Find the mean and variance for the data x i 6 10 14 18 24 28 30 f i 2 4 7 12 8 4 3

The given data in tabular form is as follows. x i f i f i x i x i - x ¯ x i - x ¯ 2 f i x i - x ¯ 2 6 2 12 - 13 169 338 10 4 40 - 9 81 324 14 7 98 - 5 25 175 18 12 216 - 1 1 12 24 8 192 5 25 200 28 4 112 9 81 324 30 3 90 11 121 363 40 760 1736 Here, N = 40 , ∑ i = 1 7 f i x i = 760 ∴ M e a n x ¯ = ∑ i = 1 7 f i x i N = 760 40 = 19 Variance = σ 2 = 1 N ∑ i = 1 7 f i x i - x ¯ 2 = 1 40 × 1736 = 43 . 4

Find the mean and variance for the given data x i 92 93 97 98 102 104 109 f i 3 2 3 2 6 3 3

The deviations and variances is obtained in tabular form for the given data as follows. x i f i f i x i x i - x ¯ x i - x ¯ 2 f i x i - x ¯ 2 92 3 276 - 8 64 192 93 2 186 - 7 49 98 97 3 291 - 3 9 27 98 2 196 - 2 4 8 102 6 612 2 4 24 104 3 312 4 16 48 109 3 327 9 81 243 22 2200 640 Here, N = 22 , ∑ i = 1 7 f i x i = 2200 ∴ x ¯ = 1 N ∑ i = 1 7 f i x i = 1 22 × 2200 = 100 Variance σ 2 = 1 N ∑ i = 1 7 f i x i - x ¯ 2 = 1 22 × 640 = 29 . 09 Therefore, variance for the given data is 29 . 09 .

Find the mean and standard deviation using short-cut method. x i 60 61 62 63 64 65 66 67 68 f i 2 1 12 29 25 12 10 4 5

The data is obtained in tabular form as follows. x i f i y i = x i - A h = x i - 64 1 y i 2 f i y i f i y i 2 60 2 - 4 16 - 8 32 61 1 - 3 9 - 3 9 62 12 - 2 4 - 24 48 63 29 - 1 1 - 29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 100 220 0 286 Let assumed mean A = 64 , h = width = 61 - 60 = 1 and N = ∑ i = 1 9 f i = 100 Therefore, Mean x ¯ = A + ∑ i = 1 9 f i y i N × h = 64 + 0 100 × 1 = 64 + 0 = 64 Variance, σ 2 = h 2 N 2 N ∑ i = 1 9 f i y i 2 - ∑ i = 1 9 f i y i 2 = 1 100 2 [ 100 × 286 - 0 ] = 2 . 86 ∴ Standard deviation ( σ ) = 2 . 86 = 1 . 691

Find the mean and variance for the following frequency distribution. Classes 0 - 30 30 - 60 60 - 90 90 - 120 120 - 150 150 - 180 180 - 210 Frequencies 2 3 5 10 3 5 2

Given data in tabular form: Class Frequency f i Mid-point x i y i = x i - A h = x i - 105 30 y i 2 f i y i f i y i 2 0 - 30 2 15 - 3 9 - 6 18 30 - 60 3 45 - 2 4 - 6 12 60 - 90 5 75 - 1 1 - 5 5 90 - 120 10 105 0 0 0 0 120 - 150 3 135 1 1 3 3 150 - 180 5 165 2 4 10 20 180 - 210 2 195 3 9 6 18 30 2 76 Let assumed mean A = 105 , h = width of class - intervals = 30 and N = ∑ i = 1 7 f i = 30 Mean, x ¯ = A + ∑ i = 1 7 f i y i N × h = 105 + 2 30 × 30 = 105 + 2 = 107 Variance σ 2 = h 2 N 2 N ∑ i = 1 7 f i y i 2 - ∑ i = 1 7 f i y i 2 = ( 30 ) 2 ( 30 ) 2 30 × 76 - ( 2 ) 2 = 2280 - 4 = 2276 Therefore, the variance for the given data is 2276 .

