 1 1 A’(x’, y’) P(a, b)

300 PRIME Opt. Maths Book - IX 1. Draw the image of a quadrilateral given in diagram under E[0 , 2]. B C D’ O A Solution: D’ C’ A’ B’ B C D’ O A Here, O is the center of enlargement quad. ABCD is an object. Then, scale factor (k) = 2. Join OA and produce to A’ where OA’ = 2OA Similarly, OB’ = 2OB OC’ = 2OC and OD = 2OD. \ Quad. A’B’C’D’ is formed which is the image of quad. ABCD 2. Find the image of a point p(–2, 3) under an enlargement (i) E1 [0, –2] (ii) E2 [(1,2), 2]. Solution: Under an enlargement of an quadrilateral given in diagram under E[0,-2]. p(x, y) → p’ (kx, ky) → p’ (–2x, – 2y) \ p(-2 , 3) → p’[–2 (–2), (–2) 3] → p’ (4, –6) Again, Under an enlargement E2 [(1, 2), 2] p(x,y) → p’[k(x – a)+a, k(y – b) + b] → p’[2(x – 1) + 1, 2(y – 2) + 2] → p’ (2x – 1, 2y – 2) \ p(–2 , 3) → p’[2(–2) – 1, 2(3) – 2] → p’ (–5, 4) Worked out Examples

PRIME Opt. Maths Book - IX 301 3. Find the image of DABC having vertices A(–2, 3), B(1,–3) and C(2, 4) under an enlargement E[0, –2]. Also plot the object and image in graph. Solution: Under enlargement about E[0,–2] p(x, y) → p’(kx , ky) → p’(–2x , –2y) \ A(–1, 3) → A’ (–2(–1), (–2) 3) = A’(2, –6) B(1, –3) → B’ (–2(1), (–2) (–3) = B’(–2, 6) C(2, 4) → C’ (–2(2), (–2) 4) = C’(–4, –8) X’ X O Y Y’ C(2, 4) B(1, –3) A’(2, –6) B’(–2, 6) C’(–4, –8) A(–1, 3) Here, DABC is an object. DA’B’C’ is the image under an enlargement E[0, –2]. 4. Find the image of quadrilateral having vertices A(2, 2), B(–1, 3), C(–2, –3) and D(3, –2) under an enlargement E(1, –1), 2). Also plot the object and image in graph. Solution: Under an enlargement E[(1, –1), 2] p(x, y) → P’[k(x – a) +a, k(y – b) + b) → P’[2(x – 1) +1, 2(y + 1) – 1) → P’(2x – 1, 2y + 1) \ A(2, 2) → A’(2 × 2 – 1, 2 × 2 + 1) =A’(3, 5) B(–1, 3) → B’(2(–1) – 1, 2 × 3 + 1) =B’(–3, 7) C(–2, –3) → C’(2(–2) – 1, 2(–3) + 1) =C’(–5, –5) D(3, –2) → D’(2(3) – 1, 2(–2) + 1) =D’(5, –3)

302 PRIME Opt. Maths Book - IX X’ X Y Y’ C(–2, –3) B(–1, 3) A(2, 2) D(3, –2) A’(3, 5) B’(–3, 7) C’(–5, –5) D’(5, –3) O Here, Quad. ABCD is an object. Quad. A’B’C’D’ is the image under an enlargement E[(1, –1), 2]. 5. Find the center of enlargement and scale factor where it maps DABC having vertices A(1, 2), B(3, 5) and C(5, –2) to DA’B’C’ having vertices A’(4, 3), B(8, 9) and C’(12, –5). Solution : Plotting the object and image given under an elargement in graph having A(1, 2) → A’(4, 3) B(3, 5) → B(8, 9) C(5, –2) → C’(12, –5), we get, X’ X Y Y’ C(5, –2) P(–2, 1) A’(4, 3) B’(8, 9) C’(12, –5) O A(1, 2) B(3, 5)

PRIME Opt. Maths Book - IX 303 Now, join AA’, BB’, and CC’, which are intersected at point p(–2, 1) Again, Scale factor (k) = ' PA PA = ' PB PB = ' PC PC = ( ) ( ) ( – ) ( – ) ` x x y y for PA x x y y for PA 2 – 1 – 2 2 1 2 2 1 2 2 1 2 + + = ( ) ( ) ( ) ( ) 1 2 2 1 4 2 3 1 – – 2 2 2 2 + + + + = 10 36 + 4 = 10 2 10 = 2 ∴ Center (a, b) = (–2, 1) Scale factor (k) = 2. Alternative method. Let An enlargement be E[a, b), k] Where, P(x, y) → P’[k(x – a) + a, k(y – b) + b] ∴ A (1, 2) → A’[k(1 – a) + a, k(2 – b)+b] B (3, 5) → B’[k(3 – a) + a, k(5 – b)+b] C(5, –2) → C’[k(5 – a) + a, k(–2 – b)+b] From given, A(1, 2) →A’(4, 3) B (3, 5) →B’(8, 9) C (5, –2) →C’(12, –5) By equating the image A’. k(1 – a) + a = 4 and k(2 – b) + b = 3 ∴ k = a a 1 4 – – ..........................(i) k = b b 2 3 – – ..........................(ii) By equating the image B’, k(3 – a) + a = 8 and k(5 – b) + b = 9 ∴ k = a a 3 8 – – .......................... (iii) k = b b 2 9 – – ..........................(iv) Solving equation (i) and (ii) a a 1 4 – – = b a 3 8 – – or, 12 – 4a – 3a + a2 = 8 – a – 8a + a2 or, 2a = –4 ∴ a = –2

304 PRIME Opt. Maths Book - IX Solving equation(ii) and (iv) b b 2 3 – – = b b 5 9 – – or, 15 – 3b –5b + b2 = 18 – 2b – 9b + b2 or, 3b = 3 \ b = 1. Putting the value of ‘a’ in equation(i), k = 1 2 4 2 + + = 2 \ Center = (–2, 1) Scale factor (k) = 2 7.4.2 Properties of enlargement: i. Object and image are similar to each other. ii. If scale factor k >1, the object is said to be enlarged in same side of the object. iii. If 0< k <1, the object is said to be reduced in same side of the object. iv. If k <0, the image will be in opposite side to the object and it will be inverted also. v. If k = 1, object and image are coincided. vi. If k = –1, object and image are equivalent but remain in opposite from the center of enlargement. vii. The scale factor is the ratio of distance of image and object from the center. i.e. k = ' PA PA where p is the center of enlargement. viii. Sides of object and image are also parallel. ix. A and A’ are always lie in a same straight line passes through the center of enlargement.

PRIME Opt. Maths Book - IX 305 Exercise 7.4 1. i) What do you mean by enlargement in transformation? ii) Write down the formula to find the image of A(x, y) under enlargement about E[(a, b), k]. iii) Find the image of a point p(a, b) under enlargement about E[0, 2], followed by Enlargement about E2[0, 3 2 ]. iv) In what condition in enlargement the object will be invariant? v) In what condition the object is reduced in enlargement? 2. Draw the image of the given objects under the center of enlargement ‘o’ and scale factor ‘k’. i) B C O (k = 2) A ii) Q R S P O E(0, 3) iii) B D C A O (k = 2) iv) P Q R O (k = 2 3 ) v) O (k = –1) A D E H B C F G 3. Find the image of a point p(–2, 4) under the following enlargements. i) E1 [0, 2] ii) E2 [0, – 2 3 ] iii) E3 [0, –2] iv) E4 [0, 2 1 ] v) E5 [0, –3] 4. Find the image of an object point A(3, –5) under the following enlargements. i) E1 [(1, 1), 2] ii) E2 [(1, 2), –2] iii) E3 [(0, 2), –3] iv) E4 [(3, 0), –1] v) E5 [(–2, 1), 3]

