Last updated on Sep 19, 2022

Latest Statistics MCQ Objective Questions

Statistics MCQ Question 1:

The mean of 25 observations is 36 . If the mean of the first 13 observations is 32 and that of the last 13 observations is 39 , the 13th observation is:

1. 22
2. 25
3. 26
4. 23

Option 4 : 23

Statistics MCQ Question 1 Detailed Solution

Given:

The mean of 25 observations is 36

The mean of the first 13 observations is 32 and that of the last 13 observations is 39

Concept used:

Mean = sum of all observation/total number of observation

Calculation:

The sum of all 25 observation = 25 × 36 = 900

Sum of first 13 observations = 13 × 32 = 416

Sum of last 13 observations = 13 × 39 = 507

∴ 13th term = (416 + 507) - 900 = 923 - 900 = 23

Statistics MCQ Question 2:

If a variable takes discrete values a + 4, a - 3.5, a - 2.5, a - 3, a - 2, a + 0.5, a + 5 and a - 0.5 where a > 0, then the median of the data set is

1. a - 2.5
2. a - 1.25
3. a - 1.5
4. a - o.75

Option 2 : a - 1.25

Statistics MCQ Question 2 Detailed Solution

Given:

The given values =  a + 4, a – 3.5, a – 2.5, a – 3, a – 2, a + 0.5, a + 5 and a – 0.5

Concept used:

If n is odd

Median = [(n + 1)/2]th observations

If n is even

Median = [(n/2)th + (n/2 + 1)th observations]/2

Calculation:

a + 4, a – 3.5, a – 2.5, a – 3, a – 2, a + 0.5, a + 5 and a – 0.5

Arrange the data in ascending order

⇒ a – 3.5, a – 3, a – 2.5, a – 2, a – 0.5, a + 0.5, a + 4, a + 5

Here, the n is 8, which is even

Median =  [(n/2)th + (n/2 + 1)th observations]/2

⇒ [(8/2) + (8/2 + 1)/2] term

⇒ 4th + 5th term

⇒ [(a – 2 + a – 0.5)/2]

⇒ [(2a – 2.5)/2]

⇒ a – 1.25

∴ The median of the data set is a – 1.25

Comprehension:

Prices of shares of X and Y are given below.

 X: 35 54 52 53 56 58 52 50 51 49 Y: 108 107 105 105 106 107 104 103 104 101

Use the data to answer the following questions.

Which one among X and Y is more stable

1. X
2. Y
3. Both are same
4. Insufficient data

Option 1 : X

Statistics MCQ Question 3 Detailed Solution

Concept:

Variance (X) = $$\frac{\sum(X_i-\bar X)^2}{n}$$

Standard Deviation = $$\sqrt{Variance }$$

Coefficient of variation for X = $$\frac{SD}{\bar X}\times 100$$ where X̅ is the means of observation.

Calculation:

For the prices of shares X:

 Xi di = (Xi - X̅ ) di2 35 35 - 48 = -13 169 24 24 - 48 = -24 576 52 52 - 48 = 4 16 53 53 - 48 = 5 25 56 56 - 48 = 8 64 58 58 - 48 = 10 100 52 52 - 48 = 4 16 50 50 - 48 = 2 4 51 51 - 48 = 3 9 49 49 - 48 = 1 1 Total = 480 Total = 980

Here, n = 10

X̅ = $$\frac{1}{n}\sum X_i$$

⇒ X̅ = $$\frac{35+54+52+53+56+58+52+50+51+49}{10}$$ = $$\frac{480}{10}=48$$

Variance (X) = $$\frac{\sum(X_i-\bar X)^2}{n}$$ = $$\frac{\sum d^2}{n}$$ = $$\frac{980}{10}=98$$

SD(X) = $$\sqrt{Var(X)}$$ = √98 = 9.9

Coefficient of variation for X = $$\frac{SD}{\bar X}\times 100$$ = $$\frac{9.9}{48}\times 100 = 20.6$$      ------(i)

For the prices of shares Y:

 Yi di = (Yi - Y̅) di2 108 108 - 105 = 3 9 107 107 - 105 = 2 4 105 105 - 105 = 0 0 105 105 - 105 = 0 0 106 106 - 105 = 1 1 107 107 - 105 = 2 4 104 104 - 105 = -1 1 103 103 - 105 = -2 4 104 104 - 105 = -1 1 101 101 - 105 = -4 16 Total = 480 Total = 40

