The age distribution of 40 children is as follows
| Age (in yr.) | 5-6 | 6-7 | 7-8 | 8-9 | 9-10 | 10-11 |
| No. of children | 4 | 7 | 9 | 12 | 6 | 2 |
Consider the following statements in respect of the above frequency distribution.
(A) The median of the age distribution is 7 yr.
(B) 70% of the children are in the age group 6-9 yr.
(C) The modal class of the children is 8-9 yr.
Which of the above statements are true?
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B) and (C) only
For grouped data, the median is found from the cumulative frequencies with \(\text{median}=l+\frac{\frac{N}{2}-cf}{f}\times h\), where \(l\) is the lower limit of the median class, \(cf\) the cumulative frequency before it, \(f\) its frequency and \(h\) the class width. The modal class is the class with the highest frequency.
Cumulative frequencies, each the running total of children up to the end of that class: 5-6: 4; 6-7: 11; 7-8: 20; 8-9: 32; 9-10: 38; 10-11: 40
(A) The median of the age distribution is 7 yr. Incorrect.
Here \(\frac{N}{2}=20\), and the cumulative frequency reaches 20 in the class 7-8, so the median works out to 8 yr:
For the median class 7-8: \(l=7,\ cf=11,\ f=9,\ h=1\)
\(\displaystyle \text{median}=7+\frac{20-11}{9}\times1=7+1=8\) yr
(B) 70% of the children are in the age group 6-9 yr. Correct.
The classes 6-7, 7-8 and 8-9 hold \(7+9+12=28\) children
As a share of all 40 children: \(\frac{28}{40}\times100=70\%\)
(C) The modal class of the children is 8-9 yr. Correct. The class 8-9 has the highest frequency, 12, so it is the modal class.