# Statistics MCQ Quiz - Objective Question with Answer for Statistics - Download Free PDF

Last updated on Dec 4, 2023

## Latest Statistics MCQ Objective Questions

#### Statistics Question 1:

The mean monthly salary paid to 77 employees in a company was Rs. 78. The mean salary of 32 of them was Rs. 45 and of the other 25 was Rs. 82. What was the mean salary of the remaining?

#### Answer (Detailed Solution Below)

#### Statistics Question 1 Detailed Solution

**Given:**

Total number of employees (n) = 77

Mean salary of all employees = Rs. 78

Number of employees with salary Rs. 45 = 32

Number of employees with salary Rs. 82 = 25

**Concept used:**

Total sum of salaries = Mean salary × Total number of employees

**Calculation:**

Total salaries for employees with salary Rs. 45

⇒ 45 × 32 = Rs. 1440

Total salaries for employees with salary Rs. 82

⇒ 82 × 25 = Rs. 2050

Total sum of salaries = 78 × 77 = Rs. 6006

Total salaries of remaining employees

⇒ Total sum of salaries - Sum of salaries for known employees

⇒ 6006 - 1440 - 2050 = Rs. 2516

Now, Mean salary of the remaining employees

⇒ 2516/(77 - 32 - 25) = 2516/20 = 125.8

**∴ The mean salary is 125.8**

#### Statistics Question 2:

If the mean of a certain set of data is 16 and variance is 4 then find the coefficient of variance.

#### Answer (Detailed Solution Below)

#### Statistics Question 2 Detailed Solution

Given

Arithmetic mean = (μ) = 16

variance = σ^{2} = 4

Formula

Standard deviation = σ = √variance

CV = Coefficient of variation = σ/μ

σ = Standard deviation

μ = Mean

Calculation

Standard deviation = σ = √4 = 2

⇒ CV = (σ/μ) × 100

⇒ CV = (2/16) × 100

**∴ Coefficient of variation is 12.5**

#### Statistics Question 3:

For a given distribution of marks mean is 35.16 and its standard deviation is 19.76. The coefficient of variation is

#### Answer (Detailed Solution Below)

#### Statistics Question 3 Detailed Solution

**Concept**:

Coefficient of variation = \(\frac{SD}{Mean}\times 100\)

**Solution:**

Given that, the distribution of marks mean is 35.16 and its standard deviation is 19.76

Then Coefficient of variation = \(\frac{SD}{Mean}\times 100\) = \(\frac{19.76}{35.16}\times 100\)

∴ The correct option is (5)

#### Statistics Question 4:

Consider the following distribution:

Marks obtained |
No. of students |

More than or equal to 0 | 63 |

More than or equal to 10 | 58 |

More than or equal to 20 | 55 |

More than or equal to 30 | 51 |

More than or equal to 40 | 48 |

More than or equal to 50 | 42 |

The frequency of the class (30 - 40) is:

#### Answer (Detailed Solution Below)

#### Statistics Question 4 Detailed Solution

**Concept:**

**Frequency of the class:** The class frequency is the number of times the items corresponding to a class interval repeat in the series.

**Calculations:**

The frequency of the class (30 - 40) is ( ≥ 30 ) - ( ≥ 40 ) = 51 - 48

= 3

∴ The frequency of the class (30 - 40) is 3

**Hence,** the correct answer is option **2)**

#### Statistics Question 5:

Standard Deviation of data -1, -2, -3, -4, -5, -6 , -7 is

#### Answer (Detailed Solution Below)

#### Statistics Question 5 Detailed Solution

Given data -1, -2, -3, -4, -5, -6, -7

Arithmetic mean AM = \(\bar X\) = (∑X)/n

= (-1 - 2 - 3 - 4 - 5 - 6 - 7)/7

= -[1 + 2 + 3 + 4 + 5 + 6 + 7]/7

= -28/7

= -4

Standard Deviation SD = \(\sigma = \sqrt {\frac{{\sum {X_i}^2}}{n} - {{\left( {\bar X} \right)}^2}} \)

∑X_{i}^{2} = (-1)^{2} + (-2)^{2} + (-3)^{2} + (-4)^{2} + (-5)^{2} + ( - 6)^{2} + (-7)^{2}

= 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 7^{2}

= 7(7 + 1)(2×7 + 1)/6

= (7 × 8 × 15)/6

= 140

SD = σ = √(140/7 - (-4)^{2})

= √(20 - 16)

= √4 = 2## Top Statistics MCQ Objective Questions

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

#### Answer (Detailed Solution Below)

#### Statistics Question 6 Detailed Solution

Download Solution PDF**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**

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 |

#### Answer (Detailed Solution Below)

#### Statistics Question 7 Detailed Solution

Download Solution PDF__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

fi = 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**

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

#### Answer (Detailed Solution Below)

#### Statistics Question 8 Detailed Solution

Download Solution PDF__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.

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

#### Answer (Detailed Solution Below)

#### Statistics Question 9 Detailed Solution

Download Solution PDF**Formula Used****∶**

- σ
^{2}= ∑(x_{i}– 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 (x_{i} – x) remains same.

