Assertion (A): If the class marks of two overlapping intervals of equal size in a distribution are 94 and 104 then the corresponding intervals are 89-99, 99-109.

Reason (R): The class mark of a class interval

= $\dfrac{\text{actual lower limit + actual upper limit}}{2}$

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Both A and R are true.

Explanation

Class mark = $\dfrac{\text{Actual lower limit + Actual upper limit}}{2}$

Class mark for the first interval:

Let the lower limit be L₁ and the upper limit be U₁.

Using the formula,

94 = $\dfrac{{L$1 + U1}}{2}

⇒ L₁ + U₁ = 94 x 2

⇒ L₁ + U₁ = 188 ……………..(1)

Class mark for the second interval:

Let the lower limit of the second interval be L₂ and the upper limit be U₂.

Using the formula,

104 = $\dfrac{{L$2 + U2}}{2}

⇒ L₂ + U₂ = 104 x 2

⇒ L₂ + U₂ = 208 ……………..(2)

Since the intervals are of equal size and overlap, we can assume that the difference between the upper limit of the first interval and the lower limit of the second interval is the size of the interval.

U₂ - L₂ = U₁ - L₁ = Interval Size

From eq (1) and (2), the difference between the two class marks is:

104 - 94 = 10

Thus, the interval size is 10.

For the first interval:

L₁ + U₁ = 188,

U₁ - L₁ = 10

Adding these two equations:

(L₁ + U₁) + (U₁ - L₁) = 188 + 10

⇒ L₁ + U₁ + U₁ - L₁ = 198

⇒ 2U₁ = 198

⇒ U₁ = $\dfrac{198}{2}$

⇒ U₁ = 99

Thus, L₁ = 188 - 99 = 89

For the second interval:

L₂ + U₂ = 208

U₂ - L₂ = 10

Adding these two equations:

(L₂ + U₂) + (U₂ - L₂) = 208 + 10

⇒ L₂ + U₂ + U₂ - L₂ = 218

⇒ 2U₂ = 218

⇒ U₂ = $\dfrac{218}{2}$

⇒ U₂ = 109

Thus, L₂ = 208 - 109 = 99

The corresponding intervals are 89−99 and 99−109.

∴ Assertion (A) is true.

The class mark of a class interval is the midpoint of the interval, given by:

Class Mark = $\dfrac{\text{Actual Lower limit + Actual Upper limit}}{2}$

∴ Reason (R) is true.

Hence, both Assertion (A) and Reason (R) are true.