Find the mean and variance for the following frequency distribution. Classes 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequencies 5 8 15 16 6

Given data in tabular form: Class Frequency f i Mid-point x i y i = x i - A h = x i - 25 10 y i 2 f i y i f i y i 2 0 - 10 5 5 - 2 4 - 10 20 10 - 20 8 15 - 1 1 - 8 8 20 - 30 15 25 0 0 0 0 30 - 40 16 35 1 1 16 16 40 - 50 6 45 2 4 12 24 50 10 68 Let assumed mean A = 25 , h = width of class - intervals = 10 and N = ∑ i = 1 5 f i = 50 Mean, x ¯ = A + ∑ i = 1 5 f i y i N × h = 25 + 10 50 × 10 = 25 + 2 = 27 Variance σ 2 = h 2 N 2 N ∑ i = 1 5 f i y i 2 - ∑ i = 1 5 f i y i 2 = ( 10 ) 2 ( 50 ) 2 50 × 68 - ( 10 ) 2 = 1 25 [ 3400 - 100 ] = 3300 25 = 132 Therefore, the variance for the given data is 132 .

Find the mean, variance and standard deviation using short-cut method Height in cms 70 - 75 75 - 80 80 - 85 85 - 90 90 - 95 95 - 100 100 - 105 105 - 110 110 - 115 No. of children 3 4 7 7 15 9 6 6 3

Given data in tabular form: Class Frequency f i Mid-point x i y i = x i - 92 . 5 5 y i 2 f i y i f i y i 2 70 - 75 3 72 . 5 - 4 16 - 12 48 75 - 80 4 77 . 5 - 3 9 - 12 36 80 - 85 7 82 . 5 - 2 4 - 14 28 85 - 90 7 87 . 5 - 1 1 - 7 7 90 - 95 15 92 . 5 0 0 0 0 95 - 100 9 97 . 5 1 1 9 9 100 - 105 6 102 . 5 2 4 12 24 105 - 110 6 107 . 5 3 9 18 54 110 - 115 3 112 . 5 4 16 12 48 60 6 254 Let assumed mean A = 92 . 5 , h = width of class intervals = 5 and N = ∑ i = 1 9 f i = 60 Mean, x ¯ = A + ∑ i = 1 9 f i y i N × h = 92 . 5 + 6 60 × 5 = 92 . 5 + 0 . 5 = 93 Variance σ 2 = h 2 N 2 N ∑ i = 1 9 f i y i 2 - ∑ i = 1 9 f i , y i 2 = ( 5 ) 2 ( 60 ) 2 60 × 254 - ( 6 ) 2 = 25 3600 ( 15204 ) = 105 . 52 ∴ Standard deviation ( σ ) = 105 . 52 = 10 . 27

The diameters of circles (in mm ) drawn in a design are given below: Diameters 33 - 36 37 - 40 41 - 44 45 - 48 49 - 52 No. of circles 15 17 21 22 25 Calculate the standard deviation and mean diameter of the circle.

The given data is not having continuous class-intervals. To make the class-intervals continuous we will subtract 0 . 5 from lower limit of each class interval and add 0 . 5 to upper limit of each class intervals. Class Frequency f i Mid-point x i y i = x i - 42 . 5 4 y i 2 f i y i f i y i 2 32 . 5 - 36 . 5 15 34 . 5 - 2 4 - 30 60 36 . 5 - 40 . 5 17 38 . 5 - 1 1 - 17 17 40 . 5 - 44 . 5 21 42 . 5 0 0 0 0 44 . 5 - 48 . 5 22 46 . 5 1 1 22 22 48 . 5 - 52 . 5 25 50 . 5 2 4 50 100 100 25 199 Here, N = 100 , h = 4 . Let the assumed mean, A , be 42 . 5 . Mean, x ¯ = A + ∑ i = 1 5 f i y i N × h = 42 . 5 + 25 100 × 4 = 43 . 5 . Variance σ 2 = h 2 N 2 N ∑ i = 1 5 f i y i 2 - ∑ i = 1 5 f i y i 2 = 16 10000 100 × 199 - ( 25 ) 2 = 16 10000 [ 19900 - 625 ] = 16 10000 × 19275 = 30 . 84 ∴ Standard deviation ( σ ) = 5 . 55

