Calculate the mean and standard deviation for the following table, if the age distribution of a group of people are given below.

$\begin{matrix} Age & 20 - 30 & 30 - 40 & 40 - 50 & 50 - 60 & 60 - 70 & 70 - 80 & 80 - 90 Number of persons & 3 & 51 & 122 & 141 & 130 & 51 & 2 \end{matrix}$

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Solution

We make the table from the given data

$\begin{matrix} Age & Mid-value X_{1} & Number of persons f_{1} & d_{1} = \frac{X_{1} - 55}{10} & d_{1}^{2} & f_{1}, d_{1} & f_{1} d_{1}^{2} 20 - 30 & 25 & 3 & - 3 & 9 & - 9 & 27 30 - 40 & 35 & 51 & - 2 & 4 & - 102 & 204 40 - 50 & 45 & 122 & - 1 & 1 & - 122 & 122 50 - 60 & 55 & 141 & 0 & 0 & 0 & 0 60 - 70 & 65 & 130 & 1 & 1 & 130 & 130 70 - 80 & 75 & 51 & 2 & 4 & 102 & 204 80 - 90 & 85 & 2 & 3 & 9 & 6 & 18 Total & N = \sum f_{i} = 500 & \sum f_{i}, d_{i} = 5 & \sum f_{i}, d_{i}^{2} = 705 \end{matrix}$

Here $A = 55, h = 10$

Therefore, Mean, $¯ ¯¯¯ ¯ X = A + (\frac{\sum f_{i} d_{i}}{\sum f_{i}}) \times h = 55 + \frac{5}{500} \times 10 = 55 + 0.1 = 55.1$

Variance, $σ^{2} = h^{2} [\frac{\sum f_{i} d_{i}^{2}}{\sum f_{i}} - {(\frac{\sum f_{i} d_{i}}{\sum f_{i}})}^{2}] = 100 [\frac{705}{500} - {(\frac{5}{100})}^{2}]$

$= 100 (\frac{352500 - 25}{250000}) = \frac{35247500}{250000} = \frac{14099}{100}$

and standard deviation , $σ = \sqrt{\frac{14099}{100}} = \frac{118.739}{10} = 11.8739$

Hence, mean $(¯ ¯¯¯ ¯ X) = 55.1$ and standard deviation $(σ) = 11.8739$

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Age	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Frequency	3	61	132	153	140	51	2

Age	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Number of Students	3	61	132	153	140	51	2

Age (in years)	10−20	20−30	30−40	40−50	50−60	60−70	70−80
No. of Persons	4	26	32	10	9	6	3

Age (in years)	10−20	20−30	30−40	40−50	50−60	60−70	70−80
No. of Persons	5	12	10	6	4	11	2

Marks	10−20	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Number of Examinees	19	3	2	49	24	2	0	1