Median, Quartiles & Mode — Assertion-Reason Type Questions

Assertion (A): For a collection of 11 arrayed data, the median is the middle number.

Reason (R): For the data 5, 9, 7, 13, 10, 11, 10, the median is 13.

1. Both A and R are correct, and R is the correct explanation of A.

2. Both A and R are correct, and R is not the correct explanation of A.

3. A is true, but R is false.

4. Both A and R are true.

Answer

For 11 arrayed data.

Median = $\dfrac{n + 1}{2}$ th term

= $\dfrac{11 + 1}{2}$

= 6th term, which will be the middle term.

∴ For a collection of 11 arrayed data, the median is the middle number.

∴ Assertion (A) is true.

Numbers = 5, 9, 7, 13, 10, 11, 10

Arranging in ascending order, we get :

5, 7, 9, 10, 10, 11, 13.

n = 7, which is odd.

Median = $\dfrac{n + 1}{2} = \dfrac{7 + 1}{2} = \dfrac{8}{2}$ = 4th term = 10.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

Assertion (A) : The first quartile of the observations 15, 14, 21, 11, 19, 10, 18 is 11.

Reason (R) : For an ungrouped data, containing n observations, median is given by $\Big(\dfrac{\text{n} + 1}{2}\Big)$ th observation, if n is odd.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Arranging the observations 15, 14, 21, 11, 19, 10, 18 in ascending order :

10, 11, 14, 15, 18, 19, 21

Here, n = 7, which is odd.

By formula,

Lower quartile (Q₁) = $\Big(\dfrac{\text{n} + 1}{4}\Big) \text{ th observation} = \dfrac{7 + 1}{4}$ th observation = 2 nd observation = 11.

∴ Assertion (A) is true.

Considering Reason (R), for an ungrouped data containing n observations, the median is given by $\Big(\dfrac{\text{n} + 1}{2}\Big)$ th observation, when n is odd.

∴ Reason (R) is true.

But Reason (R) is a statement about the median, whereas Assertion (A) is about the first quartile. So, R is not the correct explanation of A.

Hence, option 2 is the correct option.

Assertion (A) : The third quartile of the data 1, 20, 3, 15, 6, 8, 13, 5, 21, 23, 17, 10, 9, 12, 18, 21 is 18.

Reason (R) : For an ungrouped data, containing n observations, the upper quartile is given by Q₃ = $\Big(\dfrac{3\text{n}}{4} + 1\Big)$ th observation, when n is even.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Arranging the given data in ascending order :

1, 3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 21, 21, 23

Here, n = 16, which is even.

By formula,

Upper quartile (Q₃) = $\Big(\dfrac{3\text{n}}{4}\Big) \text{ th observation} = \dfrac{3 \times 16}{4}$ th observation = 12 th observation = 18.

∴ Assertion (A) is true.

Considering Reason (R), for an ungrouped data containing n observations, the upper quartile, when n is even, is given by Q₃ = $\Big(\dfrac{3\text{n}}{4}\Big)$ th observation and not $\Big(\dfrac{3\text{n}}{4} + 1\Big)$ th observation.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

Assertion (A): The difference in class marks of the modal class and the median class of the following frequency distribution table is 0.

Reason (R): Modal class and median class are always the same for a given frequency distribution.

1. Both A and R are correct, and R is the correct explanation of A.

2. Both A and R are correct, and R is not the correct explanation of A.

3. A is true, but R is false.

4. Both A and R are true.

Answer

Cumulative frequency distribution table :

Median = $\dfrac{16}{2}$ = 8th term.

The 8th term lies in the class 50-60.

∴ Median class = 50-60

Also, frequency of class 50-60 is highest.

∴ Modal class = 50-60.

∴ Assertion (A) is true.

Modal class and median class are not always the same for a given frequency distribution.

∴ Reason (R) is false.

Hence, Option 3 is the correct option.

Assertion (A) : If the median of the observations 11, 12, 14, (x − 2), (x + 4), (x + 9), 32, 38 and 47, arranged in ascending order is 24, then the value of x is 20.

Reason (R) : For an ungrouped data, containing n observations, the median is given by Median = $\dfrac{\text{n}}{2}$ th observation, where n is odd.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Arranging the given data in ascending order :

11, 12, 14, (x − 2), (x + 4), (x + 9), 32, 38, 47

Given, Median = 24.

Here, n = 9, which is odd.

By formula,

$\Rightarrow \text{Median} = \dfrac{\text{n} + 1}{2} \text{ th observation} \\[1em] \Rightarrow 24 = \dfrac{9 + 1}{2} \text{ th observation} \\[1em] \Rightarrow 24 = \dfrac{10}{2} \text{ th observation} \\[1em] \Rightarrow 24 = 5 \text{ th observation}$

⇒ 24 = x + 4

⇒ x = 24 − 4

⇒ x = 20.

∴ Assertion (A) is true.

Considering Reason (R), for an ungrouped data containing n observations, the median, when n is odd, is given by Median = $\dfrac{\text{n} + 1}{2}$ th observation and not $\dfrac{\text{n}}{2}$ th observation.

∴ Reason (R) is false.

Hence, option 3 is the correct option.