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

E M B I B E

MATHEMATICS Textbook for Class XI>Chapter 15 - Statistics>EXERCISE 15.2

EXERCISE 15.2

Author:Jharkhand Board

Jharkhand Board Mathematics Solutions for EXERCISE 15.2

Simple step-by-step solutions to EXERCISE 15.2 questions of Statistics from MATHEMATICS Textbook for Class XI. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from EXERCISE 15.2 with Hints & Solutions

EASY

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 15 - Statistics>EXERCISE 15.2>Q 3

Find the mean and variance for the data

First $10$ multiples of $3$

HARD

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 15 - Statistics>EXERCISE 15.2>Q 4

Find the mean and variance for the data

$x_{i}$	$6$	$10$	$14$	$18$	$24$	$28$	$15$
$f_{i}$	$2.25$	$4$	$7$	$12$	$8$	$4$	$3$

MEDIUM

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 15 - Statistics>EXERCISE 15.2>Q 5

Find the mean and variance for the given data

$x_{i}$	$92$	$93$	$97$	$182.25$	$102$	$104$	$109$
$f_{i}$	$(σ^{2}) = \frac{1}{n} \sum_{i = 1}^{10} {(x_{i} - \bar{x})}^{2} = \frac{1}{10} \times 742.5 = 74.25$	$2$	$3$	$2$	$6$	$3$	$3$