Question

The following table gives the frequency distribution of married women by age at marriage:

Age (in years)	Frequency
$15 - 19$	$53$
$20 - 24$	$140$
$25 - 29$	$98$
$30 - 34$	$32$
$35 - 39$	$12$
$40 - 44$	$9$
$45 - 49$	$5$
$50 - 54$	$3$
$55 - 59$	$2$
$60$ and above	$2$

Calculate the median and interpret the results.

Accepted Answer

The given frequency distribution is not continuous. We convert it into continuous by subtracting the conversion factor from the lower limit and adding it to the upper limit of each class interval.

Here, the conversion factor $= \frac{20 - 19}{2} = \frac{1}{2} = 0.5$

The cumulative frequency table is as below:

Age (in years)
Class Interval Frequency $25 - 29$ Cumulative Frequency $(c . f .)$

$32$ $14.5 - 19.5$ $53$ $53$

$20 - 24$ $19.5 - 24.5$ $140$ $140 + 53 = 193$

$25 - 29$ $24.5 - 29.5$ $98$ $98 + 193 = 291$

$30 - 34$ $29.5 - 34.5$ $32$ $32 + 291 = 323$

$35 - 39$ $34.5 - 39.5$ $12$ $12 + 323 = 335$

$40 - 44$ $39.5 - 44.5$ $9$ $9 + 335 = 344$

$45 - 49$ $44.5 - 49.5$ $5$ $5 + 344 = 349$

$50 - 54$ $49.5 - 54.5$ $3$ $3 + 349 = 352$

$55 - 59$ $54.5 - 59.5$ $2$ $2 + 352 = 354$

$60$ and above $59.5$ and above $2$ $2 + 354 = 356$

Total $N = \sum f_{i} = 356$

We have, $N = 356$

So, $\frac{N}{2} = \frac{356}{2} = 178$

The cumulative frequency just greater than $178$ is $193$ , which corresponds to the class $19.5 - 24.5$

Thus, the median class is $19.5 - 24.5$ .

Here, $l = 19.5$ (the lower limit of the median class)

$f = 140$ (frequency of the median class)

$h = 5$ (width of each class)

$N = 356$ (number of observations)

and $C = 53$ (cumulative frequency of class preceding the median class)

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

$= 19.5 + \frac{178 - 53}{140} \times 5$

$= 19.5 + \frac{125}{140} \times 5 = 19.5 + 4.46 = 23.96 (approx)$

Hence, the median is $23.96$ years and nearly half of the women married between the age group $19.5 - 24.5$ .

Marks	Number of students
Less than $10$	$2$
Less than $30$	$10$
Less than $50$	$25$
Less than $2$	$2 + 354 = 356$
Less than $90$	$65$
Less than $110$	$87$
Less than $130$	$96$
Less than $150$	$100$

Marks	Number of students
Above $0$	$80$
Above $10$	$76$
Above $20$	$71$
Above $30$	$65$
Above $C = 53$	$57$
Above $= l + \frac{\frac{N}{2} - C}{f} \times h$	$43$
Above $60$	$28$
Above $70$	$15$
Above $80$	$10$
Above $90$	$8$
Above $100$	$0$

Age (in years)	Number of policy holders
Below $20$	$2$
Below $25$	$6$
Below $30$	$24$
Below $35$	$45$
Below $40$	$78$
Below $45$	$89$
Below $50$	$92$
Below $55$	$98$
Below $60$	$100$

Variable	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
Frequency	$12$	$30$	_____	$65$	_____	$25$	$18$

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	Total
Frequency	$8$	$16$	$36$	$34$	$6$	$100$

J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 3: EXERCISE

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 3: EXERCISE

Questions from J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 3: EXERCISE with Hints & Solutions

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Class interval	$100 - 120$	$120 - 140$	$140 - 160$	$160 - 180$	$180 - 200$
Frequency	$12$	$14$	$8$	$6$	$10$

Daily expenditure (in $₹$ )	$100 - 150$	$150 - 200$	$200 - 250$	$250 - 300$	$300 - 350$
Number of workers	$6$	$7$	$12$	$3$	$2$