306 PRIME Opt. Maths Book - IX 5. Find the image of the following under the given enlargements. Also plot the object and image in graph. i) Triangle having vertices A(2, 1), B(4, 5) and C(5, – 3) under the enlargement E[0, 2] ii) Triangle having vertices P(–3, 0), Q(4, 0) and R(2, 4) under the enlargement E[0, 3] iii) Triangle having vertices K(2, 4), L(–4, 8) and O(0, 0) under the enlargement E[0, 0), – 2 3 ]. iv) Quadrilateral having vertices A(2, 4), B(–4, 2), C(–2, –4) and (4, –2) under the enlargement E[0, 2 3 ]. v) Quadrilateral having vertices P(2, 1) Q(3, 4), R(5, 3) and S(6, – 2) under an enlargement E[0, –2). 6. Find the image of the followings under the given enlargements. Also plot the object and image in graph. i) Triangle PQR having vertices P(6, 2) Q(6, 6) and R(2, 4) under the enlargement E[(2, 4), 2 1 ]. ii) Triangle having vertices A(3, 4), B(–2, 6) and C(1, 6) under an enlargement E[(–1, 2), –2]. iii) Triangle having vertices P(–3, 2) Q(0, 4) and R(–1, 5) under the enlargement E[(1, 2), 2]. iv) Quadrilateral having vertices A(2, 3) B(–3, 2), C(–2, –2) and D(3, –3) under an enlargement E[(1, –1), 3]. v) Triangle having vertices A(3, 4), B(–2, 6) and C(1, –2) under an enlargement having centre (2, 2) and Scale factor –2. 7. PRIME more creative questions: i) The vertices image of a triangle are A’(8, –2) B’(6, 2) and C’(16, –6) after enlargement about centre origin and scale factor 2. Find the vertices of DABC. Also plot the object and image in graph. ii) The vertices of image of triangle PQR are P’(3, –5), Q’(5, – 1) and R’(–1, –1) under an enlargement E [(3, –1), –2]. Find the coordinate of the vertices for DPQR. Also plot the object and image in graph. iii) An enlargement gives image of an object at A(3, 2) → A’(5, 7), B(4, –2) → B’(7, 1) and C(5, 4) → C’(9, 11). Find centre of enlargement and scale factor. iv) Draw the DPQR having vertices P(2, 4), Q(4, 2) and R(6, 6) and image of DP’Q’R’ having vertices P’(–4, –8), Q’ (–8, –4) and R’ (–12, –12). Then find the centre of enlargement and Scale factor. v) If an enlargement transferred a triangle having vertices A(3, 2), B(4, –5) and C(5, 0) to the triangle having vertices A’(–3, –7), B’(–5, 7) and C’(–7, –3). Find the centre of enlargement and scale factor. vi) What is enlargement? Write down its properties. Where is it applicable in our daily life? Discuss and prepare a report in group.

PRIME Opt. Maths Book - IX 307 Answer 1. Show to your teacher. 2. Show to your teacher. 3. i. (–4, 8) ii. (3, –6) iii. (4, –8) iv. (–1, 2) v. (6, –12) 4. i. (5, –11) ii. (–3, 16) iii. (–9, 23) iv. (3, 5) v. (13, –17) 5. i. A’(4, 2), B’(8, 10), C’(10, –6) ; graph. ii. P’(–9, 0), Q’(12, 0), R’(6, 12) ; graph. iii. K’(–3, –6), L’(6, –12), O’(0, 0) ; graph. iv. A’(3, 6), B’(–6, 3), C’(–3, –6), D’(6, –3) ; graph. v. P’(–4, –2), Q’(–6, –8), R’(–10, –6), S’(–12, 4) ; graph. 6. i. P’(4, 3), Q’(4, 5), R’(2, 4) ; graph. ii. A’(–9, –2), B’(1, –6), C’(–5, –6) ; graph. iii. P’(–7, 6), Q’(–1, 10), R’(–3, 12) ; graph. iv. A’(4, 11), B’(–11, 8), C’(–8, –4), D’(7, –7) ; graph. v. A’(0, –2), B’(10, –6), C’(4, 10) ; graph. 7. i. A(4, –1), B’(3, 1), C’(8, –3) ; graph. ii. P(3, 1), Q(2, –1), R(5, –1) ; graph. iii. Centre = (1, –3), Scale factor = 2 iv. Centre = O(0, 0), Scale factor = 2 ; graph. v. Centre = (1, –1), Scale factor = –2 vi. Show to your teacher. 8. Project work Prepare a list of formula in a chart paper and paste at the project board kept in your classroom.

308 PRIME Opt. Maths Book - IX Tramsformation Unit Test - 1 Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. Which transformations are isometric and which are non- isometric? Name them. 2. a. Find the image a point A(2, –3) under reflection about x-axis followed by translation about T = 1 2 < F. b. If A(3, –2) is the image of a point ‘A’ under rotation about – 90° with centre origin followed by reflection about y – 2 = 0. Find the co-ordinate of point A. c. If a point P(2, 4) is translated to P’(4, –2) under a translation ‘T’. Find the translation vector. 3. a. Find the image of a triangle having vertices A(2, 4), B(–1, 1) and C(3, –3). Find the image of it under reflection about x + 2 = 0 followed by rotation about –270° with centre origin. Also plot the object and images in graph. 4. a. Find the image of quadrilateral having vertices (–3, 3), (–2, –1), (1, –3) and (2, 4) under translation about AB followed by enlargement about E[(3, 2), 2]. Also plot the object and image in graph. b. Find the centre of enlargement and scale factor where the vertices of triangle are transferred as A(2,1) → A’(3,4), B(–2,–1) → B’(–5,0) and C(–1,3)→C’(–3,8).

PRIME Opt. Maths Book - IX 309 Unit 8 Statistics Specification Grid Table K(1) U(2) A(4) HA(5) TQ TM Periods No. of Questions – 1 2 – 3 10 12 Weight – 2 4 – K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total Question, TM = Total Marks Objectives : At the end of the lesson • Students are able to understand and find the central tendency, its types and procedure. • Students are able to find the disparsion. • Student are able to find the partition values like median, quartiles, decides and percentiles. • Students are able to find the value of dispersion like mean deviation, quartile deviation and standard deviation for individual and discrete observations. Materials Required: • Chart paper. • Formula chart of central tendency. • Secondary data of different field like marks, population, cost etc • Formula chart and model of measure of dispersion.

310 PRIME Opt. Maths Book - IX 8.1 Statistics Enjoy the recall The collection of information like marks, ages, incomes, productions etc in a certain area to analyses the informations is called data. • The number of the members involve in the collected data is called frequency. • Distribution of the information in a table to analyses is called tabulation. • Diagrammatic representation of the tabulated data called diagrammatic interpretation. • The collected data should be taken in ascending or in descending order for systemic analysis. • The observation lies at the middle of the data is called median. i.e. median = size of N 2 1 th + a k item. • The average value of the collected data is called arithmetic mean i.e. AM( ) X = N x N Σ Σfx = . • The central value of the observations like mean, median and mode are called the measures of central tendency. The quartiles, deciles and percentiles of observation are called the partitions. 8.1.1 Partition Values : Let us take an example of the marks obtained by 7 students of grade 9: 8, 12, 24, 18, 14, 20, 16. Here, Taking the marks in ascending order as 8, 12, 14, 16, 18, 20, 24. Where 16 divides the observations in two equal parts which is called median. Also, 12, 16 and 20 divides the observations in four equal parts which are called quartiles. Here, the observations 12, 16 and 20 divides the observations in different equal parts which are called the partition values. Mean, mode and median are the central values of a frequency distribution where as median, quartiles, deciles and percentiles are the values of partitions. 8.1.2 Quartiles: The partition values which divide the arranged (ascending or decending) data into four equal parts are called quartiles. • There are three quartiles which divide the data into four equal parts called first quartile (Q1 ), median (M2 ) and third quartile (Q3 )

