16

You visited us 16 times! Enjoying our articles? Unlock Full Access!

Mean of Continuous Series

Calculate the...

Question

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

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

Open in App

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 $(σ) = 11.8739$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Calculate standard deviation of the following data, using step-deviation method.

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

Q. Given the following data, calculate coefficient of variation:

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

Q. From the following data of the ages of different persons, determine the modal age.

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

Q. From the following data of the ages of different persons, prepare less than and more than cumulative frequency distribution.

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

Q. Calculate standard deviation of marks secured by 100 examinees in the examination:

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Mean of Grouped Data

STATISTICS

Watch in App

explore_more_icon

Explore more

Mean of Continuous Series

Standard XI Statistics

Join BYJU'S Learning Program

CrossIcon