## Statistics Class 10 MCQ Questions with Answers

Question 1.

## In the formula \(\bar{x}=a+\frac{\sum x_{i} d_{i}}{\sum f_{i}}\) for finding thc mean of grouped data d;s are the deviations from a of:

(A) lower limits of the classes

(B) upper limits of the classes

(C) mid-points of the classes

(D) frequencies of the class marks

Answer:

(B) upper limits of the classes

Explanation:

In the given formula, a is assumed mean from class marks (aq) and d_{i} = x_{1} – a Therefore, d{ is the deviation of class mark (mid-value) from the assumed mean ‘a’.

Question 2.

## While computing mean of grouped data, we assume 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

Answer:

(B) centred at the class marks of the classes

Explanation:

In grouping the data from ungrouped data, all the observations between lower and upper limits of class marks are taken in one group then mid-value or class mark is taken for further calculation. Therefore, frequencies or observations must be centred at the class marks of the classes.

Question 3.

## If x_{i}‘s are the mid-points of the class intervals of grouped data f_{i} s are the corresponding frequencies

and \( \bar{x}\) is the mean, then \( \sum\left(f_{i} x_{i}-\bar{x}\right)\) is equal to :

(A) 0

(B) -1

(C) 1

(D) 2

Answer:

(A) 0

From equation (i) and (ii), we have

Question 4.

## In the formula \( \bar{x}\) = a + h \( \frac{\Sigma f_{i} l l_{i}}{\sum f_{i}}\) , for finding the mean of grouped frequency distribution

(A) \(\frac{x_{i}+a}{h}\)

(B) h (x^{1} – a)

(C) \(\frac{x_{i}-a}{i}\)

(D) \( \frac{a-x_{i}}{h}\)

Answer:

(C) \(\frac{x_{i}-a}{i}\)

Explanation:

\( \bar{x}\) = a + h\(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

u_{1} = \(\frac{x_{i}-a}{h}\)

Question 5.

## The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its :

(A) mean

(B) median

(C) mode

(D) All of these

Answer:

(B) median

Explanation:

The point of intersection of the less than type and of the more than type cumulative frequency curves gives the median on abscissa as on x-axis we take the upper or lower limits, respectively and on y-axis we take cumulative frequency.

Question 6.

## 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 median class and modal class is :

(A) 15

(B) 25

(C) 30

(D) 35

Answer:

(B) 25

Explanation:

Class | Frequency | Cumulative frequency |

0-5 | 10 | 10 |

5 – 10 | 15 | 25 |

10 – 15 | 12 | 37 |

15 – 20 | 20 | 57 |

20 – 25 | 9 | 66 |

The modal class is the class having the maximum frrquency 20 belongs to class (15 – 20).

Hrer, n = 66

So, \( \frac{n}{2}\) = \(\frac{66}{2}\) = 33

33 lies in the class 10 – 15.

Therefore, 10 – 15 is the median class.

So, sum of lower limits of (15 – 20) and (10 -15) is (15 – 10) = 25

Question 7.

## 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) 7

(B) 17.5

(C) 18

(D) 18.5

Answer:

(B) 17.5

Explanation:

Class | Frequency | Cumulative frequency |

0.5-5.5 | 13 | 13 |

5.5-11.5 | 10 | 23 |

11.5-17.5 | 15 | 38 |

17.5-23.5 | 8 | 46 |

23.5-29.5 | 11 | 57 |

The median of 57 (odd) observations = \(\frac{(57+1)}{2}\) = \(\frac{58}{2}\) = 29th term

29th term lies in class 11.5 – 17.5.

So, upper, limit is 17. 5

Question 8.

## For the following distribution :

Marks | Number of students |

Below 10 | 3 |

Below 20 | 12 |

Below 30 | 27 |

Below 40 | 57 |

Below 50 | 75 |

Below 60 | 80 |

## the modal class is :

(A) 10-20

(B) 20-30

(C) 30-10

(D) 50-60 .

