Data = 37, 41, 56, 62, 70, 74, 81, 89, 95 and 90.

Assertion(A): Median = 72.

Reason(R): If number of data(n) is odd, the median = $\Big(\dfrac{n + 1}{2}\Big)^{th}$ term.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Given, data = 37, 41, 56, 62, 70, 74, 81, 89, 95 and 90.

Arrange the data in ascending order: 37, 41, 56, 62, 70, 74, 81, 89, 90, 95

Number of data = 10

If the number of data points (n) is odd, the median is the $\Big(\dfrac{n+1}{2}\Big)^{th}$ term

If the number of data points (n) is even, the median is the $\Big(\dfrac{\Big(\dfrac{n}{2}\Big)^{th} + \Big(\dfrac{n}{2} + 1\Big)^{th}}{2}\Big)$

Here, n = 10

$\text{Median }= \Big(\dfrac{\Big(\dfrac{10}{2}\Big)^{th} + \Big(\dfrac{10}{2} + 1\Big)^{th}}{2}\Big)\\[1em] = \Big(\dfrac{5^{th} + (5 + 1)^{th}}{2}\Big)\\[1em] = \Big(\dfrac{5^{th} + 6^{th}}{2}\Big)\\[1em] = \Big(\dfrac{70 + 74}{2}\Big)\\[1em] = \Big(\dfrac{144}{2}\Big)\\[1em] = 72.$

∴ Both A and R are true and R is incorrect reason for A.

Hence, option 4 is the correct option.