Calculate the mean deviation about median age for the age distribution of 100 persons given below: (12)

12. Calculate the mean deviation about median age for the age distribution of $\small 100$ persons given below:

Age (in years) $\small 16-20$ $\small 21-25$ $\small 26-30$ $\small 31-35$ $\small 36-40$ $\small 41-45$ $\small 46-50$ $\small 51-55$

Number $\small 5$ $\small 6$ $\small 12$ $\small 14$ $\small 26$ $\small 12$ $\small 16$ $\small 9$

[Hint Convert the given data into continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval]

Answers (1)

Age (in years)	Number $f_i$	Cumulative Frequency c.f.	Mid Points $x_i$	$\|x_i - M\|$	$f_i\|x_i - M\|$
15.5-20.5	5	5	18	20	100
20.5-25.5	6	11	23	15	90
25.5-30.5	12	23	28	10	120
30.5-35.5	14	37	33	5	70
35.5-40.5	26	63	38	0	0
40.5-45.5	12	75	43	5	60
45.5-50.5	16	91	48	10	160
50.5-55.5	9	100	53	15	135
					$\sum f_i\|x_i - M\|$ =735

Now, N = 100, which is even.

The class interval containing $\left (\frac{N}{2} \right)^{th}$ or $50^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median $= l + \frac{\frac{N}{2}- C}{f}\times h$

Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

Therefore, Median $= 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and

$\sum f_i|x_i - M|$ = 735

$\therefore$ $M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|$

$= \frac{735}{100} = 7.35$

Hence, the mean deviation about the median is 7.35

Posted by

HARSH KANKARIA

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Age (in years)	$\small 16-20$	$\small 21-25$	$\small 26-30$	$\small 31-35$	$\small 36-40$	$\small 41-45$	$\small 46-50$	$\small 51-55$
Number	$\small 5$	$\small 6$	$\small 12$	$\small 14$	$\small 26$	$\small 12$	$\small 16$	$\small 9$

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12. Calculate the mean deviation about median age for the age distribution of persons given below: Age (in years) Number [Hint Convert the given data into continuous frequency distribution by subtracting from the lower limit and adding to the upper limit of each class interval]

Answers (1)

Posted by

HARSH KANKARIA

Crack CUET with india's "Best Teachers"

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