NCERT Solutions for Chapter: Statistics, Exercise 2: EXERCISE 15.2

Author:NCERT

NCERT Mathematics Solutions for Exercise - NCERT Solutions for Chapter: Statistics, Exercise 2: EXERCISE 15.2

Attempt the free practice questions on Chapter 13: Statistics, Exercise 2: EXERCISE 15.2 with hints and solutions to strengthen your understanding. MATHEMATICS Textbook for Class XI solutions are prepared by Experienced Embibe Experts.

Questions from NCERT Solutions for Chapter: Statistics, Exercise 2: EXERCISE 15.2 with Hints & Solutions

EASY

11th Uttar Pradesh Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 13 - Statistics>EXERCISE 15.2>Q 3

Find the mean and variance for the data

First $10$ multiples of $3$

HARD

11th Uttar Pradesh Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 13 - Statistics>EXERCISE 15.2>Q 4

Find the mean and variance for the data

$x_{i}$	$12$	$- 4.5$	$14$	$18$	$24$	$28$	$30$
$f_{i}$	$2$	$4$	$21$	$4.5$	$8$	$4$	$3$

MEDIUM

11th Uttar Pradesh Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 13 - Statistics>EXERCISE 15.2>Q 5

Find the mean and variance for the given data

$x_{i}$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
$f_{i}$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

HARD

11th Uttar Pradesh Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 13 - Statistics>EXERCISE 15.2>Q 6

Find the mean and standard deviation using short-cut method.

$x_{i}$	$60$	$61$	$62$	$63$	$64$	$65$	$66$	$67$	$68$
$f_{i}$	$2$	$1$	$12$	$29$	$25$	$12$	$10$	$4$	$5$

HARD

11th Uttar Pradesh Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 13 - Statistics>EXERCISE 15.2>Q 7

Find the mean and variance for the following frequency distribution.

Classes	$0 - 30$	$30 - 60$	$60 - 90$	$90 - 120$	$120 - 150$	$150 - 180$	$180 - 210$
Frequencies	$2$	$3$	$5$	$10$	$3$	$5$	$2$

HARD

11th Uttar Pradesh Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 13 - Statistics>EXERCISE 15.2>Q 8

Find the mean and variance for the following frequency distribution.

Classes	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$
Frequencies	$5$	$8$	$15$	$16$	$6$

HARD

11th Uttar Pradesh Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 13 - Statistics>EXERCISE 15.2>Q 9

Find the mean, variance and standard deviation using short-cut method

Height in cms	$70 - 75$	$75 - 80$	$80 - 85$	$85 - 90$	$90 - 95$	$95 - 100$	$100 - 105$	$105 - 110$	$110 - 115$
No. of children	$3$	$4$	$7$	$7$	$15$	$9$	$6$	$6$	$3$

HARD

11th Uttar Pradesh Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 13 - Statistics>EXERCISE 15.2>Q 10

The diameters of circles (in $mm$ ) drawn in a design are given below:

Diameters	$33 - 36$	$37 - 40$	$41 - 44$	$45 - 48$	$49 - 52$
No. of circles	$15$	$17$	$21$	$22$	$25$

Calculate the standard deviation and mean diameter of the circle.

NCERT Solutions for Chapter: Statistics, Exercise 2: EXERCISE 15.2

NCERT Mathematics Solutions for Exercise - NCERT Solutions for Chapter: Statistics, Exercise 2: EXERCISE 15.2

Questions from NCERT Solutions for Chapter: Statistics, Exercise 2: EXERCISE 15.2 with Hints & Solutions

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