PRIME Opt. Maths Book - IX 311 • First quartile (Q1 ) cuts the data in two parts, making 25% in the left and 75% in the right, which is called lower quartile too. • Median (Md ) cuts the data making in two parts, 50% in the left and 50% in the right which is called median too. • Third quartile Q3 cuts the data in two parts making 75% in the left and 25% in the right which is called upper quartile. • Size of N 4 1 + th ` j item is the first quartile, N 2 4 1 + th ` j item is the second quartile and N 3 4 1 + th ` j item is the third quartile. • Items for different quartiles to be determined from cumulative frequency table. Examples 1. Find the 3 quartiles of the marks obtained by 10 students in mathematics 44, 68, 95, 98, 86, 70, 32, 50, 58, 72. Solution: The marks obtained by 10 students in ascending order are: 32, 44, 50, 58, 68, 70, 72, 72, 86, 95, 98. No. of students (n) = 10 First quartile (Q1 ) = Size of n 4 1 + th ` j item = Size of 4 10 1 + th ` j item = Size of 2.75th item =2nd + 0.75 (3rd – 2nd) = 44 + 0.75 (50 – 44) = 44 + 4.5 = 48.5 Second quartile (Q2 ) = Size of n 2 4 1 + th ` j item = Size of 5.5th item. = 5th + 0.5 (6th – 5th) = 68 + 0.5(70 – 68) = 68 + 1 = 69 Third quartile (Q3 ) = Size of 3 n 4 1 + th ` j item = Size of 8.25th item = 8th + 0.25 (9th – 8th) = 86 + 0.25 (95 – 86) = 86 + 2.25 = 88. 25

312 PRIME Opt. Maths Book - IX 2. Find the first and third quartile from the given observations. x 24 18 12 48 40 36 30 f 9 7 3 2 8 10 6 Solution: The observations in ascending order with c.f. table. x f c.f 12 3 3 18 7 3 + 7 = 10 24 9 10 + 9 = 19 30 6 19 + 6 = 25 36 10 25 + 10 = 35 40 8 35 + 8 = 43 48 2 43 + 2 = 45 N = 45 First quartile lies in = Size of N 4 1 + th item = Size of 4 45 1 th + item = Size of 11.5th item 19 is the just greater than 11.5 in c.f. \ First quartile (Q1 ) = 24. Again Third quartile lies in, = Size of 3 N 4 1 + th ` j item = Size of 3 × 11.5th item = Size of 34.5th item 35 is the just greater than 34.5 in c.f. \ Third quartile (Q3 ) = 36. 8.1.3 Deciles: The partition values which divide the given data into ten equal parts are called deciles. • There are 9 deciles D1 , D2 , D3 , .................... D9 which divide the data into ten equal parts. • 5th decile cuts the data in two equal half 50% in the left and 50% in the right which is called median of the data. • Size of N 10 1 th + a k item gives the first decile (D1 ) and 9 N 10 1 th + a k item gives the last decile (D9 ) • Cumulative frequency distribution table to be constructed for discrete observations.

PRIME Opt. Maths Book - IX 313 Example: 3 3. Find 4th and 9th deciles from the observations 15, 21, 24, 30, 36, 40, 44, 48, 50, 54. Solution: The observations in ascending order are 15, 21, 24, 30, 36, 40, 44, 48, 50, 54. No. of observations (n) = 10. Then, 4th decile (D4 ) = Size of 4 n 10 1 th + a k item = Size of 4 10 10 1 th + a k item = Size of 4.4th item = 4th + 0.4 (5th – 4th) = 30 + 0.4 (36 – 30) = 30 + 0.4 × 6 = 32.4. 9th decile (D9 ) = Size of 9 n 10 1 th + a k item = Size of 9.9th item = 9th item + 0.9 (10th – 9th) = 50+ 0.9 (54 – 50) = 50 + 3.6 = 53.6 4. Find the 6th and 8th deciles of the marks obtained by the students of grade IX. x 10 18 28 34 40 48 52 f 5 7 9 15 8 6 4 Solution : The observations in ascending order with c.f. table. x f c.f 10 5 5 18 7 5 + 7 = 12 28 9 12 + 9 = 21 34 15 21 + 15 = 36 40 8 36 + 8 = 44 48 6 44 + 6 = 50 52 4 50 + 4 = 54 N = 54 Now, 6th deciles lies in, = Size of 6 N 10 1 th + a k item = Size of 6 × 10 55 th item = Size of 33rd item 36 is just greater than 33 in c.f. \ D6 = 34

314 PRIME Opt. Maths Book - IX Again, 8th decile lies in, = Size of 8 N item 2 1 th + a k = Size of 44th item 44 is equal to he 44 in c.f. \ D8 = 40 8.1.4 Percentiles: The partition values which divide the data in 100 equal parts are called percentiles. • There are 99 percentiles from P1 , P2 , P3 , ................, P99 which divide the data into 100 equal parts. • P50 divides the data 50% in the left and 50% in the right which is called median of the data. • Size of N 100 1 th + a k item given the first percentile (P1 ) and 99 N 100 1 th + a k item gives the last percentile (P99) of the data. • Cumulative frequency table have to be used to calculate percentiles for discrete frequency distribution. 5. Find 25th percentile and 90th percentile of the observations 54, 58, 62, 70, 76, 84, 90, 94, 98, 106, 110 and 116. Solution: The given observations in ascending order are, 54, 58, 62, 70, 76, 84, 90, 94, 98, 106, 110, 116. No. of observation (N) = 11 Now, 25th percentiles (P25) = Size of 25 n 100 1 th + a k item = Size of 25 100 11 1 th + a k item = Size of 3rd item = 62 Again, 90th percentile (P90) = Size of 90 n 100 1 th + a k item = Size of 10.8th item = 10th + 0.8(11th – 10th) = 110 + 0.8(116 – 110) = 110 + 4.8 = 114.8

PRIME Opt. Maths Book - IX 315 6. Find the 32nd percentile and 80th percentile of the observation given below. x 18 28 32 40 46 52 60 f 2 6 9 13 8 5 3 Solution: The observation in ascending order are with c.f table. x f c.f 18 2 6 + 2 = 8 32 6 8 + 9 = 17 40 13 17 + 13 =30 46 8 30 + 8 = 38 52 5 38 + 5 = 43 60 2 43 + 5 = 45 N = 45 Now, 32nd percentile lies in, = Size of 32 N 100 1 th + a k item. = Size of 14.72th item. 17 is just greater than 4.72 in c.f, \ P32 = 32 Again, 80th percentile lies in, = Size of 80 N 100 1 th + a k item. = Size of 36.8th item. 38 is just greater then 36.8 in c.f. \ P80 = 46

316 PRIME Opt. Maths Book - IX Exercise 8.1 1. Answer the following questions. i) What is median? What step have to be done to calculate median of the individual observation? ii) What is quartile? What steps have to be done to calculate 3rd quartile from discrete observations? iii) What is decile? What steps have to be done to calculate 6th decides of the individuals observations? iv) What is percentiles? What steps have to be done to calculate 75th percentile from the discrete observation? v) What is partition value? Write down the types of such values. 2. Find the followings from the given observations. i) All the quartiles from the marks obtained by the students of grade IX 18, 12, 16, 8, 20, 14, 10, 24, 22, 26, 28 ii) First and third quartiles from: 112, 108, 104, 100, 124, 120, 116, 96 iii) Find quartiles from the given observations 60, 64, 68, 70, 75, 79, 82, 84, 92, 96 iv) Find first and last quartiles of the observations given below. 210, 200, 140, 150, 180, 170, 160, 130, 120, 110, 100, 190, 220 v) Find the lower and upper quartile of the following production of crops given in metric tonn. 220, 230, 240, 260, 270, 210, 200, 290, 280, 310, 300, 250 3. i. Find the first quartile from the given frequency distribution table. x 28 36 42 48 52 60 64 f 4 5 7 12 9 6 3 ii) Find the third quartile of data given in table. x 55 65 50 40 45 f 4 5 7 12 9 iii) Find the lower quartile of the observations given below. marks 120 110 100 160 140 150 130 f 13 8 4 3 8 9 15 iv) Find the second quartile of the marks obtained by the students out of 20. Age 12 14 16 18 20 f 2 5 9 6 3