Answer:

(C) 30-10

Explanation:

Marks | Number of students | f_{i} |

0-10 | 3 – 0=3 | 3 |

10-20 | 12 – 3=9 | 9 |

20-30 | 27 – 12=15 | 15 |

30-40 | 57 – 27=30 | 30 |

40-50 | 75 – 57=18 | 18 |

50-60 | 80 – 75=5 | 5 |

Modal class has maximum frequency (30) in class 30 – 40.

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of of A Reason (R). Mark the correct choice as.

(A) Both A and R are true and R is the correct explanation

(B) Both A and R are true but R is NOT the correct explanation of A

(C) A is true but R is false

(D) A is false and R is True

Question 1.

## Assertion (A): Consider the following data:

Class | Frequency |

65-85 | 4 |

85 -105 | 5 |

105 -125 | 13 |

125 -145 | 20 |

145 -165 | 14 |

165-185 | 7 |

185 – 205 | 4 |

## The difference of the upper limit of the median class and the lower limit of the modal class is 20.

Reason (R): The median class and modal class of grouped data always be different.

Answer:

(C) A is true but R is false

Explanation:

In case of assertion :

Class | Frequency | Cumulative frequency |

65-85 | 4 | 4 |

85-105 | 5 | 9 |

105-125 | 13 | 22 |

125-145 | 20 | 42 |

145-165 | 14 | 56 |

165-185 | 7 | 63 |

185-205 | 4 | 67 |

Hence, n= 67 (odd)

so, Median = \(\frac{67+1}{2}\) = 34

34 lies in class 125 – 145 and upper limit is 145.

Now, the maximum frequency is 20 and it lies in class 125 – 145 (modal class).

Lower limit of modal class = 125

Hence, the reuired difference 145 – 125 = 20.

∴ Assertion is correct.

In case of reason :

The median andmodal class may be same. If modal class is median class which is not always possible as the number of trequencies may be maximum in any class.

So, given dtatement is not true.

∴ Reason is incorrect.

Hence, assertion is correct but reason is incorrect.

Question 2.

## Assertion (A): Consider the data:

Class | 4-7 | 8-11 | 12-15 | 16-19 |

Frequency | 5 | 4 | 9 | 10 |

## The mean of the above data is 12.93.

Reason (R): The following table gives the number of pages written by Sarika for completing her own book for 30 days:

Class | 16-18 | 19-21 | 22-24 | 25-27 | 28-30 |

Frequency | 1 | 3 | 4 | 9 | 13 |

## The mean of the above date is 26

Answer:

(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:

In case of assertion :

Class marks of these classes are same ,so no need to convert given date continuous.

Class | Class marks(x_{i}) |
marks(x_{i}) d_{i} = (x_{i} – a) |
Frequency (f_{i}) |
f_{i}d_{i} |

4-7 | 5.5 | -4 | 5 | -20 |

8-11 | 9.5 = a | 0 | 4 | 0 |

12-15 | 13.5 | +4 | 9 | 36 |

16-19 | 17.5 | +8 | 10 | 80 |

Σf_{i} = 28 |
Σf_{i}d_{i} = 96 |

a = Assumed mean, di = Deviation fro mean

Mean, \( \bar{x}\) = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\) = 9.5 + \(\frac{96}{28}\) = 9.5 + 3.43

∴\( \bar{x}\) = 12.93

∴ Hence, the mean = 12.93

∴Assertion is correct.

In case of reason:

No need to change the class intervals of continuous intervals as class marks of continuous and discontinuous classes are same. di is deviation from assumed mean.

Class interval | Midvalue | d_{i} =(x_{i} – a) |
Number of days | f_{i}d_{i} |

16-18 | 17 | -6 | 1 | -6 |

19-21 | 20 | -3 | 3 | -9 |

22-24 | a = 23 | 0 | 4 | 0 |

25-27 | 26 | 3 | 9 | 27 |

28-30 | 29 | 6 | 13 | 78 |

Σf_{i} = 30 |
Σf_{i}d_{i} = 90 |

a = Assumed mean = 23

Mean, \( \bar{x}\) = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\) = 23 + \(\frac{90}{30}\) = 23 + 3 = 26

∴\( \bar{x}\) = 26

Hence, the mean of pages written per day is 26.

