# Statistics – Exercise 15.2 – Class XI

1. the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12

Solution:

Given data is 6, 7, 10, 12, 13, 4, 8, 12

 xi (xi –  x) (xi –  x)2 6 -3 9 7 -2 4 10 -1 1 12 3 9 13 4 16 4 -5 25 8 -1 1 12 3 9 74

1. Find the mean and variance for the first n natural numbers

Solution:

1. Find the mean and variance for the first 10 multiples of 3

Solution:

The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Here, number of observations, n = 10

 xi (xi – Mean.) (xi-Mean.)2 3 -13.5 182.25 6 -10.5 110.25 9 -7.5 56.25 12 -4.5 20.25 15 -1.5 2.25 18 1.5 2.25 21 4.5 20.25 24 7.5 56.25 27 10.5 110.25 30 13.5 182.25 742.5

1. Find the mean and variance for the data:
 xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3

Solution:

 xi fi fixi xi-mean (xi – mean)2 fi(xi-Mean.)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

N=40

1. Find the mean and variance for the data:
 xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3

Solution:

 xi fi fixi xi-mean (xi – mean)2 fi(xi-Mean.)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

N=22

1. Find the mean and standard deviation for the data:
 xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5

Solution:

 xi fi fi=(xi-64)/I yi2 fiyi fiyi2 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

N=100

1. Find the mean and standard deviation for the data:
 classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2

Solution:

 Class frequency midpoint xi yi=(xi-105)/30 yi2 fiyi fiyi2 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

N=30

1. Find the mean and variance for the data:
 Classes 0-10 10-20 20-30 30-40 40-50 frequency 5 8 15 16 6

Solution:

 Class frequency midpoint xi yi=(xi-105)/30 yi2 fiyi fiyi2 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

N=50

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

Solution:

 Height in cms No. of children Midpoint xi yi = (xi-92.5)/5 yi2 fiyi fiyi2 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

1. The diameters of circles(in mm) drawn in a design are given below:
 Diameters No. of children 33-36 15 37-40 17 41-44 21 45-48 22 49-52 25

Solution:

 Diameters No. of children Midpoint xi yi = (xi-92.5)/5 yi2 fiyi fiyi2 33-36 15 34.5 -2 4 -30 60 37-40 17 38.5 -1 1 -17 17 41-44 21 42.5 0 0 0 0 45-48 22 46.5 1 1 22 22 49-52 25 50.5 2 4 50 100 100 25 199