PRIME Opt. Maths Book - IX 317 v) Find the upper quartile of the weight of children given below. Height 20 10 25 45 15 40 35 30 f 12 4 9 3 7 6 8 5 4. Find the following deciles from the observation given below. i) Find the 4th and 9th deciles of the production of cropes in metric tonn of a place. 124, 136, 130, 120, 140, 160, 154, 150, 170, 164, 146 ii) Find 6th decile of the weight of students given below. 21, 25, 30, 35, 40, 44, 50, 54, 60 iii) Find the 7th decile of the observations: 60, 21, 27, 35, 42, 46, 39, 12, 15, 18, 50, 54, 24, 30 iv) Find the 6th decile of the observation given below: x 44 34 24 14 74 64 54 f 15 12 10 7 12 8 6 v) Find the 3th deciles of the marks obtained by the grade IX student in mathematics in the third terminal examination given below. Marks 210 250 290 370 330 f 31 37 45 32 34 vi) Find the 8th deciles of the observations given in table. Age 15 10 5 25 20 30 35 f 9 7 3 6 11 5 2 5. Find the following percentiles from the given observations. i) Find the 15th percentiles and 80th percentiles of the observations. 12, 20, 30, 36, 44, 52, 60, 68, 76, 84, 90, 98, 100 ii) Find the 20th and 90th percentiles of the observations given below. 105, 120, 100, 115, 110, 140, 135, 130, 125 iii) Find 28th and 67th percentiles of the number of students of different schools given below in hilly region. 230, 240, 250, 220, 210, 220, 300, 290, 280, 270, 260 iv) Find the 8th percentile of the observations given below. x 25 35 45 55 65 75 85 f 90 100 110 120 105 95 79 iv) Find the 49th percentiles of the production of crops of districts given in metric tonn: x 200 300 400 500 600 700 800 f 3 7 9 13 20 12 8 v) Find the 85th percentiles of the observations given in table. x 120 110 100 160 140 130 85 f 24 18 12 15 24 30 16

318 PRIME Opt. Maths Book - IX Answer 1. Discuss with your subject teacher. 2. i) 12, 18, 24 ii) 101, 110, 119 iii) 67, 77, 86 iv) 125, 160, 195 v) 222.5, 255, 287.5 3. i) 42 ii) 55 iii) 120 iv) 16 v) 35 4. i) 139.2, 168.8 ii) 44 iii) 44 iv) 15 v) 250 5. i) 21,91.6 ii) 105, 140 iii) 223.6, 270.4 iv) 25 v) 140 vi) 600 6. i) 25 ii) 20 iii) 50 iv) 50 v) 15 6. PRIME more creative questions: i) If 3rd quartile of the observations taken in order 24, 30, 36, 40, 44, 48, 2x + 3, 3x – 20, 60 is 54, find the value of ‘x’. ii) If 7th decile of the observations taken in order 52, 60, 68, 76, 80, 84, 88, 4x + 12, 5x – 4, 100, 106 is 93.6, find the value of ‘x’. iii) If 60th percentile of the observations taken in order 114, 116, 118, 120, 122, 2x + 24, 126, 128, 130, 132 is 124, find the value of ‘x’. iv) If 75th percentiles of the observations taken in order 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, x +2, 2x – 44, 60, 64 is 53, find the value of ‘x’. v) If lower quartile of the observations taken in order 13, 2x – 12, 24, 28, 32, 36, 40 is 18, find the value of ‘x’.

PRIME Opt. Maths Book - IX 319 8.2 Measure of dispersion: The measure of central values mean, median and mode are learned in previous classes as well as in compulsory mathematics in the topic of measure of central tendency. But central values fail to represent the characteristic of data when the values are scattered widely and deviated more irregularly from central values. The study of scatteredness of the values of variable from its central value is called the measurement of dispersion. Let us take two types of observations having same mean and median. Data A 30 36 40 44 50 40 40 Data B 10 25 40 45 60 40 40 Here the mean and median of both types of observations are same which are 40 for each. But observations are dispersed closely in ‘A’ and dispersed far in ‘B’. It results distribution of data in example A is better than in example B. This type of analysis is called dispersion. The measurement of scatteredness of the observation from their central values mean, median and mode to know the detail of distribution is called the measure of dispersion. In grade IX, we discuss the following measure of dispersions for individual and discrete series. • Range (It is given in compulsory mathematics) • Quartile deviation • Mean deviation • Standard deviation 8.2.1 Quartile deviation: It is defined as the half of the range of quartiles of the given observations out of three quartiles Q1 , Q2 and Q3 which is also called semi-interquartile range (quartile deviation). Range of quartiles out of Q1 , Q2 and Q3 is Q3 – Q1 (which is inter-quartile range) and half of its value is called semi-inter-quartile range (Quartile deviation). i.e. Q.D. = Q Q 2 3 2 – where, Q1 = First quartile Q3 = Third quartile Range = Q3 – Q1 for quartiles. Coefficient of quartle deviation is calculated by using the formula,

320 PRIME Opt. Maths Book - IX Coefficient of Q.D. = Q Q Q Q– 3 1 3 1 + Note : Only first and third quartiles are involved in it. Steps for the calculation of Quartile deviation • Arranging the observation in ascending order. • Finding of total number of observations N = ∑f. • Finding of Q1 and Q3 of the observations. Q1 = size of N 4 1 + th ` j item Q3 = size of 3 N 4 1 + th ` j item • Find of corresponding observations for such sizes which are Q1 and Q3 respectively. • Find quartile deviation as, Q.D. = 2 1 (Q3 – Q1 ) Coefficient of Q.D. = Q Q Q Q– 3 1 3 1 + • Corresponding observations have to be taken in discrete observations from cumulative frequency table. Table : X f C.f. Q1 Q3 N 4 1 + th ` j 3 N 4 1 + th ` j N = ∑f Q1 lies in size of N 4 1 + th ` j item. Q3 lies in size of 3 N 4 1 + th ` j item. Then, Q.D. = 2 1 (Q3 – Q1 ) [Semi-interquartile range] Interquartile range = Q3 – Q1 Coefficient of Q.D. = Q Q Q Q– 3 1 3 1 +

PRIME Opt. Maths Book - IX 321 1. If first quartile and third quartile of the observations are 10 and 30 respectively. Find the quartile deviation and its coefficient. Solution : First quartile (Q1 ) = 10 Third quartile (Q3 ) = 30 Then, Quartile deviation (Q.D.) = 2 1 (Q3 – Q1 ) = 2 1 (30 – 10) = 10 Coefficient of Q.D. = Q Q Q Q– 3 1 3 1 + = 30 10 30 – 10 + = 0.5 2. If quartile deviation and its coefficient are respectively 30 and 0.6. Find th first and third quartiles. Solution : Quartile deviation (Q.D.) = 30 Coefficient of Q. D. = 0.6 We have, Q.D. = 2 1 (Q3 – Q1 ) or, 30 × 2 = Q3 – Q1 or, Q1 = Q3 – 60 .................... (i) Again, Coefficient of Q.D. = Q Q Q Q– 3 1 3 1 + or, 0.6 = Q Q Q Q– 3 1 3 1 + or, 10 6 = Q Q Q Q– 3 1 3 1 + or, 3Q3 + 3Q1 = 5Q3 – 5Q1 or, 8(Q3 – 60) = 2Q3 or, 6Q3 = 480 \ Q3 = 80 Putting the value of Q3 in equation (i) Q1 = 80 – 60 = 20 \ Q1 = 20 Q2 = 60 Worked out Examples

