IIT-JEE Advanced 2023 Maths Paper 1 Problem 15

Problem

Consider the given data with frequency distribution.

xi	3	8	11	10	5	4
fi	5	2	3	2	4	4

Match each entry in List-I to the correct entries in List-II.

List-I	List-II
(P) The mean of the above data is	(1) 2.5
(Q) The median of the above data is	(2) 5
(R) The mean deviation about the mean of the above data is	(3) 6
(S) The mean deviation about the median of the above data is	(4) 2.7
	(5) 2.4

Solution

Let’s pick each item from List-I, one at a time.

Option P

To find the mean, we’ll apply the formula, i.e. x̄ = ∑f_ix_i/∑f_i.

This comes out to be (3 x 5 + 8 x 2 + 11 x 3 + 10 x 2 + 5 x 4 + 4 x 4)/(5 + 2 + 3 + 2 + 4 + 4), which simplies to 120/20 or 6.

So, (P) matches with (3).

Option Q

Since there are 20 observations, the median will be the average of the 10th and 11th observation, when arranged in increasing order.

Using the table, here are the first 12 observations: 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, …

So, the median is (5 + 5)/2 or 5, which means (Q) matches with (2).

Option R

Again, we’ll apply the formula, i.e. ∑f_i|x_i – x̄|/∑f_i.

Skipping the calculations here. The above expression simplifies to 54/20 or 2.7.

So, (R) matches with (4).

Option S

Same here. The required deviation equals ∑f_i|x_i – m|/∑f_i, where m is the median.

Skipping the calculations here. The above expression simplifies to 48/20 or 2.4.

So, (S) matches with (5).

Comments

Boring – direct application of formulas. And calculation intensive. Calculations for (S) could have been saved, given that there was only one option (in the paper) where the previous three entries matched.