ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency MCQs

MCQs

Question 1

If the classes of a frequency distribution are 1-10, 11-20, 21-30, …, 51-60, then the size of each class is

(a) 9

(b) 10

(c) 11

(d) 5.5

Sol :

In the classes 1-10, 11-20, 21-30, …, 51-60,

the size of each class is 10. (b)

Question 2

If the classes of a frequency distribution are 1-10, 11-20, 21-30,…, 61-70, then the upper limit of the class 11-20 is

(a) 20

(b) 21

(c) 19.5

(d) 20.5

Sol :

In the classes of distribution, 1-10, 11-20, 21-30, …, 61-70,

upper limit of 11-20 is 20-5 as the classes after adjustment are

0.5-10.5, 10.5-20.5, 20.5-30.5, … (d)

Question 3

If the class marks of a continuous frequency distribution are 22, 30, 38, 46, 54, 62, then the class corresponding to the class mark 46 is

(a) 41.5-49.5

(b) 42-50

(c) 41-49

(d) 41-50

Sol :

The class marks of distribution are 22, 30, 38, 46, 54, 62,

then classes corresponding to these class marks 46 is

46.4 – 4 = 42, 46 + 4 = 50

(Class intervals is 8 as 30 – 22 = 8, 38 – 30 = 8

i.e:, 42 – 50 (b)

Question 4

If the mean of the following distribution is 2.6,

x_i	1	2	p	4	5
f_i	3	3	1	1	2

then the value of P is

(a) 2

(b) 3

(c) 2.6

(d) 2.8

Sol :

Mean = 2.6

x_i	1	2	p	4	5	Total
f_i	3	3	1	1	2	10
f_ix_i	3	6	p	4	10	23+p

Mean$=\frac{\sum f_{i} x_{i}}{\sum f}=\frac{23+p}{10}$

=2.6

$23+p=2 \cdot 6 \times 10=26 $

$\Rightarrow p=26-23=3$

Ans (b)

Question 5

The measure of central tendency of statistical data which takes into account all the data is

(a) mean

(b) median

(c) mode

(d) range

Sol :

A measure of central tendency of statistical data is mean. (a)

Question 6

In a grouped frequency distribution, the mid-values of the classes are used to measure which of the following central tendency?

(a) median

(b) mode

(c) mean

(d) all of these

Sol :

In a grouped frequency distribution,

the mid-values of the classes are used to measure Mean (c)

Question 7

In the formula: $\bar{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{i}}$ for finding the mean of the grouped data, d’is are deviations from a (assumed mean) of

(a) lower limits of the classes

(b) upper limits of the classes

(c) mid-points of the classes

(d) frequencies of the classes

Sol :

The formula $\bar{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{i}}$ is the finding of mean of the grouped data, d’is are mid-points of the classes

Question 8

In the formula: $\bar{x}=a+c\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right)$, for finding the mean of grouped frequency distribution $\text{u}_{i}$

(a) $\frac{y_{i}+a}{c}$

(b) $c\left(y_{i}-a\right)$

(c) $\frac{y_{i}-a}{c}$

(d) $\frac{a-y_{i}}{c}$

Sol :

In $\bar{x}=a+c\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right)$

for finding the mean of grouped frequency, $\mathrm{u}_{\mathrm{i}}$ is $\frac{y_{i}-a}$

Ans (c)

Question 9

While computing mean of grouped data, we assumed that the frequencies are

(a) evenly distributed over all the classes

(b) centred at the class marks of the classes

(c) centred at the upper limits of the classes

(d) centred at the lower limits of the classes

Sol :

For computing mean of grouped data,

we assumed that frequencies are centred at class marks of the classes.

Ans (b)

Question 10

Construction of a cumulative frequency distribution table is useful in determining the

(a) mean

(b) median

(c) mode

(d) all the three measures

Sol :

Construction of a cumulative frequency distribution table

is used for determining the median,

Ans (b)

Question 11

The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below:

Class	13.8-14	14-14.2	14.2-14-4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	5	71	48	20

The number of athletes who completed the race in less than 14.6 seconds is

(a) 11

(b) 71

(c) 82

(d) 130

Sol :

Time taken in seconds by 150 athletes to run a 110 m hurdle race as given in the sum,

the number of athletes who completed the race in less then 14.6 second is

2 + 4 + 5 + 71 = 82 athletes.

Ans (c)

Question 12

Consider the following frequency distribution:

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

The upper limit of the median class is

(a) 17

(b) 17.5

(c) 18

(d) 18.5

Sol :

From the given frequency upper limit of median class is 17.5

as total frequencies 13 + 10 + 15 + 8 + 11 = 57

$\frac{57+1}{2}=\frac{58}{2}=29$

and 13 + 10 + 15 = 28 where class is 12-17

But actual class will be 11.5-17.5

Upper limit is 17.5

Ans (b)

Question 13

Daily wages of a factory workers are recorded as:

Daily wages(in ₹)	131-136	137-142	143-148	149-154	155-160
No. of workers	5	27	20	18	12

The lower limit of the modal class is

(a) Rs 137

(b) Rs 143

(c) Rs 136.5

(d) Rs 142.5

Sol :

In the daily wages of workers of a factory are 131-136, 137-142, 142-148, …

which are not a proper class

So, proper class will be 130.5-136.5, 136.5-142.5, 142.5-148.5, …

Lower limit of a model class is 136.5 as 136.5-142.5 is the modal class. (c)

Question 14

For the following distribution:

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

The sum of lower limits of the median class and modal class is

(a) 15

(b) 25

(c) 30

(d) 35

Sol :

From the given distribution

Sum of frequencies = 10 + 15 + 12 + 20 + 9 = 66

and median is $\frac{66}{2}=33$

Median class will be 10-15 and modal class is 15-20

Sum of lower limits = 10 + 15 = 25 (b)

Question 15

Consider the following data:

Class	65-85	85-105	105-125	125-145	145-165	14
Frequency	4	5	13	20	14	14

The difference of the upper limit of the median class and the lower limit of the modal class is

(a) 0

(b) 19

(c) 20

(d) 38

Sol :

From the given data

Total frequencies = 4 + 5 + 13 + 20 + 14 + 7 + 4 = 67

Median class $\frac{67+1}{2}=34$

which is (4 + 5 + 13 + 20) 125-145 and modal class is 125-145

Difference of upper limit of median class and the lower limit of the modal class

= 145 – 125 = 20 (c)

Question 16

An ogive curve is used to determine

(a) range

(b) mean

(c) mode

(d) median

Sol :

An ogive curve is used to find median.

Ans (d)

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