Statistics — Multiple Choice Questions

The marks obtained by 17 students in a mathematics test (out of 100) are given below :

91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 66, 68, 48, 49

The range of data is

1. 46

2. 54

3. 90

4. 100

Answer

Range = Upper limit - Lower limit = 100 - 46 = 54.

Hence, Option 2 is the correct option.

The class mark of the class 90 - 120 is

1. 90

2. 105

3. 115

4. 120

Answer

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

= $\dfrac{90 + 120}{2} = \dfrac{210}{2}$

= 105.

Hence, Option 2 is the correct option.

In a frequency distribution, the mid-value of a class is 10 and the width of the class is 6. The lower limit of the class is

1. 6

2. 7

3. 8

4. 12

Answer

Lower limit of class = Mid value - $\dfrac{\text{Width}}{2}$

= 10 - $\dfrac{6}{2}$

= 10 - 3

= 7.

Hence, Option 2 is the correct option.

The width of each of 5 continuous classes in a frequency distribution is 5 and the lower limit of the lowest class is 10. The upper limit of the highest class is

1. 15

2. 25

3. 35

4. 40

Answer

Let x and y be the upper and lower class limit of frequency distribution.

Given,

Width of class = 5

∴ x - y = 5

⇒ x - 10 = 5

⇒ x = 5 + 10 = 15.

Upper class limit of highest class = No. of continuous class × Width + Lower class limit of lowest class

= 5 × 5 + 10

= 25 + 10

= 35.

Hence, Option 3 is the correct option.

The class marks of a frequency distribution are given as follows :

15, 20, 25, .........

The class corresponding to the class mark 20 is

1. 12.5 - 17.5

2. 17.5 - 22.5

3. 18.5 - 21.5

4. 19.5 - 20.5

Answer

As the class marks are 15, 20, 25, ......... which are at equal gaps, so the classes are of equal size.

∴ size of class = difference between two consecutive class marks
= 20 - 15 = 5.

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

∴ Lower limit of class corresponding to the class mark 20 = class mark - half of class size
= 20 - 2.5 = 17.5

∴ Upper limit of class corresponding to the class mark 20 = class mark + half of class size
= 20 + 2.5 = 22.5

∴ Class corresponding to the class mark 20 is 17.5 - 22.5

Hence, Option 2 is the correct option.

In the class intervals 10 - 20, 20 - 30, the number 20 is included in

1. 10 - 20

2. 20 - 30

3. both the intervals

4. none of these intervals

Answer

The number 20 will be included in 20 - 30 interval.

Hence, Option 2 is the correct option.

A grouped frequency distribution table with class intervals of equal size using 250 - 270 (270 not included in this interval) as one of the class intervals is constructed for the following data :

268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236.

The frequency of class 310 - 330 is

1. 4

2. 5

3. 6

4. 7

Answer

Elements from above data in the class 310 - 330 are :

310, 310, 320, 319, 318, 316.

Hence, the frequency of class 310 - 330 is 6.

Hence, Option 3 is the correct option.

The mean of x - 1, x + 1, x + 3 and x + 5 is

1. x + 1

2. x + 2

3. x + 3

4. x + 4

Answer

By formula,

$\text{Mean } = \dfrac{\text{Sum of observations}}{\text{No. of observations}}$

Sum of observations = (x - 1) + (x + 1) + (x + 3) + (x + 5) = 4x + 8.

$\text{Mean } = \dfrac{4x + 8}{4} \\[1em] = \dfrac{4(x + 2)}{4} \\[1em] = x + 2.$

Hence, Option 2 is the correct option.

The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is

1. 28

2. 30

3. 35

4. 38

Answer

By formula,

$\text{Mean } = \dfrac{\text{Sum of observations}}{\text{No. of observations}} \\[1em] 30 = \dfrac{\text{Sum of observations}}{5} \\[1em] \text{Sum of observations} = 150$

Let excluded number be x.

Given, on excluding x the mean is 28.

$\therefore \dfrac{150 - x}{4} = 28 \\[1em] \Rightarrow 150 - x = 112 \\[1em] \Rightarrow x = 150 - 112 \\[1em] \Rightarrow x = 38.$

Hence, Option 4 is the correct option.