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

__Alternate Method__

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

#### Answer (Detailed Solution Below)

#### Statistics Question 10 Detailed Solution

Download Solution PDF__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

What is the standard deviation of the observations

\(-\sqrt{6}, -\sqrt{5},- \sqrt{4}, -1, 1, \sqrt{4}, \sqrt{5}, \sqrt{6} \ ?\)

#### Answer (Detailed Solution Below)

#### Statistics Question 11 Detailed Solution

Download Solution PDF**Concept:**

**Standard deviation:**

The standard deviation of the observation set \(\rm \{x_i,i=1,2,3,\cdots\}\) is given as follows:

\(\rm \sigma=\sqrt{\dfrac{\sum\left(x_i-\mu\right)^2}{N}}\)

Where \(\rm N=\mbox{size of the observation set}\) and \(\rm \mu=\mbox{mean of the observations}\) .

**Calculations:**

First, we will calculate the mean of the given observations.

\(\begin{align*} \mu &= \dfrac{-\sqrt6-\sqrt5-\sqrt4-1+1+\sqrt4+\sqrt5+\sqrt6}{8}= 0 \end{align*}\)

Therefore, the numerator inside the square root term of the standard deviation formula will simply be equal to \(\rm (x_i-\mu)^2=x_i^2\) .

Now we observe that \(\rm N=8\) .

Therefore, the standard deviation is given as follows:

\(\begin{align*} \sigma &= \sqrt{\dfrac{\left(-\sqrt6\right)^2+\left(-\sqrt5\right)^2+\left(-\sqrt4\right)^2+\left(-1\right)^2+\left(1\right)^2+\left(\sqrt4\right)^2+\left(\sqrt5\right)^2+\left(\sqrt6\right)^2}{8}}\\ &= \sqrt{\dfrac{32}{8}}\\ &= \sqrt4\\ &= 2 \end{align*}\)

Therefore, the standard deviation of the given observations is 2.

The mean of four numbers is 37. The mean of the smallest three of them is 34. If the range of the data is 15, what is the mean of the largest three?

#### Answer (Detailed Solution Below)

#### Statistics Question 12 Detailed Solution

Download Solution PDF**Calculation:**

Let the numbers be x_{1, }x_{2}, x_{3}, x_{4}.

The mean of four numbers x_{1, }x_{2}, x_{3}, x_{4} = 37

The sum of four numbers x_{1, }x_{2}, x_{3}, x_{4 }= 37 × 4 = 148.

The mean of the smallest three numbers x_{1, }x_{2}, x_{3} = 34

The sum of the smallest three numbers x_{1, }x_{2}, x_{3} = 34 × 3 = 102.

∴ The value of the largest number x_{4 }= 148 – 102 = 46.

The Range (Difference between largest and smallest value) x_{4} – x_{1} = 15.

∴ Smallest number x_{1} = 46 – 15 = 31.

Now,

The sum of x_{2}, x_{3 }= Total sum – (sum of smallest and largest number).

⇒ 148 – (46 + 31)

⇒ 148 – 77

⇒ 71

Now,

The mean of the Largest three numbers x_{2}, x

_{3}, x

_{4 }= (71 + 46)/3 = 117/3 = 39

If the mean of a frequency distribution is 100 and the coefficient of variation is 45%, then what is the value of the variance?

#### Answer (Detailed Solution Below)

#### Statistics Question 13 Detailed Solution

Download Solution PDF**Concept:**

Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)

Variance = (Standard Deviation)^{2}

**Calculation:**

Given coefficient of variation = 45% = 0.45

And mean = 100

As Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)

0.45 = \(\rm\text{Standard Deviation} \over100\)

Standard Deviation = 100 × 0.45

SD = 45

∴ Variance = 45^{2} = **2025**

The data given below shows the marks obtained by various students.

Marks |
Number of students |

10 – 12 |
6 |

12 – 14 |
8 |

14 – 16 |
5 |

16 – 18 |
7 |

18 - 20 |
4 |

What is the mean marks (Correct up to two decimal places) of given data?

#### Answer (Detailed Solution Below)

#### Statistics Question 14 Detailed Solution

Download Solution PDF\(\bar x\left( {mean} \right)\; = \;\frac{{\sum fx}}{n}\)

⇒ n = total frequency

\(\sum fx = Sum\;of\;the\;product\;of\;mid - interval\;values\;and\;their\;corresponding\;frequency\;\;\)

Mid value of 10 – 12 = (10 + 12)/2 = 11

Mid value of 12 – 14 = (12 + 14 )/2 = 13

Mid value of 14 – 16 = (14 + 16 )/2 = 15

Mid value of 16 – 18 = (16 + 18 )/2 = 17

Mid value of 18 – 20 = (18 + 20 )/2 = 19

\(\Rightarrow \;Mean\; = \;\frac{{11\; \times \;6\; + \;13\; \times \;8\; + \;15\; \times \;5\; + \;17\; \times \;7\; + \;19\; \times \;4}}{{6\; + \;8\; + \;5\; + \;7\; + \;4}} = \;\frac{{440}}{{30}}\)

⇒ Mean = 14.67

∴The mean marks of the given data are 14.67

If the mode of the following data is 7, then the value of k in the data set 3, 8, 6, 7, 1, 6, 10, 6, 7, 2k + 5, 9, 7, and 13 is:

#### Answer (Detailed Solution Below)

#### Statistics Question 15 Detailed Solution

Download Solution PDF__Concept:__

Mode is the value that occurs most often in the data set of values.

__Calculation:__

Given data values are 3, 8, 6, 7, 1, 6, 10, 6, 7, 2k + 5, 9, 7, and 13

In the above data set, values 6, and 7 have occurred more times i.e., 3 times

But given that mode is 7.

So, 7 should occur more times than 6.

Hence the variable 2k + 5 must be 7

⇒ 2k + 5 = 7

⇒ 2k = 2

∴ k = 1