322 PRIME Opt. Maths Book - IX 3. Find the quartile deviation of the observations 15, 24, 18, 12, 36, 33, 30 27, 21. Solution : The given observations in ascending order are: 12, 15, 18, 21, 24, 27, 30, 33, 36 Number of observations (N) = 9 Now, First quartile (Q1 ) = size of N 4 1 + th ` j item = size of 4 10 th ` j item = size of 2.5th item = 2nd + 0.5 (3rd – 2nd) = 15 + 0.5(18 – 15) = 15 + 1.5 = 16.5 Third quartile (Q3 )= size of 3 N 4 1 + th ` j item = size if 3 × 2.5th item = size of 7.5th item = 7th + 0.5(8th – 7th) = 30 + 0.5(33 – 30) = 31.5 Then, Quartile deviation (Q.D.) = 2 1 (Q3 – Q1 ) = 2 1 (31.5 16.5) = Q Q Q Q– 3 1 3 1 + × 15 = 7.5 Coefficient of Q.D. = Q Q Q Q– 3 1 3 1 + = . . . . 31 5 16 5 31 5 1– 6 5 + = 48 15 = 0.3125

PRIME Opt. Maths Book - IX 323 4. Find quartile deviation and its coefficient of the marks obtained by students in grade IX. Marks 35 30 24 12 18 40 48 f 15 12 7 3 5 4 4 Solution : The observations in ascending order: Marks f c.f. 12 3 3 18 5 3 + 5 = 8 24 7 8 + 7 = 15 30 12 15 + 12 = 27 35 15 27 + 15 = 42 40 4 42 + 4 = 46 48 4 46 + 4 = 50 N = 50 Here, First quartile lies in, = size of N 4 1 + th ` j item = size of 4 50 1 th + a k item = size of 12.75th item 15 is just greater than 12.75 in c.f. \ Q1 = 24 Third quartile lies in, = size of 3 N 4 1 + th ` j item = size of 38.25th item 42 is just greater than 38.25 in c.f. \ Q3 = 35 Then, Quartile deviation Q.D. = 2 1 (Q3 – Q1 ) = 2 1 (35 – 24) = 2 1 × 1 = 5.5 Coefficient of Q.D. = Q Q Q Q– 3 1 3 1 + = 35 24 35 – 24 + = 59 11 = 0.1864

324 PRIME Opt. Maths Book - IX Exercise 8.2 1. Find the followings. i) If first and third quartiles of the observations are 20 and 60. Find the quartile deviation and its coefficient. ii) The upper and lower quartiles of the observations are 25 and 55 respectively. Find the semi-inter-quartile range and its coefficient. iii) The first quartile of the observations having inter-quartile range 20 is 10, find the coefficient of quartile deviation. iv) The quartile deviation and its coefficient are respectively 30 and 0.6. Find the first and third quartiles of the observations. v) The inter-quartile range and coefficient of quartile deviation are 100 and 0.5 respectively find the two extremes quartiles. 2. Find the quartile deviation and its coefficient of the following observations. i) 28, 24, 18, 12, 6, 52, 48, 42, 38, 34, 56 ii) 42, 48, 44, 46, 58, 56, 52, 50, 38 iii) 120, 110, 100, 140, 130, 160, 150,170 iv) 200, 180, 160, 260, 240, 220, 300, 280 v) 10, 20, 30, 100, 90, 80, 70, 60, 50, 40, 120, 130, 110 3. Find semi-interquartile range and its coefficient of the followings. i) x 12 18 24 30 36 42 f 3 7 12 9 6 4 ii) Marks 44 50 25 30 36 40 f 6 4 5 8 10 7 iii) Age 15 20 28 32 40 46 50 f 2 4 7 8 6 3 1 iv) Height 100 140 120 180 160 f 18 42 60 45 34 v) Weight 34 42 22 30 60 56 50 f 16 20 10 14 9 11 15

PRIME Opt. Maths Book - IX 325 4. PRIME more creative questions i) What is quartile deviation? Also define coefficient of quartile deviation with their calculation formula. ii) If Q1 of the observations taken in order 12, m – 5, m – 3, 21, 24, 27, 31, 35, 36 is 16, find the quartile deviation and its coefficient by finding ‘m’. iii) If third quartile of the observations taken in order 100, 110, 120, 130, 140, P + 50, 2P – 40 and 170 is 157.5, find the ‘P’. Also find Q.D. and its coefficient. iv) If 3rd quartile is three times the first quartile and Q.D. is 10, find the value of quartiles and coefficient of quartile deviation. v) collect the marks obtained by the students of your class in first term examination. Construct discrete frequency distribution table and find the value of inter quartile range and coefficient of quartile deviation. Answer 1. i) 20, 0.5 ii) 15, 0.375 iii) 0.5 iv) 20, 80 v) 50, 150 2. i) 15, 0.454 ii) 5.5, 0.113 iii) 22.5, 0.1667 iv) 45., 0.195 v) 35, 0.5 3. i) 6, 0.2 ii) 7, 0.189 iii) 6, 0.176 iv) 20, 0.142 v) 10, 0.25 4. i) Show to your teacher ii) 20, 8.5, 0.346 iii) 100, 22.5, 0.1667 iv) 10, 30, 0.5 v) Show to your teacher

326 PRIME Opt. Maths Book - IX 8.3 Mean deviation (Average deviation) It is defined as the average of positive deviation of all the observations from their central values mean and median. It is calculated by taking the positive difference of the observations from mean and median and their arithmetic mean. • For individual observations: Mean deviation from mean (M.D. from X) = N Σ X X– Mean deviation from median (M.D. from, Md ) = N Σ X M– d • For discrete observations: M.D. (X) = N Σ f X–X M.D. (Md ) = N Σ f X–Md • Where X is the arithmetic mean and Md is the median of the observations. • Coefficient of mean deviation is calculate by using, Coefficient of M.D. = . . X M D (from mean) Coefficient of M.D. = . . M M D d (from median) 1. Find mean deviation and its coefficient of 18, 12, 24, 20, 36 from mean. Solution : The observations taken in ascending order, X X |X – X| 12 22 10 18 4 20 2 24 2 36 14 ∑X = 110 ∑|X – X| = 32 No. of observations (N) = 5 Mean (X) = N X 5 Σ 110 = = 22 Worked out Examples

PRIME Opt. Maths Book - IX 327 Here, M.D. ^ h X = N R X X – = 5 32 = 6.4 Coefficient of M.D. ^ h X = . . X M D = . 22 6 4 = 0.29 2. Find mean deviation from median and its coefficient of: 108, 112, 116, 122, 126, 130, 136 Solution : The observation in ascending order, X Md X M– d 108 122 14 112 10 116 6 122 0 126 4 130 8 136 14 R X M– d = 56 Here, No of observations (N) = 7 Median (Md ) = size of N 2 1 th + a k item = size of 2 7 1 th + a k item = size of 4th item = 122 Then, M.D. (Md ) = N R X M– d = 7 56 Coefficient of M.D. (Md ) = . . M M D d = 122 8 = 0.065

328 PRIME Opt. Maths Book - IX 3. Find mean deviation from mean and its coefficient of: x 15 22 30 35 40 f 6 10 12 8 4 Solution : X f f × x X X X – f X – X 15 6 90 27.75 12.75 76.5 22 10 220 5.75 57.5 30 12 360 2.25 27 35 8 280 7.25 58 40 4 160 12.25 49 N = 40, ∑fx = 1110 Rf X – X = 268 Here, Mean ( X ) = N Rfx = 40 1110 = 27.75 Then, Mean deviation (M.D.) = N Rf X – X = 40 268 = 6.7 Coefficient of M.D. = X MD = . . 27 75 6 7 = 0.245 4. Find mean deviation from median and its coefficient of: Marks 18 24 30 36 42 50 f 6 8 15 7 5 1 Solution : X f c.f. Md X M– d f X – Md 18 6 6 30 12 72 24 8 14 6 48 30 13 27 0 0 36 7 34 6 42 42 1 40 12 60 50 1 40 20 20 N = 40 Rf X – Md =

PRIME Opt. Maths Book - IX 329 Here, Median lies in = size of N 2 1 th + a k item = size of 2 40 1 th + a k item = size of 20.5th item = 27 is just greater than 20.5 \ Median (Md ) = 30 Then, Mean deviation (M.D.) = N Rf X–Md = 6 5. 242 = 6.5 Coefficient of M.D. = . .( ) M M D M d d = . 30 6 5 = 0.216 4. If mean of the observations taken in order 90, 100, 110, P, 125, is 108, find mean deviation from mean and its coefficient. Solution : X X X X – 90 108 18 100 8 110 2 P 7 125 17 ∑x = 425 + P R X X – = 52 Here, No of observations (N) = 5 Mean ( X ) = N Rx or, 108 = P 5 425 + or, 540 = 425 + P \ P = 115 Again, M.D. from mean (M.D.) = N R X X – = 5 52 = 10.4 Coefficient of M.D. = . .( ) X M D X = . 108 10 4 = 0.096