If the mean of x₁, x₂ is 7.5, and the mean of x₁, x₂, x₃ is 8, then the value of x₃ is

1. 9

2. 8

3. 7.5

4. 6

Answer

By formula,

$\text{Mean} = \dfrac{\text{Sum of observations}}{\text{No. of observations}}$

Given, Mean of x₁, x₂ = 7.5

$\therefore 7.5 = \dfrac{x_1 + x_2}{2} \\[1em] \Rightarrow x_1 + x_2 = 15 ...........(i)$

Given, Mean of x₁, x₂ and x₃ = 8

$\therefore 8 = \dfrac{x_1 + x_2 + x_3}{3} \\[1em] \Rightarrow x_1 + x_2 + x_3 = 24 ...........(ii)$

Subtracting equation (i) from (ii), we get :

⇒ x₁ + x₂ + x₃ - (x₁ + x₂) = 24 - 15

⇒ x₁ - x₁ + x₂ - x₂ + x₃ = 9

⇒ x₃ = 9.

Hence, Option 1 is the correct option.

If each observation of the data is increased by 5, then their mean

1. remains the same

2. becomes 5 times the original mean

3. is decreased by 5

4. is increased by 5

Answer

Let x₁, x₂, x₃ be the observations.

So, their mean (M) = $\dfrac{x_1 + x_2 + x_3}{3}$

Since, each observations is increased by 5, so observations will be

x₁ + 5, x₂ + 5, x₃ + 5.

$\text{Mean (M}_1) = \dfrac{x_1 + 5 + x_2 + 5 + x_3 + 5}{3} \\[1em] = \dfrac{x_1 + x_2 + x_3 + 15}{3} \\[1em] = \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{15}{3} \\[1em] = \dfrac{x_1 + x_2 + x_3}{3} + 5.$

So, M₁ = M + 5.

Hence, Option 4 is the correct option.

The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be

1. 50.5

2. 51

3. 51.5

4. 52

Answer

By formula,

Mean = $\dfrac{\text{Sum of observations}}{\text{No. of observations}}$

Given,

Mean of 100 observations is 50.

$\therefore 50 = \dfrac{\text{Sum of observations}}{100} \\[1em] \text{Sum of observations} = 50 \times 100 = 5000.$

Since, observation which was 50 is replaced by 150.

So,

Sum of observation = 5000 - 50 + 150 = 5100.

New mean = $\dfrac{5100}{100}$ = 51.

Hence, Option 2 is the correct option.

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are respectively :

1. upper limits of the classes

2. lower limits of the classes

3. class marks of the classes

4. upper limits of preceding classes

Answer

We know, class marks are the mean of lower and upper limit of the class intervals.

Hence, Option 3 is the correct option.

Median of the numbers 4, 4, 5, 7, 6, 7, 7, 3, 12 is

1. 4

2. 5

3. 6

4. 7

Answer

Arranging given data in ascending order :

3, 4, 4, 5, 6, 7, 7, 7, 12.

No. of observation (n) = 9.

Here, n is odd.

By formula,

$\text{Median} = \dfrac{(n + 1)}{2} \text{th observation} \\[1em] = \dfrac{9 + 1}{2} \text{th observation} \\[1em] = \dfrac{10}{2} \text{th observation} \\[1em] = 5 \text{th observation} \\[1em] = 6.$

Hence, Option 3 is the correct option.

The median of the data

78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is

1. 45

2. 49.5

3. 54

4. 56

Answer

Arranging given data in ascending order :

22, 34, 39, 45, 54, 54, 56, 68, 78, 84.

No. of observation (n) = 10.

Here, n is even.

By formula,

$\text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \Big(\dfrac{n}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2} \text{th observation} + \Big(\dfrac{10}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \dfrac{\text{5th observation + 6th observation}}{2} \\[1em] = \dfrac{54 + 54}{2} \\[1em] = \dfrac{108}{2} \\[1em] = 54.$

Hence, Option 3 is the correct option.

In a data, 10 numbers are arranged in ascending order. If the 8th entry is increased by 6, then the median increases by

1. 0

2. 2

3. 3

4. 6

Answer

Since, number of observations (n) = 10, which is even.

By formula,

$\text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \Big(\dfrac{n}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2} \text{th observation} + \Big(\dfrac{10}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \dfrac{\text{5th observation + 6th observation}}{2}$

Since, 5th and 6th observation is not changed at all, so there is no change in median.

Hence, Option 1 is the correct option.

Consider the following two statements.

Statement 1: A histogram consists of a set of adjacent rectangles whose bases are equal to class size, and heights are equal to class frequencies.

Statement 2: In a bar graph, the breadth of a rectangle has no significance, whereas in a histogram, the breadth of a rectangle is meaningful and it represents the class size.

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

In a histogram, there are no gaps between the bars, indicating the continuous nature of the data.

The width of each bar (its base) represents the class interval or class size.

The height of each bar represents the frequency (or count) of observations falling within that specific class interval.

∴ Statement 1 is true.

In a bar graph, the bars have arbitrary width and does not represent any numerical quantity.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is correct option.