# Class 11 Maths NCERT Solutions for Chapter 15 Statistics Exercise 15.2

### Statistics Exercise 15.2 Solutions

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

Solution

6, 7, 10, 12, 13, 4, 8, 12

The following table is obtained.

 xi 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

2. Find the mean and variance for the first natural numbers
Solution
The mean of first n natural numbers is calculated as follows.
Mean = (Sum of all observations)/(Number of observations)

3. 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

The following table is obtained.
 xi 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

4. 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
The data is obtained in tabular form as follows.
 xi fi fixi 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

5. 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
The data is obtained in tabular form as follows.
 xi fi fi xi 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

6. Find the mean and standard deviation using short-cut method.
 xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5
Solution
The data is obtained in tabular form as follows.
 xi fi yi2 fi yi fi yi2 60 2 -4 15 -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

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
Solution
 Class Frequency fi Mid – point xi yi2 fi yi fi yi2 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

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
Solution
 Class Frequency fi Mid – point xi yi2 fi yi fi yi2 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

9. Find the mean, variance and standard deviation using 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
 Class Interval Frequency fi Mid – point xi yi2 fi yi fi yi2 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

10. 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
 Class Interval Frequency fi Mid – point xi fi2 fi yi fi yi2 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.