∴Reason is correct.

Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 3.

## Assertion (A): If the median of a series exceeds the mean by 3, then mode exceeds mean by 10.

Reason (R): If made = 12.4 and mean = 10.5, then the median is 11.13.

Answer:

(D) A is false and R is True

Explanation:

In case of assertion:

Given, median = mean + 3

Since, Mode = 3 Median – 2 Mean

= 3 (Mean + 3) – 2 Mean

⇒ Mode = Mean + 9

Hence, mode exceed mean by 9.

∴ Assertion is correct.

In case of reason:

Median = \(\frac{1}{3}\) Mode + \(\frac{2}{3}\) Mean

= \(\frac{1}{3}\) (12 . 4) + \(\frac{2}{3}\) (10.5)

\(\frac{12.4}{3}\) + \(\frac{21}{3}\)

= \(\frac{12.4+21}{3}\) = \(\frac{33.4}{3}\)

= \(\frac{33.4}{3}\) = 11.13

∴ Reason is correct.

Hence, Assertion is incorrect but reason is correct.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.

I. Read the following text and answer the following question on the basis of the same:

COVID-19 Pandemic The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans.

The following tables shows the age distribution of case admitted during a day in two different hospitals

Question 1.

## The averge age for which maximum cases occurred is

(A) 32.24

(B) 34.36

(C) 36.82

(D) 42.24

Answer.

(C) 36.82

Explanation:

Since, highest frequency is 23. So, modal class is 35 – 45

Now, Mode = l +\(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}} \) × h

Here, l =35, h = 10, f_{i} = 23, f0 = 21, f2 = 14,

⇒ Mode = 35 + \(\frac{23-21}{46-21-14}\) × 10

= 35 + \(\frac{2}{11}\) × 10

= 35 + \(\frac{20}{11}\)

= 35 + 1.81

= h 36.818 = 36.82

Question 2.

## The upper limit of modal class is

(A) 15

(B) 25

(C) 35

(D) 45

Answer:

(D) 45

Question 3.

## The mean of the given data is

(A) 26.2

(B) 32.4

(C) 33.5

(D) 35.4

Answer:

(D) 35.4

Explanation:

Age (in years) | Class marks – (x_{i}) |
frequency (f_{i}) |
Deviation d_{i}= (x_{i} – a) |
f_{i}d_{i} |

5-15 | 10 | 6 | -3 | -15 |

15-25 | 20 | 11 | 6 | 66 |

25-35 | 30 | 21 | 16 | 336 |

35-45 | 40 | 23 | 26 | 598 |

45-55 | 50 | 14 → a | 1 | 18 |

55-65 | 60 | 5 | 46 | 46 |

Σf_{i} = n = 80 |
Σf_{i}d_{i} = 1,716 |

Now, MEan (\( \bar{x}\)) a + \( \bar{x}\)

= 14 + \( \bar{x}\)

= 14 + 21.45

= 35. 45

Refer to table 2

Question 4.

## The mode of the given data is

(A) 41.4

(B) 48.2

(C) 55.3

(D) 64.6

Answer:

(A) 41.4

Question 5.

## The median of the given data is

(A) 32.7

(B) 40.2

(C) 42.3

(D) 48.6

Answer:

(B) 40.2

Explanation:

Age (in years) | frequency (f_{i}) (No. of cases) |
C.f. |

5-15 | 8 | 8 |

15-25 | 16 | 24 |

25-35 | 10 | 34 |

35-45 | 42 (frequency) | 76 (Nearest to \(\frac{n}{2}\))) |

45-55 | 24 | 100 |

55-65 | 12 | 112 |

Σf_{i} = n = 12 |

Now, \(\frac{n}{2}\) = \(\frac{112}{2}\) =56

l = 35 (lower mlimit of median class)

Cf = 34 (Preceding to median class)

Here, Madian = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h

= 35 + \(\left(\frac{56-34}{42}\right)\) × 10

= 35 + \(\left(\frac{22}{42}\right)\) × 10

= 35 + \(\left(\frac{11}{21}\right)\) × 10

= 35 + \(\frac{110}{21}\)

= 40.25

II. Read the following text and answer the following question on the basis of the same:

Electricity Energy ConsumptionElectricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption).