330 PRIME Opt. Maths Book - IX Exercise 8.3 1. Find mean deviation from mean and its coefficient of the followings. i) Weight kg : 24, 30, 36, 40, 50 ii) Rainfall (mm) : 85, 115, 120, 140, 135, 125 iii) Temperature (°C) : 41, 25, 30, 18, 26, 45, 32, 35, 31, 27 iv) Income (In Rs. 1000) : 24, 28, 29, 23, 36, 35, 25, 18 v) Marks (Out of 20) : 19, 17, 20, 18, 15, 12, 9, 16, 14, 10 2. Find mean deviation from median and its coefficient of the followings: i) Temperature (°C) : 17, 10, 15, 7, 13, 9, 6, 18, 11, 14, 1 ii) Weight (kg) : 32, 35, 42, 30, 33, 37, 40, 43 iii) Height (ft) : 5.4, 5.2, 5.6, 5.8, 8, 4.8 iv) Marks (out of 25) : 24, 23, 25, 18, 20, 16, 21 v) In come (in Rs. 1000) : 41, 25, 30, 18, 20, 26, 43, 32, 35, 31, 27 3. Find the mean deviation from mean and its coefficient of the followings. i) Marks 5 8 15 18 20 22 f 6 5 8 10 6 5 ii) Age 12 18 24 30 36 f 4 5 8 9 4 iii) x 14 10 5 25 20 f 5 6 2 4 3 iv) Cost (Rs.) 50 45 60 70 75 55 No fo shops 12 4 7 3 6 8 v) Weight 65 70 48 50 60 45 55 f 4 1 5 4 3 6 2 4. Find mean deviation from median and its coefficient of the followings. i) Marks 48 44 38 25 36 40 No of Students 2 4 15 9 12 8

PRIME Opt. Maths Book - IX 331 ii) Rainfall (mm) 15 35 25 65 55 45 75 No of Places 1 7 5 6 4 14 3 iii) Height (cm) 50 60 75 82 90 95 Numbers 10 12 8 5 3 2 iv) x 10 18 14 16 12 f 6 10 16 23 5 v) x 12 16 20 24 28 32 36 f 7 8 10 13 6 4 2 5. PRIME more creative questions: i) If mean of the observations taken in order 2, 4, 6, m, 10, 12 is 7, find the mean deviation from mean of the observations: ii) If median of the observations taken in order is 110, find mean deviation from median of the observations 90, 100, P + 5, P + 15, 125, 130 iii) If mean of the given observations is 12, find the mean deviation from mean and its coefficient of the observations. Marks 6 8 10 12 14 16 18 20 f 1 14 25 27 m 9 4 2 iv) Differentiate between mean deviation from mean and mean deviation from median. Also write down their calculating formula. v) Collect the marks obtained by the students of grade IX in first terminal examination of your school in optional mathematics and calculate mean deviation from mean and from median by constructing discrete frequency distribution table.

332 PRIME Opt. Maths Book - IX Answer 1. i) 7.2, 0.2 ii) 13.33, 0.111 iii) 5.8, 0.187 iv) 4.75, 0.1773 v) 3, 0.2 2. i) 4, 0.3636 ii) 4, 0.111 iii) 0.3, 0.0566 iv) 2.57, 0.1224 v) 6, 0.2 3. i) 4.75, 0.3167 ii) 6.05, 0.248 iii) 5.5, 0.3667 iv) 8, 0.139 v) 7.04, 0.132 4. i) 4.02, 0.1057 ii) 13, 0.288 iii) 12.25, 0.204 iv) 1.8, 0.1125 v) 5.36, 0.223 5. i) 3 ii) 12.5 iii) 2.24, 0.187 iv) Show to your teacher. v) Show to your teacher.

PRIME Opt. Maths Book - IX 333 8.4 Standard deviation (Root mean square deviation) It is defined as the positive square root of the mean of square of deviation taken from the mean. i.e. S.D.(σ) = ( ) N X X – 2 Σ We calculate it as the positive square root the mean of the square of the deviation from the mean (or assume mean). So it is called the root mean square deviation too. For standard deviation, S.D.(σ) = ( ) N X X – 2 Σ coefficient of standard deviation = . . X S D variance(σ2 ) = (S.D.)2 Coefficient of variation (C.V.) = X σ × 100% Where ‘σ’ is the Greek letter sigma. i) For individual observations: • Direct method (using mean) S.D. (σ) = ( ) N X X – 2 Σ , where the table is, X X X – X (X – X)2 ∑X = ∑(X – X)2 = • Direct method (without mean) S.D. (σ) = N X N X – 2 2 Σ Σ` j , where the table is, X X2 ∑X = ∑X2 = • Indirect method (using assume mean) [shortcut method]

334 PRIME Opt. Maths Book - IX S.D. (σ) = N d N d – 2 2 R Ra k , where the table is, X d = X – A d2 ∑d = ∑d2 = A = assume mean taken from X. ii) Discrete observations: • Direct method (using mean) S.D. (σ) = ( ) N f X – X 2 R , where the table is X f fx X X – X(X – X)2 f(X – X)2 N = ∑fx = ∑f(X – X)2 • Direct method (without mean) S.D. (σ) = N fx N fx – 2 2 Σ Σa k , where the table X f fx x2 fx2 N = ∑fx = ∑fx2 • Indirect method (using assume mean) [shortcut method] S.D. (σ) = N fd N fd – 2 2 Σ Σa k , where the table is X f d = X – A fd d2 fd2 N = ∑fx = ∑fx2 A = assume mean

PRIME Opt. Maths Book - IX 335 1. Find the standard deviation of the observations by using mean of 32, 35, 40, 45, 48. Also find the coefficient of variation. Solution : X X X – X (X – X)2 32 44 –12 144 35 –9 81 40 –4 16 45 1 1 48 4 16 ∑x = 220 ∑(X – X)2 = 258 Here, No. of observations (N) = 5 Mean (X) = N Rx = 5 220 = 44 (σ) = ( – ) N X X 2 R = 5 258 = 51.6 = 7.18 Again, Coefficient of variation (C.V.) = X σ × 100% = . 44 7 18 × 100% = 16.32% 2. Find C.V. of the given observations by using assume mean (direct method). 12, 15, 18, 20, 24, 18, 32, 35 Solution : X A d = X – A d2 12 20 –8 64 15 –5 25 18 –2 4 20 0 0 24 4 16 28 8 64 32 12 144 35 15 225 ∑d = 24, ∑d2 = 542 Worked out Examples

336 PRIME Opt. Maths Book - IX No. of observations (N) = 8 Here, S.D. (σ) = N d N d – 2 2 R Ra k = 8 542 8 24 – 2 a k = 67.75 – 9 = 7.665 Mean (X) = A + N Rd = 20 + 8 24 = 23 \ C.V. = X σ × 100% = . 23 7 665 × 100% = 33.32% 3. Find standard deviation and its coefficient of the observation given below. Height (cm) 20 25 35 44 50 No. of persons 8 6 12 5 9 Solution : x f fx X (X – X) (X – X)2 f(X – X)2 20 8 160 35 –15 225 1800 25 6 150 –10 100 600 35 12 420 0 0 0 44 5 220 9 81 405 50 9 450 15 225 2025 N = 40, ∑fx = 1400 ∑f(X – X)2 = 4830 Here, Mean (X) = N Rfx = 40 1400 = 35 S.D. (σ) = ( ) N f X – X 2 R = 40 4830 = 120 7. 5 = 10.9886 Coefficient of S.D. = X σ = . 35 10 9886 = 0.3139

