Mean MCQ Quiz - Objective Question with Answer for Mean - Download Free PDF

Last updated on Apr 11, 2025

Latest Mean MCQ Objective Questions

Mean Question 1:

The mean of the series x1, x2...xn is x̅ If xn is replaced by k, then what is the new mean?

  1. x̅ - xn + k
  2. \(\rm \frac{n\bar x-\bar x+k}{n}\)
  3. \(\rm \frac{\bar x-x_n-k}{n}\)
  4. \(\rm \frac{n\bar x-x_n+k}{n}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \frac{n\bar x-x_n+k}{n}\)

Mean Question 1 Detailed Solution

Explanation:

Given:

\(\overline x = \frac{x_1+x_2+x_3+......... x_n}{n}\)

⇒ \(x_1+x_2+X_3+......x_n= n\overline x \)

⇒ Mean when xn is replaced by k

⇒ New mean = \(\frac{x_1+x_2+x_3+......... x_{n-1}+k}{n}\)

\(\frac{n\overline x -x_n+k}{n}\)

∴ Option (d) is correct

Mean Question 2:

The mean of n observations

1, 4, 9, 16 ...., n2 is 130. What is the value of n?

  1. 18
  2. 19
  3. 20
  4. 21

Answer (Detailed Solution Below)

Option 2 : 19

Mean Question 2 Detailed Solution

Explanation:

Mean  = \(\frac{1+ 4+ 9 ... n^2}{n}\)

⇒ 130 = \(\frac{(n(n+1)(2n+1)}{6n}\)

⇒ 780 = 2n+ 3n + 1

⇒ 2n2 + 3n – 779 =0

⇒ 2n2 + 41n – 38n – 779 = 0

⇒ n(2n + 41) – 19(2n + 41) = 0 

⇒ (2n + 41) (n – 19) = 0

⇒ x = 19 ..... ( n = \(-\frac{41}{2}\) is not possible)

∴ Option (b) is correct.

Mean Question 3:

Find the mean deviation about the mean for the data

4, 7, 8, 9, 10, 12, 13, 17 

  1. 3
  2. 24
  3. 10
  4. 8

Answer (Detailed Solution Below)

Option 1 : 3

Mean Question 3 Detailed Solution

Calculation

Arithmetic mean x̅ of 4, 7, 8, 9, 10, 12, 13, 17 is

x̅ = \(\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10\)

∑|xi - x̅| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24

∴ Mean deviation about mean

\(\text { M.D. }(\bar{x})=\frac{\sum\left|x_{i}-x\right|}{n}=\frac{24}{8}=3\)

Hence option 1 is correct

Mean Question 4:

The marks obtained by 8 students in a mathematics test are:  

15, 20, 25, 25, 30, 35, 40, 50.  

If the arithmetic mean \(( \mu ) \), median ( \({Md} \) ), and mode\( ( {Mo} ) \) are related by the equation:  

\(2\mu + 3{Md} = 5{Mo}, \)

find the value of \(\mu + {Md} + {Mo} \) .

(Provide your answer rounded to two decimal places.)  

 

Answer (Detailed Solution Below) 85.00

Mean Question 4 Detailed Solution

Solution:

1. Given Data:  

Marks: 15, 20, 25, 25, 30, 35, 40, 50 .  

2. Calculate Mean ( \(\mu \)):  

\(\mu = \frac{{Sum \quad of \quad observations}}{{Number\quad of \quad observations}} \)
  

\(\mu = \frac{15 + 20 + 25 + 25 + 30 + 35 + 40 + 50}{8} \)


\(\mu = \frac{260}{8} = 32.5 \)

For the median:  

Arrange the data in ascending order (already done).  

Since there are 8 observations (even number),

the median is the average of the 4th and 5th values:  

\({Md} = \frac{25 + 30}{2} = 27.5 \)

The mode is the value that occurs most frequently in the dataset.

Here: \( {Mo} = 25 \quad \)(occurs twice, more frequent than others).

 Find \( \mu + {Md} + {Mo} \) =  32.5 + 27.5 + 25 = 85.0

Hence 85.00 is the answer.

Mean Question 5:

The means of two samples of size 30 and 40 are 35 and 42 respectively. Then the mean of the combined sample of size 70 is

  1. 36
  2. 37
  3. 38
  4. 39
  5. 40

Answer (Detailed Solution Below)

Option 4 : 39

Mean Question 5 Detailed Solution

Concept Used:

The mean of the combined sample is given by:

\(\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}\)

Calculation:

Given:

Sample 1: n₁ = 30, mean x̄₁ = 35

Sample 2: n₂ = 40, mean x̄₂ = 42

\(\bar{x} = \frac{(30 \times 35) + (40 \times 42)}{30 + 40}\)

\(\bar{x} = \frac{1050 + 1680}{70}\)

\(\bar{x} = \frac{2730}{70}\)

\(\bar{x} = 39\)

∴ The mean of the combined sample is 39.

