Frequency Distribution — Competency Focused Questions

Age of teachers of a school varies from 24 years to 72 years. Rishabh has to make a grouped frequency distribution table of age of teachers. Which among the following class intervals of age will be most appropriate to make meaningful conclusion in less time and effort?

1. 24 - 26, 26 - 28, 28 - 30, and so on

2. 24 - 30, 30 - 36, 36 - 42, and so on

3. 24 - 48, 48 - 72

4. none of these

Answer

Age of teachers varies from 24 years to 72 years

So, Range = 72 - 24 = 48

By checking the option (ii)

Class size = 6

Total classes = $\dfrac{48}{6}$

= 8 classes

∴ 8 class intervals are suitable to check the age of teachers.

Hence, option 2 is the correct option.

The distribution of marks of students of a class in a test is shown in the table below.

Which of the following statements are correct?

(i) We can definitely say that no one scored 60 marks.

(ii) The number of students who got at least 30 marks is 12.

1. only (i)

2. only (ii)

3. both (i) and (ii)

4. none of these

Answer

(i) We know that given table is in exclusive form. So in exclusive form generally the upper limit of class is excluded from that group and moved to next one.

However, last class interval of the given table is 45 - 60.

So, we can strongly say that no one scored 60 marks.

∴ Statement (i) is correct.

(ii) The number of students who got at least 30 marks are

30 – 45 ⟶ 8 students

45 – 60 ⟶ 4 students

Total = 8 + 4 = 12.

∴ Statement (ii) is Correct.

Hence, option 3 is the correct option.

Range of 14, 12, 17, 18, 16 and p is 20. If p>0, then p= ?

1. 10

2. 26

3. 32

4. 35

Answer

Given numbers 14, 12, 17, 18, 16 and p

Maximum Value = 18

Minimum value = 12

Range = Maximum value - Minimum value = 18 - 12 = 6 (excluding p)

But the range given is 20, So p must be either maximum value or minimum value

If p is the maximum value, then

p - 12 = 20

P = 32

If p is the minimum value, then

18 - p = 20

p = -2

Since p > 0

∴ p = 32

Hence, option 3 is the correct option.

The class mark of a class-interval is 'a'. If its lower limit is a-b, then its upper limit is :

1. a + b

2. ab

3. b-a

4. 2b

Answer

Class mark = a

Lower limit = a − b

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

$a = \dfrac{(a - b) + \text{Upper limit}}{2}$

2a = (a - b) + Upper limit

Upper limit = 2a − (a − b)

= 2a - a + b

= a + b.

Hence, option 1 is the correct option.