Advertisements
Advertisements
Question
Determine the mean and standard deviation for the following distribution:
Marks | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
Frequency | 1 | 6 | 6 | 8 | 8 | 2 | 2 | 3 | 0 | 2 | 1 | 0 | 0 | 0 | 1 |
Chart
Sum
Solution
`x` | `f_i` | `f_ix_i` | `d_i = x_i - barx` | `f_i d_i` | `f_i d_ix^2` |
2 | 1 | 2 | – 4 | – 4 | 16 |
3 | 6 | 18 | – 3 | – 18 | 54 |
4 | 6 | 24 | – 2 | – 12 | 24 |
5 | 8 | 40 | – 1 | – 8 | 8 |
6 | 8 | 48 | 0 | 0 | 0 |
7 | 2 | 14 | 1 | 2 | 2 |
8 | 2 | 16 | 2 | 4 | 8 |
9 | 3 | 27 | 3 | 9 | 27 |
10 | 0 | 0 | 4 | 0 | 0 |
11 | 2 | 2 | 5 | 10 | 50 |
12 | 1 | 12 | 6 | 6 | 36 |
13 | 0 | 0 | 7 | 0 | 0 |
14 | 0 | 0 | 8 | 0 | 0 |
15 | 0 | 0 | 9 | 0 | 0 |
16 | 1 | 16 | 10 | 10 | 100 |
N = 40 | `sumf_ix_i` = 239 | `sumf_i d_i` = – 1 | `sumf_i d_i^2` = 325 |
Mean `barx = (sumf_ix_i)/N = 239/40` = 5.9 = 6
∴ S.D. = `sigma = sqrt((sumf_i d_i^2)/N - ((sumf_i d_i)/N)^2`
= `sqrt(325/40 - ((-1)/40)^2`
= `sqrt(8.125 - 0.000625)`
= `sqrt(8.124375)`
= 2.85
Here, the required mean = 6 and M.D. = 2.85
shaalaa.com
Standard Deviation - by Short Cut Method
Is there an error in this question or solution?