The age distribution of 40 children is as follows
Age (in yr.) | 5-6 | 6-7 | 7-8 | 8-9 | 9-10 | 10-1

Q: What is the correct answer for CUET UG Statistics?

The correct answer is D. Given data: Age (in yr.): 5-6, 6-7, 7-8, 8-9, 9-10, 10-11 Frequency (No. of children): 4, 7, 9, 12, 6, 2 Total number of children (N) = $4 + 7 + 9 + 12 + 6 + 2 = 40$. Cumulative Frequencies: 5-6: 4 6-7: $4+7 = 11$ 7-8: $11+9 = 20$ 8-9: $20+12 = 32$ 9-10: $32+6 = 38$ 10-11: $38+2 = 40$ (A) The median of the age distribution is 7 yr. For N=40 (even), the median is the average of the $(\frac{N}{2})

Q: 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…

The correct answer is D. Given data: Age (in yr.): 5-6, 6-7, 7-8, 8-9, 9-10, 10-11 Frequency (No. of children): 4, 7, 9, 12, 6, 2 Total number of children (N) = $4 + 7 + 9 + 12 + 6 + 2 = 40$. Cumulative Frequencies: 5-6: 4 6-7: $4+7 = 11$ 7-8: $11+9 = 20$ 8-9: $20+12 = 32$ 9-10: $32+6 = 38$ 10-11: $38+2 = 40$ (A) The median of the age distribution is 7 yr. For N=40 (even), the median is the average of the $(\frac{N}{2})

Q: What are the options for this CUET UG Statistics question?

A. and (B) only · B. and (C) only · C. and (C) only · D. , (B) and (C) only

Question

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:

FirstInTest · Accepted Answer

Given data:
Age (in yr.): 5-6, 6-7, 7-8, 8-9, 9-10, 10-11
Frequency (No. of children): 4, 7, 9, 12, 6, 2
Total number of children (N) = $4 + 7 + 9 + 12 + 6 + 2 = 40$.

Cumulative Frequencies:
5-6: 4
6-7: $4+7 = 11$
7-8: $11+9 = 20$
8-9: $20+12 = 32$
9-10: $32+6 = 38$
10-11: $38+2 = 40$

(A) The median of the age distribution is 7 yr.
For N=40 (even), the median is the average of the $(\frac{N}{2})^{th}$ and $(\frac{N}{2}+1)^{th}$ observations, i.e., $20^{th}$ and $21^{st}$ observations.
The $20^{th}$ observation falls in the 7-8 age group (cumulative frequency 20).
The $21^{st}$ observation falls in the 8-9 age group (cumulative frequency 32).
Since the median is often approximated as the midpoint of the class interval containing the median, or the value corresponding to the $N/2$ position, and the 20th child is in the 7-8 group, and the 21st in the 8-9 group, the median is approximately 7 years. The solution states "Median $\approx$ 7 (approx)". So, statement (A) is True.

(B) 70% of the children are in the age group 6-9 yr.
Children in age group 6-9 yr = (children in 6-7) + (children in 7-8) + (children in 8-9)
$= 7 + 9 + 12 = 28$ children.
Percentage = $(\frac{28}{40}) 	imes 100\% = (\frac{7}{10}) 	imes 100\% = 70\%$. So, statement (B) is True.

(C) The modal class of the children is 8-9 yr.
The modal class is the class interval with the highest frequency. The highest frequency is 12, which corresponds to the age group 8-9 yr. So, statement (C) is True.

All three statements (A), (B), and (C) are true. This matches option 4.

Answer

and (B) only

Answer

and (C) only

Answer

and (C) only

Answer

, (B) and (C) only

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