Hence option 4 is correct

Top Mean MCQ Objective Questions

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

Answer (Detailed Solution Below)

Option 2 : 35.70

Mean Question 6 Detailed Solution

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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 ∑X = 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

A random sample of 24 people is classified in the following table according to their ages:

Age

Frequency

10 – 20

4

20 – 30

6

30 – 40

8

40 – 50

2

50 - 60

4


What is the mean age of this group of people?

  1. 25
  2. 33.3
  3. 16.67
  4. 54.54

Answer (Detailed Solution Below)

Option 2 : 33.3

Mean Question 7 Detailed Solution

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Concept:

  • \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)


Calculation:

Age

Frequency (f)

x

xf

10 – 20

4

15

60

20 – 30

6

25

150

30 – 40

8

35

280

40 – 50

2

45

90

50 - 60

4

55

220

 

\(\sum {\rm{f}} = 24\)

 

\(\sum {\rm{xf}} = 800\)

 

We know that, \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

\(\therefore {\rm{Mean}} = \frac{{800}}{{24}} = 33.3\)

What is the mean of first 99 natural numbers ?

  1. 100
  2. 50.5
  3. 50
  4. 99

Answer (Detailed Solution Below)

Option 3 : 50

Mean Question 8 Detailed Solution

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Concept:

Suppose there are ‘n’ observations {\({\rm{\;}}{{\rm{x}}_{\rm{1}}},{{\rm{x}}_{{\rm{2\;}}}},{{\rm{x}}_{{\rm{3\;}}}}, \ldots ,{{\rm{x}}_{{\rm{n\;}}}}\)}

Mean \(\left( {{\rm{\bar x}}} \right) = \frac{{{\rm{\;}}({{\rm{x}}_1} + {{\rm{x}}_{2{\rm{\;}}}} + {{\rm{x}}_{3{\rm{\;}}}} + \ldots + {{\rm{x}}_{{\rm{n\;}}}})}}{{\rm{n}}}\) \( = {\rm{\;}}\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)

Sum of the first n natural numbers = \(\rm \frac{n(n+1)}{2}\)

 

Calculation:

To find:  Mean of the first 99 natural numbers

As we know, Sum of first n natural numbers = \(\rm \frac{n(n+1)}{2}\)

Now, Mean = \(\rm \dfrac { \frac{99(99+1)}{2}}{99}\)

\(\rm \frac{(99+1)}{2}=50\)

Let the average of three numbers be 16. If two of the numbers are 8 and 10, what is the remaining number?

  1. -30
  2. 18
  3. 12
  4. 30

Answer (Detailed Solution Below)

Option 4 : 30

Mean Question 9 Detailed Solution

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Concept:

\(\rm Mean\ of\ 'n'\ observations=\dfrac{Sum\ of\ observations}{n}\).

 

Calculation:

Here n = 3. Let's say that the third number is x.

∴ \(\rm 16=\dfrac{x+8+10}{3}\)

⇒ x + 18 = 48

⇒ x = 30.

A random sample of 20 people is classified in the following table according to their ages:

Age

Frequency

15 – 25

2

25 – 35

4

35 – 45

6

45 – 55

5

55 - 65

3


What is the mean age of this group of people?

  1. 41.0
  2. 41.5
  3. 42.0
  4. 42.5

Answer (Detailed Solution Below)

Option 2 : 41.5

Mean Question 10 Detailed Solution

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Concept:

\({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

Calculation:

Age

Frequency (f)

x

xf

15 – 25

2

20

40

25 – 35

4

30

120

35 – 45

6

40

240

45 – 55

5

50

250

55 - 65

3

60

180

 

\(\sum {\rm{f}} = 20\)

 

\(\sum {\rm{xf}} = 830\)

 

We know that, \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

\(\therefore {\rm{Mean}} = \frac{{830}}{{20}} = 41.5\)

If the difference between mean and mode is 48 and median is 12, Find mean.

  1. 38
  2. 36
  3. 42
  4. 28

Answer (Detailed Solution Below)

Option 4 : 28

Mean Question 11 Detailed Solution

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Concept:
Mean - mode = 3(Mean - median)

Calculation:

Given:

The difference between mean and mode = 48

And median = 12

Mean - mode = 3(Mean - median)

⇒ 48 = 3(mean - 12)

⇒ 48 = 3 mean - 36

⇒ 3 mean = 48 + 36

⇒ 3 mean = 84

⇒ mean = 84/3

⇒ mean = 28

∴ Mean is 28

What is the sum of deviations of the variate values 73, 85, 92, 105, 120 from their mean?