Here, n = 10

Y̅  = $$\frac{1}{n}\sum Y_i$$

⇒ Y̅  = $$\frac{108+107+105+105+106+107+104+103+104+101}{10}$$ = $$\frac{1050}{10}=105$$

Variance (Y) = $$\frac{\sum(Y_i-\bar Y)^2}{n}$$ = $$\frac{\sum di^2}{n}$$ = $$\frac{40}{10}=4$$

SD(Y) = $$\sqrt{Var(Y)}$$ = = √4 = 2

Coefficient of variation for Y $$\frac{SD}{\bar Y}\times 100$$ = $$\frac{2}{105}\times 100 = 1.90$$      ------(ii)

Comparing the equation (i) and (ii), coefficient of variation of Y is smaller than that of X.

∴ X is more stable than Y.

Comprehension:

Prices of shares of X and Y are given below.

 X: 35 54 52 53 56 58 52 50 51 49 Y: 108 107 105 105 106 107 104 103 104 101

Use the data to answer the following questions.

Standard devaition of Y is

1. 1.90
2. 105
3. 4
4. 2

Option 4 : 2

Statistics MCQ Question 4 Detailed Solution

Concept:

Variance (X) = $$\frac{\sum(X_i-\bar X)^2}{n}$$

Standard Deviation = $$\sqrt{Variance }$$

Coefficient of variation for X = $$\frac{SD}{\bar X}\times 100$$ where X̅ is the means of observation.

Calculation:

 Yi di = (Yi - Y̅) di2 108 108 - 105 = 3 9 107 107 - 105 = 2 4 105 105 - 105 = 0 0 105 105 - 105 = 0 0 106 106 - 105 = 1 1 107 107 - 105 = 2 4 104 104 - 105 = -1 1 103 103 - 105 = -2 4 104 104 - 105 = -1 1 101 101 - 105 = -4 16 Total = 480 Total = 40

Here, n = 10

For the prices of shares Y:

Y̅  = $$\frac{1}{n}\sum Y_i$$

⇒ Y̅ $$\frac{108+107+105+105+106+107+104+103+104+101}{10}$$ = $$\frac{1050}{10}=105$$

Variance (Y) = $$\frac{\sum(Y_i-\bar Y)^2}{n}$$ = $$\frac{\sum di^2}{n}$$ = $$\frac{40}{10}=4$$

SD(Y) = $$\sqrt{Var(Y)}$$

⇒ SD(Y) = √4 = 2

∴ Standard deviation of Y is 2.

Comprehension:

Prices of shares of X and Y are given below.

 X: 35 54 52 53 56 58 52 50 51 49 Y: 108 107 105 105 106 107 104 103 104 101

Use the data to answer the following questions.

Variance of X is

1. 98
2. 9.9
3. 48
4. 20.6

Option 1 : 98

Statistics MCQ Question 5 Detailed Solution

Concept:

Variance (X) = $$\frac{\sum(X_i-\bar X)^2}{n}$$

Standard Deviation = $$\sqrt{Variance }$$

Coefficient of variation for X = $$\frac{SD}{̅ X}\times 100$$ where X̅ is the means of observation.

Calculation:

 Xi di = (Xi - X̅ ) di2 35 35 - 48 = -13 169 24 24 - 48 = -24 576 52 52 - 48 = 4 16 53 53 - 48 = 5 25 56 56 - 48 = 8 64 58 58 - 48 = 10 100 52 52 - 48 = 4 16 50 50 - 48 = 2 4 51 51 - 48 = 3 9 49 49 - 48 = 1 1 Total = 480 Total = 980

Here, n = 10

For the prices of shares X:

⇒ X̅ = $$\frac{1}{n}\sum X_i$$

⇒ X̅ = $$\frac{35+24+52+53+56+58+52+50+51+49}{10}$$ = $$\frac{480}{10}=48$$

Variance (X) = $$\frac{\sum(X_i-\bar X)^2}{n}$$ = $$\frac{\sum d^2}{n}$$ = $$\frac{980}{10}=98$$

∴ The Variance of X is 98.

Top Statistics MCQ Objective Questions

Statistics MCQ Question 6

What is the mean of the range, mode and median of the data given below?