Tariff : LT – Residential | Bill Number : 384756′ |

Type of Supply : Single Passes | Connected lead : 3 kW |

Mater Reading : 31-11-13 Date | Mater Reading : 65700 |

Previous Reading : 31-10-13 Date | Previous Mater : 65500 Reading |

Units consumed : 289 |

A survey is conducted for 56 families of a Colony A. The following tables gives the weekly consumption of electricity of these families.

Weekly consumption (in units) | 0-10 | 10-20 | 20-30 | 30 – 40 | 40-50 | 50-60 |

No. of families | 16 | 12 | 18 | 6 | 4 | 0 |

Weekly consumption (in units) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

No. of families | 0 | 5 | 10 | 20 | 40 | 5 |

Refer to data received from Colony A

Question 1.

## The median weekly consumption is

(A) 12 units

(B) 16 units

(C) 20 units

(D) None of these

Answer:

(C) 20 units

Explanation:

Weekly consumption (in units) | frequency (f_{i}) (No. of families) |
C.f. |

0 – 10 | 8 | 16 |

10-20(Medianclass) | 12 (frequency) | 28(Nearest to \(\frac{n}{2}\)) |

20-30 | 18 | 46 |

30-40 | 6 | 52 |

40-50 | 4 | 56 |

50-60 | 0 | 56 |

Σf_{i} = n = 56 |

Now, \(\frac{n}{2}\) = \(\frac{56}{2}\) = 28

l = 10, Cf = 16 ,f = 12, h = 10

Here,

Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h

= 10 + \(\left(\frac{28-16}{12}\right)\) × 10

= 10 + \(\left(\frac{12}{12}\right)\) × 10

= 10 + 10

= 20

Hence, median weekiy consumption = 20 units.

Question 2.

## The mean weekly consumption is

(A) 19.64 units

(B) 22.5 units

(C) 26 units

(D) None of these

Answer:

(A) 19.64 units

Question 3.

## The modal class of the above data is I

(A) 0 – 10

(B) 10 – 20

(C) 20 – 30

(D) 30-40

Answer:

(C) 20 – 30

Refer to data received from Colony B

Question 4.

## The modal weekly consumption is

(A) 38.2 units

(B) 43.6 units

(C) 26 units

(D) 32 units

Answer:

(B) 43.6 units

Question 5.

## The mean weekly consumption is

(A) 15.65 units

(B) 32.8 units

(C) 38.75 units

(D) 48 units

Answer:

(C) 38.75 units

III. Read the following text and answer the following question on the basis of the same:

The weights (in kg) of 50 wrestlers are recorded in the following table :

Question 1

## What is the upper limit of modal class.

(A) 120

(B) 130

(C) 100

(D) 150

Answer:

(B) 130

Explanation:

Modal Class = 120 – 130

Upper limit = 130

Question 2.

## What is the mode class frequency of the given data

(A) 21

(B) 50

(C) 25

(D) 80

Answer:

(A) 21

Explanation:

Mode class frequency of the given data is 21.

Question 3.

## How many wrestlers weights have more than 120 kg weight?

(A) 32

(B) 50

(C) 16

(D) 21

Answer:

(A) 32

Explanation:

No. of wrestlers with more than 120 kg weight = 21 + 8 + 3 = 32

Question 4.

## What is the class mark for class 130 – 140?

(A) 120

(B) 130

(C) 135

(d) 150

Answer:

(C) 135

Explanation:

For class mark of 130 -140,

= \(\frac{130+140}{2}\)

= \(\frac{270}{2}\) = 135

Question 5.