  1. -2
  2. -1
  3. 0
  4. 5

Answer (Detailed Solution Below)

Option 3 : 0

Mean Question 12 Detailed Solution

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Concept:

For the observations \(\rm x_1 ,x_2 ,x_3 ,..........,x_n\)

Mean (\(\rm\overline x\)) = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\)

Deviation of the observation from the mean = xi - \(\rm\overline x\)

where i ∈ [1, n]

Calculation:

Mean = \(\rm {73+85+92+105+120\over 5}\)

\(\rm\overline x\) = \(\rm {475\over 5}\) = 95

Value Mean Deviation of the value
73 95 73 - 95 = -22
85 95 85 - 95 = -10
92 95 92 - 95 = -3
105 95 105 - 95 = 10
120 95 120 - 95 = 25

 

The sum of the deviation = -22 +(-10) + (-3) + 10 + 25 = 0

The mean of six numbers is 47. If one number is excluded, their mean becomes 41. The excluded number is

  1. 77
  2. 78
  3. 60
  4. 45

Answer (Detailed Solution Below)

Option 1 : 77

Mean Question 13 Detailed Solution

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Concept:

Mean = (Sum of observations) / (Total number of observations)

 

Calculation:

Let the six numbers be a, b, c, d, e, f

So, Mean = \(\rm \frac{a+b+c+d+e+f}{6}=47\)

\(\rm ⇒ a+b+c+d+e+f=282\)          ....(1)

Let, the excluded number be a, 

So mean of remaining five numbers = \(\rm \frac{b+c+d+e+f}{5}=41\)

\(\rm ⇒ b+c+d+e+f=205\)                ....(2)

∴ a + 205 = 282              (from (1) and (2))

⇒ a = 77

Hence, option (1) is correct.

If the mean of a set of observations x1, x2, x3,...,x10 is 50, then the mean of x1 + 5, x2 + 10, x3 + 15,..., x10 + 50 is

  1. 77.5
  2. 87.5
  3. 72.5
  4. 82.5

Answer (Detailed Solution Below)

Option 1 : 77.5

Mean Question 14 Detailed Solution

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Concept:

\({\rm{Mean\;}} = {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\), where xi is observations and n is total number of observation.

Sum of the first n natural numbers: \(\sum {\rm{n}} = 1 + 2 + 3 + \ldots + {\rm{n}} = {\rm{\;}}\frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{2}\)

Calculation:

Given: Mean of a set of observations x1, x2, x3,..., x10 is 50

Here n = 10,

\({\rm{Mean\;}} = {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}} = {\rm{\;}}\frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3} + \ldots + {{\rm{x}}_{10}}}}{{10}} = 50\)

⇒ x1 + x2 + x3 +...+ x10 = 500                      …. (1)

 

Now find out the mean of new observations x1 + 5, x2 + 10, x3 + 15,..., x10 + 50

\({\rm{\;Mean\;}} = {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}} = {\rm{\;}}\frac{{\left( {{{\rm{x}}_1} + 5} \right) + \left( {{{\rm{x}}_2} + 10} \right) + ({{\rm{x}}_3} + 15) + \ldots + ({{\rm{x}}_{10}} + 50)}}{{10}}\)

 \( = {\rm{\;}}\frac{{\left( {{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3} + \ldots + {{\rm{x}}_{10}}} \right) + \left( {5 + 10 + 15 + \ldots + 50} \right)}}{{10}} \)

From equation (1),

\(= \frac{{500 + \;5\left( {1 + 2 + 3 + \ldots + 10} \right)}}{{10}} \)

\(= 50 + \frac{1}{2} \times \frac{{10 \times 11}}{2} \)

\(= 50 + 27.5 \)

= 77.5

Find the value of ‘n’ if the mean of the set of the numbers 8, 5, n, 10, 15, 21 is given as 11.

  1. 5
  2. 7
  3. 4
  4. 6

Answer (Detailed Solution Below)

Option 2 : 7

Mean Question 15 Detailed Solution

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CONCEPT:

The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations. It is denoted by the symbol  \(\bar x\), read as ‘x bar’.

Mean \(\rm \bar x = \frac{{Sum\;of\;all\;the\;observations}}{{Total\;number\;of\;observations}}\)

CALCULATION:

Given set of number is 8, 5, n, 10, 15, 21.

So \(\rm \bar x = \frac{{Sum\;of\;all\;the\;observations}}{{Total\;number\;of\;observations}} = 11\)

∴ \(11 = \frac{{8\; + \;5\; + \;n\; +\; 10\; + \;15\; + \;21}}{6}\)  ⇒ n = 7

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