5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

1. 10
2. 12
3. 8
4. 9

Option 4 : 9

Statistics MCQ Question 6 Detailed Solution

Given:

The given data is 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Concept used:

The mode is the value that appears most frequently in a data set

At the time of finding Median

First, arrange the given data in the ascending order and then find the term

Formula used:

Mean = Sum of all the terms/Total number of terms

Median = {(n + 1)/2}th term when n is odd

Median = 1/2[(n/2)th term + {(n/2) + 1}th] term when n is even

Range = Maximum value – Minimum value

Calculation:

Arranging the given data in ascending order

2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19

Here, Most frequent data is 4 so

Mode = 4

Total terms in the given data, (n) = 15 (It is odd)

Median = {(n + 1)/2}th term when n is odd

⇒ {(15 + 1)/2}th term

⇒ (8)th term

⇒ 6

Now, Range = Maximum value – Minimum value

⇒ 19 – 2 = 17

Mean of Range, Mode and median = (Range + Mode + Median)/3

⇒ (17 + 4 + 6)/3

⇒ 27/3 = 9

∴ The mean of the Range, Mode and Median is 9

Statistics MCQ Question 7

Find the mean of given data:

 class interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 9 13 6 4 6 2 3

1. 39.95
2. 35.70
3. 43.95
4. 23.95

Option 2 : 35.70

Statistics MCQ Question 7 Detailed Solution

Formula used:

The mean of grouped data is given by,

$$\bar X\ = \frac{∑ f_iX_i}{∑ f_i}$$

Where, $$u_i \ = \ \frac{X_i\ -\ a}{h}$$

Xi = mean of ith class

f= frequency corresponding to ith class

Given:

 class interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 9 13 6 4 6 2 3

Calculation:

Now, to calculate the mean of data will have to find ∑fiXi and ∑fi as below,

 Class Interval fi Xi fiXi 10 - 20 9 15 135 20 - 30 13 25 325 30 - 40 6 35 210 40 - 50 4 45 180 50 - 60 6 55 330 60 - 70 2 65 130 70 - 80 3 75 225 ∑fi = 43 ∑Xi  = 315 ∑fiXi = 1535

Then,

We know that, mean of grouped data is given by

$$\bar X\ = \frac{∑ f_iX_i}{∑ f_i}$$

$$\frac{1535}{43}$$

= 35.7

Hence, the mean of the grouped data is 35.7

Statistics MCQ Question 8

If mean and mode of some data are 4 & 10 respectively, its median will be:

1. 1.5
2. 5.3
3. 16
4. 6

Option 4 : 6

Statistics MCQ Question 8 Detailed Solution

Concept:

Mean: The mean or average of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set.

Mode: The mode is the value that appears most frequently in a data set.

Median: The median is a numeric value that separates the higher half of a set from the lower half.

Relation b/w mean, mode and median:

Mode = 3(Median) - 2(Mean)

Calculation:

Given that,

mean of data = 4 and mode of  data = 10

We know that

Mode = 3(Median) - 2(Mean)

⇒ 10 = 3(median) - 2(4)

⇒ 3(median) = 18

⇒ median = 6

Hence, the median of data will be 6.

Statistics MCQ Question 9

If the standard deviation of 0, 1, 2, 3 ______ 9 is K, then the standard deviation of 10, 11, 12, 13 _____ 19 will be:

1. K + 1
2. K
3. K + 4
4. K + 8

Option 2 : K

Statistics MCQ Question 9 Detailed Solution

Formula Used

• σ2 = ∑(xi – x)2/n
• Standard deviation is same when each element is increased by the same constant

Calculation:

Since each data increases by 10,

There will be no change in standard deviation because (xi – x) remains same.

∴ The standard deviation of 10, 11, 12, 13 _____ 19 will be will be K.

Alternate Method

Statistics MCQ Question 10

Find the median of the given set of numbers 2, 6, 6, 8, 4, 2, 7, 9

1. 6
2. 8
3. 4
4. 5

Option 1 : 6

Statistics MCQ Question 10 Detailed Solution

Concept:

Median: The median is the middle number in a sorted- ascending or descending list of numbers.

Case 1: If the number of observations (n) is even

$${\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}$$

Case 2: If the number of observations (n) is odd

$${\rm{Median\;}} = {\rm{value\;of\;}}{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)^{{\rm{th}}}}{\rm{\;observation}}$$

Calculation:

Given values 2, 6, 6, 8, 4, 2, 7, 9

Arrange the observations in ascending order:

2, 2, 4, 6, 6, 7, 8, 9

Here, n = 8 = even

As we know, If n is even then,

$${\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}$$

$$\rm \frac{4^{th} \;\text{observation}+5^{th} \;\text{observation}}{2}$$

$$\frac{6+6}{2} =6$$

Hence Median = 6