## Which method is more suitable to find the mean of the above data ?

(A) Direct method

(B) Assumed mean method

(C) Step-Deviation method

(D) None of these

Answer:

(C) Step-Deviation method

IV Read the following text and answer the following question on the basis of the same:

The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows.

Question 1.

## What is the modal class of the given data?

(A) 85 – 100

(B) 100 – 115

(C) 115 – 130

(D) 130 – 145

Answer:

(A) 85 – 100

Explanation:

Modal class is the class with highest frequency i.e., 85 – 100

Question 2.

## What is the value of class interval for the given data set?

(A) 10

(B) 15

(C) 5

(D) 20

Answer:

(B) 15

Explanation:

The value of class interval = 100 – 85 = 15

again =115-100 = 15

and = 130-115 = 145 – 130

Question 3.

## What is the median class of the given data?

(A) 85-100

(B) 100-115

(C) 115 – 130

(D) 130 – 145

Answer:

(B) 100-115

Explanation:

N = Number of observations = 33

Median of 33 observations = 17.5^{th} observation, which in class 100 – 115

Question 4.

## What is the median of bowling speed?

(A) 109.17 km/hr (Approx)

(B) 109.71 km/hr (Approx)

(C) 107.17 km/hr (Approx)

(D) 109.19 km/hr (Approx)

Answer:

(A) 109.17 km/hr (Approx)Explanation:

Median = l + \(\frac{\left(\frac{n}{2}-c . f\right)}{f}\) × h

l = 100, f = 9, c.f. = , h = 100 – 85 = 15

Median = l + \(\frac{\left(\frac{n}{2}-c . f\right)}{f}\) × h

= 100 + \(\frac{\left(\frac{33}{2}-11\right)}{9}\) × 15

= \(\frac{100+(16.5-11)}{9 \times 15}\)

= 100 + \(\frac{5.5 \times 15}{9}\)

= 100 + \(\frac{82.5}{9}\)

= 100 + 9.166

= 109.17 km/h (Approx)

Hence, the median bowling sppeed is 109. 17 km/h (Approx)

Question 5.

## What is the sum of lower limit of modal class and upper limit of median class?

(A) 100

(B) 200

(C) 300

(D) 400

Answer:

(B) 200

Explanation:

Lower limit of modal class = 85 and upper limit of median class =115 sum = 85 + 115 200V. Read the following text and answer the following question on the basis of the same:

100 m RACE A stopwatch was used to find the time that it took a group of students to run 100 m.

Question 1.

## Estimate the mean time taken by a student to finish the race.

(A) 54

(B) 63

(C) 43

(D) 50

Answer:

(C) 43

Explanation:

Time (in sec) | x | f | cf | fx |

0 – 20 | 10 | 8 | 8 | 80 |

20 – 40 | 30 | 10 | 18 | 300 |

40 – 60 | 50 | 13 | 31 | 650 |

60 – 80 | 70 | 6 | 37 | 420 |

80 – 100 | 90 | 3 | 40 | 270 |

Total | 40 | 1720 |

Mean = \(\frac{1720}{40}\) = 43

Question 2.

## What will be the upper limit of the modal class?

(A) 20

(B) 40

(C) 60

(D) 80

Answer:

(C) 60

Explanation:

Modal class = 40-60

Upper limit = 60

Question 3.

## The construction of cumulative frequency table is useful in determining the

(A) Mean

(B) Median

(C) Mode

(D) All of the above

Answer:

(B) Median

Explanation:

The construction of c.f. table is useful in determining the median.

Question 4.

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

(A) 60

(B) 100

(C) 80

(D) 140

Answer:

(C) 80

Explanation:

Median class = 40-60

Modal class = 40-60

Therefore, the sum of the lower limits of median and modal class = 40 + 40 = 80

Question 5.

## How many students finished the race within 1 minute?

(A) 18

(B) 37

(C) 31

(D) 8

Answer:

(C) 31

Explanation:

Number of students who finished the race within 1 minute = 8 + 10 + 13 = 31