PRIME Opt. Maths Book - IX 337 4. Find standard deviation and its coefficient of : x 85 90 100 124 130 142 f 6 5 9 15 10 5 Solution : X f fx x2 fx2 85 6 510 7225 43350 90 5 450 8100 40500 100 9 900 10000 90000 124 15 1860 15376 230640 130 10 1300 16900 169000 142 5 710 20164 100820 N = 50, ∑fx = 5730, ∑fx2 = 674310 Here, S.D. (σ) = N fx N fx – 2 2 Σ Σa k = 13486.20 – (114.6)² = 13486.20 – 13133.16 = 353 0. 4 = 18.789 Mean (X) = N Rfx = 50 5730 = 114.60 \ Coefficient of S.D. = X σ = . . 114 60 18 789 = 0.1639 5. If X = 12, ∑fx2 = 4880 and standard deviation of discrete series is 10, find the total number of participates. Solution : X = 12 = N Rfx ∑fx2 = 4880 S.D. = 10 We have, S.D. = N fx N fx – 2 2 Σ Σa k or, 10 = N 4880 – 12 2 ^ h or, 100 = N 4880 – 144 [a SBS] or, 244N = 4880 or N = 244 4880 \ N = 20

338 PRIME Opt. Maths Book - IX Exercise 8.4 1. Find the followings: i) If ∑(X – X)2 = 5000 and N = 50 of individual observations, find the S.D. and coefficient of variation where x = 4000. ii) If standard deviation of discrete observations is 4 where ∑ f(X – X)2 = 480, find the total number of participants. iii) If ∑fx = 200, ∑fx2 = 3280 and ∑f = 20, find the standard deviation and its coefficient. iv) If standard deviation of the observations having ∑fx2 = 22250 out of 40 students is 12.5, find the value of arithmetic mean of the discrete observations. Also find its C.V. v) What is standard deviation? Write down about the coefficient of variation with calculating formula of them. 2. Find the standard deviation and its coefficient following individual observations: i) Temperature (°C) : 30, 28, 35, 25, 42, 20 ii) Weight (kg) : 50, 30, 40, 20, 10 iii) Marks (out of 100) : 80, 75, 85, 95, 90, 70, 65 iv) Rainfall (mm) : 28, 34, 38, 42, 48, 52, 56, 60, 64, 68 v) Height (cm) : 40, 45, 48, 52, 56, 62, 64, 33 3. Find the standard deviation and C.V. of the following observations. i) Height (cm) 10 20 25 30 35 40 f 1 5 10 12 8 4 ii) Marks 15 20 24 28 40 No. of students 4 8 10 5 3 iii) Wages 270 250 260 240 300 320 No of workers 6 8 12 15 7 2 iv) Age 12 24 36 48 60 72 f 6 4 6 12 7 5 v) X 8 16 24 32 40 f 2 3 4 5 6 4. PRIME more creative questions: i) Construct discrete frequency distribution table and find standard deviation of the observations : 20, 20, 20, 10, 10, 30, 40, 30, 30, 30, 30, 20, 40, 40, 50, 30, 40, 50, 10, 40.

PRIME Opt. Maths Book - IX 339 ii) Construct discrete frequency distribution table and find C.V. of the observations : 15, 15, 10, 20, 20, 20, 25, 25, 25, 30, 30, 30, 10, 15, 20, 25, 25, ,25, 25, 30, 25, 30, 15, 25, 30, 20, 20. iii) Construct discrete frequency distribution table and find varience of the observations : 130, 140, 120, 120, 140, 140, 140, 130, 160 ,150, 140, 150, 160, 170, 170, 130, 120, 140, 140, 150, 160, 150, 140, 130, 150. iv) If mean of the observations is 15, find the varience of the observations: 5, 10, x, x + 5 & 25. v) If mean of the observations is 22, find the standard deviation of the data. Marks 10 15 20 25 30 f 2 4 p 8 5 5. Project work Collects the marks obtained by the students of classes V, VI, VII, VIII and IX in first terminal examination in mathematics of your school and constructs the discrete frequency distribution of each classes. Also compare the standard deviation and C.V. of each classes. Answer 1. i) 10, 12.5% ii) 30 iii) 8, 0.8 iv) 20, 62.5% v) Show to your teacher. 2. i) 7.047, 0.2349 ii) 14.14, 0.47 iii) 10, 0.125 iv) 12.657, 0.258 v) 9.987, 0.1997 3. i) 6.538, 22.54% ii) 6.582, 27.42% iii) 22.659, 8.66% iv) 18.72, 43.04% v) 10.58, 37.79% 4. i) 12.03 ii) 27.27% iv) 196.16 iv) 50 v) 6 5. Show to your teacher.

340 PRIME Opt. Maths Book - IX Statistics Unit Test Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. What do you mean by measure of dispersion? 2. a. If Σ x x – = 80 of the 8 observations where ∑x = 200, find the coefficient of mean deviation. b. If ∑fx2 = 4000 and ∑fx = 1000 of 50 students, find the standard deviation. c. If first quartile is 12 and coefficient of quartile deviation is 0.4, find the third quartile. 3. a. Find mean deviation from median and its coefficient of: Marks 24 18 12 40 32 36 No. of students 12 10 8 4 9 7 b. If quartile deviation and its coefficient are respectively 8 and 0.4 of a data. Find the lower and upper quartiles of the data. 4. Find the root mean square deviation of the observations given below. Also find the coefficient of variation. Age 14 18 25 36 40 No. of people 5 10 14 5 6

PRIME Opt. Maths Book - IX 341 Proposed Syllabus with Grid for First Terminal Examination S.N. Containts Topics K-1 U-2 A-4 HA-5 TQ TM Periods 1 Algebra i. Relation ii. function 2 2 2 1 7 18 8 2 Matrices i. Introduction ii. Addition 2 2 1 – 5 10 6 3 Co-ordinate Geometry i. Distance Formula ii. Locus iii. Section Formula 2 2 2 1 7 19 12 4 Trigonometry i. Measurement of A angle ii. Trigonometric Ratios iii. Conversion of TR 2 4 3 1 9 27 18 5 Transformation i. Reflection ii. Rotation 1 2 1 1 5 14 6 6 Statistics Partition Values 1 1 2 – 4 11 6 Total Questions 10 13 11 4 38 Total Marks 10 26 44 20 100 56 K = Knowledge, U = Understanding, A = Application, HA = Higher ability Model Question Set for First Terminal Examination Group : A [10 × 1 = 10] 1. a. What do you mean by antecedent and consequence is an ordered pair? b. Find element of image of a function f(x) = 2x – 1 where pre-image is 1. 2. a. What do you mean by scalar matrix? b. If A = 3 4 2 2 1 – –3 < F, find a13 + a22. 3. a. What do you mean be points of trisection of a line segment AB? b. Is (1, 2) a point in the locus of a point having equation 3x – y = 1? 4. a. Express 12° 15’ 5’’ into seconds. b. Write down the trigonometric ratios of Tanq and Secq from the given right angled DABC. 5. a. What is the image of a point A(2, 1) under reflection about x = 1? b. Write down the calculating formula of quartile deviation and its coefficient for the individual observations. Group : B [13 × 2 = 26] 6. a. If ordered pairs (2x – y, 4 – y) and (4, y), are the equal ordered pairs, find the value of ‘x’ and ‘y’. B A q C

