Frequency Distribution — Multiple Choice Questions

If in a grouped frequency distribution, the classes are 20 - 30, 30 - 40, 40 - 50, ..., then the observation 40 is included in the class :

1. 30 - 40

2. 40 - 50

3. 50 - 60

4. 20 - 30

Answer

Since the given class-intervals are in exclusive form.

So, in exclusive form lower limit is included but the upper limit is excluded.

So 40 is included in the class-interval, 40-50.

Hence, option 2 is the correct option.

If in a grouped frequency distribution, the class intervals are 20 - 29, 30 - 39, 40 - 49, ...., then the observation 29.4 is included in the class :

1. 20 - 29

2. 30 - 39

3. 40 - 49

4. 29 - 30

Answer

Given class intervals are in inclusive form but the observation is in decimal form (29.4).

So convert the inclusive form into exclusive form.

Gap between the classes = 30 - 29 = 1

Adjustment factor = 0.5

So new class boundaries become

19.5 - 29.5, 29.5 - 39.5, 39.5 - 49.5, ...,

29.4 lies between the class-interval 19.5 - 29.5.

19.5 - 29.5 is the exclusive form of 20 - 29.

Hence, option 1 is the correct option.

The classes of a frequency distribution are 30 -34, 35 - 39, ..., 50 - 54. The lower boundary of the class 35 - 39 is :

1. 30.5

2. 35

3. 34.5

4. 39

Answer

Gap between the classes = 35 - 34 = 1

Adjustment factor = $\dfrac{1}{2}$ = 0.5

New class boundaries :

29.5 - 34.5, 34.5 - 39.5, ..., 49.5 - 54.5.

Thus lower limit of class 35 - 39 is 34.5

Hence, option 3 is the correct option.

In a grouped frequency distribution, the class interval are 40 - 49, 50 - 59, ..., 80 - 89. The upper boundary of the class (50 - 59) is :

1. 59

2. 59.5

3. 49.5

4. 60

Answer

Gap between the classes = 50 - 49 = 1

Adjustment factor = 0.5

Upper boundary for the class interval (50 - 59) = Upper limit + 0.5

= 59 + 0.5

= 59.5

Hence, option 2 is the correct option.

If 0 - 9, 10 - 19, 20 - 29, ...., are the classes of a grouped frequency distribution, then the width of each class is :

1. 9

2. 9.5

3. 10

4. 9 or 10

Answer

Gap between classes = 10 - 9 = 1

Adjustment factor = 0.5

So the new class boundaries become

-0.5 - 9.5, 9.5 - 19.5, 19.5 - 29.5, ...,

Class width = Upper boundary - Lower boundary

Considering second class :

Class width = 19.5 - 9.5 = 10

Similarly, for rest of the class intervals.

Hence, option 3 is the correct option.

If 10, 15, 20, ... are respectively the mid-value of the classes of a grouped frequency distribution, then the class whole mid-value is 25 is :

1. 20 - 27

2. 22.5 - 27.5

3. 21.5 - 28.5

4. 21 - 28

Answer

Difference between the mid values = 15-10 = 5

So, the class width = 5

For the class whose mid-value 25 is

Lower limit = 25 - $\dfrac{\text{Class width}}{2}$ = 22.5

Upper limit = 25 + $\dfrac{\text{Class width}}{2}$ = 27.5

So the Class is 22.5 - 27.5.

Hence, option 2 is the correct option.

The width of each class interval of a grouped frequency distribution is 8. The lower boundary of the class whose mid-value is 10 is :

1. 2

2. 4.5

3. 5.5

4. 6

Answer

Class width = 8

Half of class width = $\dfrac{8}{2}$ = 4

Mid-value = 10

So, Lower boundary = 10 - 4

= 6

Hence, option 4 is the correct option.

The width of each class of a frequency distribution is 5. The upper boundary of the class whose class mark is 25 is :

1. 22

2. 22.5

3. 27

4. 27.5

Answer

Class width = 5

Class mark = 25

Half of class width = $\dfrac{5}{2}$ = 2.5

Upper boundary = 25 + 2.5 = 27.5

Hence, option 4 is the correct option.

Class marks of a distribution are: 47, 52, 57, 62, 67, 72, 77, 82. Class size of the given distribution is :

1. 5

2. 6

3. 7

4. 2

Answer

Class size = Difference between the consecutive class marks.

Class size = 52 - 47 = 5

57 - 52 = 5

Therefore class size = 5

Hence, option 1 is the correct option.

Mid value for the class interval 19.5 - 29.5 is :

1. 14.5

2. 24.5

3. 15.5

4. 25.5

Answer

Class interval = 19.5-29.5

Mid value = $\dfrac{\text{Lower limit} + \text{Upper limit}}{2}$

= $\dfrac{19.5 + 29.5}{2}$

= $\dfrac{49.5}{2}$

= 24.5

Hence, option 2 is the correct option.