342 PRIME Opt. Maths Book - IX b. If a relation R = {(x, y) : x + y = 8, x, y∈N}, find R in ordered Paris form. c. If aij = 2i + j is the general element of a matrix, find the matrix of order 2 × 2. 7. a. If A = 2 3 1 1 – < F and B = 1 1 2 – 3 – < F, find the matrix 3A – 2B. b. Find the co-ordinate of a point on y-axis which is 5 units distant from a point (4, – 1) 8. a. In what ratio does x-axis cuts the line joining the points (1, 4) and (5, – 4)? b. Convert 45g 24’ 42’’ into centesimal measurement. 9. a. If nSinA = m, prove that Sec2 A – Tan2 A = 1 b. Prove that Cos A 1 – Sin A 4 4 = 1 + 2Tan2 A. c. Find the length of an arc of a circle which subtends an angle of 60° at the centre of the circle of radius 4.2 cm. 10. a. What is transformation? Write down the name of geometrical transformations. b. Find the image of a point A(3, –2) under rotation about –90° with centre origin followed by reflection about x + y = 0. c. If 42 is the 7th decile of the observations taken in order 12, 16, 20, 24, 28, 32, 36, 2m, m + 25, 48, 52. Find the value of ‘m’. Group : C [11 × 4 = 44] 11. If a relation R = {(x, y) : 2x + y ≤ 6} domain A = {1, 2, 3, 4}, find R from A × A. Also show in arrow diagram. 12. If f(x + 2) = f(x) + f(2), prove that f(0) = 0 and f(–2) = –f(2). Also prove that f(4) = 2f(2) 13. If A + B = 3 2 3 4 – < F and A – B = 1 4 1 –2 < F find the matrices A and B. 14. Find the equation of locus of a point which moves so that its distance from (1, 2) is equidistance from the point (2, 3). 15. Find the co-ordinate of fourth vertex of a parallelogram where three of the vertices are A(1, 2), B(3, 5) and C(5, 4). 16. Find the angles of a triangle in degrees where the angles are x 5 2 c a k , x 9 8 g a k and x 75 r C ` j respectively. 17. Prove that : 1 + Secq – Tanq 1 + Secq + Tanq = – Cos 1 Sin i i 18. If 2Sin2 q + Cosq – 1 = 0, find the value of Cosq. 19. Find the image of DABC having vertices A(1, –2), B(3, 3) and C(5, 0) under an enlargement about E[(0, 1), 2]. Also plot the object and images in graph. 20. Find the 60th percentile of the observations. Class 0-10 10-20 20-30 30-40 40-50 f 9 13 24 20 14 21. Find quartile deviation and its coefficient of the observations 15, 35, 25, 20, 30, 10.

PRIME Opt. Maths Book - IX 343 Group : D [4 × 5 = 20] 22. If f(x) = 3x – 2, range = {1, 4, 7, 10}, find the domain. Also show the function in arrow diagram. 23. Find the co-ordinate of a point which is equidistant from the points (–2, 0), (5, –1) and (–1, 7). 24. Prove that : Cot Tan i – 1 i + Tan – Cot i 1 i = Secq.Cosecq + 1 25. Find the co-ordinate of image of DABC having vertices A(–1, 1), B(2, 4) and C(3, –3) under reflection about x + 2 = 0 followed by rotation about 180° with centre (0, 2). Also plot the object and images in graph. Proposed Syllabus with Grid for Second Terminal Examination S.N. Containts Topics K-1 U-2 A-4 HA-5 TQ TM Periods 1 Algebra i. Polynomial ii. Sequence & Series 2 2 2 1 7 19 6 2 Matrices i. Multiplication ii. Transpose 2 2 1 – 5 10 12 3 Co-ordinate Geometry i. Equation of Straight line ii. Perpendicular dis. iii. Area of Triangle 2 2 2 1 7 19 20 4 Trigonometry i. Standard Angles ii. Certain Angles iii. Conversion of TR 2 4 4 – 10 26 16 5 Transformation i. Translation ii. Enlargement 1 2 1 1 5 14 10 6 Statistics i. Quartile Deviation ii. Mean Deviation iii. Standard Deviation 1 1 1 1 4 12 12 7. First Term Review 6 Total Questions 10 13 11 4 38 Total Marks 10 26 44 20 100 80 K = Knowledge, U = Understanding, A = Application, HA = Higher ability

344 PRIME Opt. Maths Book - IX Model Question Set for Second Terminal Examination Group : A [10 × 1 = 10] 1. a. Define the term cartesian product A × B. b. Evaluate n n( ) 1 n 1 3 + = / 2. a. Find (AT)T where A = a c b d < F b. If A = 1 3 2 –2 < F and I is an identity matrix of order 2 × 2, find IA. 3. a. Write down the formula of finding perpendicular distance from a point to the straight line. b. If (1, 2) lies in the locus of equation 2x + my = 8, find the value of ‘m’. 4. a. Find the value of Tan(675)° b. If 5SinA = 3, find the value of CosA. 5. a. What is the image of A(–2, 1) under translation about T = 3 1 < F ? b. Write down the calculation formula of standard deviation and its C.V. by using mean of the observations. Group : B [13 × 2 = 26] 6. a. If (32x + 1, 16) = (27, 23x – 8), find the value of ‘x’ and ‘y’. b. If Sn = 3n + 2, find the value of t5 . c. If A + B = 2 3 4 –2 < F and B = 1 1 2 4 – – < F, find the matrix A and its transpose. 7. a. Which matrix pre-multiplies to 2 3 0 1 < F results the matrix [7 1]? b. Find the equation of straight line AB from the given diagram where OC = 2 units. 8. a. Prove that the points (3, 4), (7, 7) and (11, 10) are collinear points. b. Find the central angle of a circle in degrees having radius 8.4cm which subtends by an arc of length 8.8 cm. 9. a. Prove that : Sec²q + Cosec²q = Tanq + Cotq. b. Prove that : Cot Tan 1 60 1 30 ° – ° + = 2 – 3 c. If Secq – Cosecq = 0, find the value of Cosecq. 10. a. Find the image of a point A(2, –3) under reflection about y = 0 followed by rotation about +90° with centre (1, 2). b. Find the reflection axis which transferred a point A(3, 5) to A’(1, 5) c. If 25th percentile of the observations 24, 30, 2p+6, 42, 48, 54, 60, 66, 72 respectively taken in order is 33, find the value of ‘p’. 30° X’ X Y’ A O B C Y

PRIME Opt. Maths Book - IX 345 Group : C [11 × 4 = 44] 11. If f(x) = 2Sinx + 1, domain = {0°, 30°, 60°, 90°} find range. Also show the function in arrow diagram. 12. If p(x) = 2x3 – 3x + 5 – x2 and q(x) = 3x3 + 5x – 2 + 3x2 , answer the questions given below. i. Find p(x) + q(x) ii. Write down the degree of the polynomial obtained after addition. iii. Write down the types of the polynomial according to degree. 13. If A = 3 1 2 –1 < F and B = 2 3 1 1 – < F, prove that (AB)T = BT AT. 14. Prove that the points P(2, 3), Q(2, –4) and R(9, –4) are the vertices of an isosceles right angled triangle. 15. Find the points of trisection of the line joining the points (–5, –5) and (25, 10). 16. Prove that : Sin2 16 r + Sin2 16 3r + Sin2 16 5r + Sin2 16 7r = 2 17. Prove that : (3 – 2Sin2 q)(3Tan2 q – 1) = (3 + Tan2 q)(4Sin2 q – 1) 18. If 8Cosq + 15Sinq = 17, find the value of Cotq. 19. The angles of a triangle in degrees, grades and radians respectively are in the ratio 288:280:p. Find the angles of the triangle in degrees. 20. Find the centre of enlargement and scale factor which transform the points A(3, 3) → A’(5, 4), B(4, –1) → B’(7, –4) and C(6, 1) → C’(11, 0). 21. Find the mean deviation from median and its coefficient of: x 12 18 24 30 36 40 f 3 7 12 15 8 5 Group : D [4 × 5 = 20] 22. Answer the questions given below by the study of given diagrams. i) Add one more diagram in the given pattern. ii) Calculate the nth term of the sequence of number of dots used there in the diagrams. iii) Write down in sigma notation. 23. Find the equation of locus of a point which moves in such a way that from the points A(3, 0) and (–3, 0) as AP2 + BP2 = AB2 . 24. Find the image of a triangle having vetices A(1, –2), B(3, 1) and C(5, –4) under a translation AB followed by enlargement about E[(1, 2), –2]. Also plot them in graph. 25. Find the standard deviation and C.V. of : x 15 24 33 42 51 60 f 3 5 8 12 9 3