# Class 10 Maths Chapter 14 Statistics MCQs

**Class 10 Maths Chapter 14 MCQs (Statistics)** are provided here, online. Students can practice these objective type questions, which are prepared as per the CBSE syllabus (2021 – 2022) and NCERT curriculum. These multiple-choice questions will help students to score good marks in board exams. All the multiple choice questions are presented here, chapter-wise, with their answers and detailed explanations. Also, click here to get chapter-wise MCQs for class 10 Maths.

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## Class 10 Maths MCQs for Statistics

**1. If x _{1}, x_{2}, x_{3},….., x_{n} are the observations of a given data. Then the mean of the observations will be:**

(a) Sum of observations/Total number of observations

(b) Total number of observations/Sum of observations

(c) Sum of observations +Total number of observations

(d) None of the above

Answer: **(a) Sum of observations/Total number of observations**

Explanation: The mean or average of observations will be equal to the ratio of sum of observations and total number of observations.

x_{mean}=x_{1}+x_{2}+x_{3}+…..+x_{n}/n

**2. If the mean of frequency distribution is 7.5 and ∑f _{i} x_{i} = 120 + 3k, ∑f_{i} = 30, then k is equal to:**

(a) 40

(b) 35

(c) 50

(d) 45

Answer: **(b) 35**

Explanation: As per the given question,

X_{mean }= ∑f_{i} x_{i} /∑f_{i}

7.5 = (120+3k)/30

225 = 120+3k

3k = 225-120

3k= 105

k=35

**3. The mode and mean is given by 7 and 8, respectively. Then the median is:**

(a) 1/13

(b) 13/3

(c) 23/3

(d) 33

Answer: **(c) 23/3**

Explanation: Using Empirical formula,

Mode = 3Median – 2 Mean

3Median = Mode + 2Mean

Median = (Mode + 2Mean)/3

Median = [7 + 2(8)]/3 = (7 + 16)/3 = 23/3

**4. The mean of the data: 4, 10, 5, 9, 12 is;**

(a) 8

(b) 10

(c) 9

(d) 15

Answer: **(a) 8**

Explanation: mean = (4 + 10 + 5 + 9 + 12)/5 = 40/5 = 8

**5. The median of the data 13, 15, 16, 17, 19, 20 is:**

(a) 30/2

(b) 31/2

(c) 33/2

(d) 35/2

Answer: **(c) 33/2**

Explanation: For the given data, there are two middle terms, 16 and 17.

Hence, median = (16 + 17)/2 = 33/2

**6. If the mean of first n natural numbers is 3n/5, then the value of n is:**

(a) 3

(b) 4

(c) 5

(d) 6

Answer: **(c) 5**

Explanation: Sum of natural numbers = n(n + 1)/2

Given, mean = 3n/5

Mean = sum of natural numbers/n

3n/5 = n(n + 1)/2n

3n/5 = (n + 1)/2

6n = 5n + 5

n = 5

**7. If AM of a, a+3, a+6, a+9 and a+12 is 10, then a is equal to;**

(a) 1

(b) 2

(c) 3

(d) 4

Answer: **(d) 4**

Explanation: Mean of AM = 10

(a + a + 3 + a + 6 + a + 9 + a + 12)/5 = 10

5a + 30 = 50

5a = 20

a = 4

**8. The class interval of a given observation is 10 to 15, then the class mark for this interval will be:**

(a) 11.5

(b) 12.5

(c) 12

(d) 14

Answer: **(b) 12.5**

Explanation: Class mark = (Upper limit + Lower limit)/2

= (15 + 10)/2

= 25/2

= 12.5

**9. If the sum of frequencies is 24, then the value of x in the observation: x, 5,6,1,2, will be;**

(a) 4

(b) 6

(c) 8

(d) 10

Answer: **(d) 10**

Explanation:

Given,

∑f_{i} = 24

∑f_{i} = x + 5 + 6 + 1 + 2 = 14 + x

24 = 14 + x

x = 24 – 14 = 10

**10. The mean of following distribution is:**

x_{i} |
11 | 14 | 17 | 20 |

f_{i} |
3 | 6 | 8 | 7 |

(a) 15.6

(b) 17

(c) 14.8

(d) 16.4

Answer: **(d) 16.4**

Explanation:

x_{i} |
f_{i} |
f_{i}x_{i} |

11 | 3 | 33 |

14 | 6 | 84 |

17 | 8 | 136 |

20 | 7 | 140 |

∑f_{i} = 24 |
∑f_{i} x_{i }= 393 |

x_{mean} = ∑f_{i} x_{i}/∑f_{i} = 393/24 = 16.4

**11. Construction of a cumulative frequency table is useful in determining the**

(a) mean

(b) median

(c) mode

(d) all the above three measures

Answer: **(b) median**

Construction of a cumulative frequency table is useful in determining the median.

**12. 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 the three above

Answer: (b) median

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 median.

**13. While computing mean of grouped data, we assume that the frequencies are**

(a) centred at the class marks of the classes

(b) evenly distributed over all the classes

(c) centred at the upper limits of the classes

(d) centred at the lower limits of the classes

Answer: **(a) centred at the class marks of the classes**

While computing the mean of grouped data, we assume that the frequencies are centred at the class marks of the classes.

**14. Consider the following frequency distribution of the heights of 60 students of a class:**

Height (in cm) |
150 – 155 | 155 – 160 | 160 – 165 | 165 – 170 | 170 – 175 | 175 – 180 |

Number of students |
15 | 13 | 10 | 8 | 9 | 5 |

**The sum of the lower limit of the modal class and upper limit of the median class is**

(a) 310

(b) 315

(c) 320

(c) 330

Answer: **(b) 315**

Explanation:

Height (in cm) | 150 – 155 | 155 – 160 | 160 – 165 | 165 – 170 | 170 – 175 | 175 – 180 |

Number of students | 15 | 13 | 10 | 8 | 9 | 5 |

Cumulative frequency | 15 | 28 | 38 | 46 | 55 | 60 |

N/2 = 60/2 = 30

Cumulative frequency nearer and greater than 30 is 38 which corresponds to the class interval 160 – 165.

Thus, median class = 160 – 165

Upper limit of median class = 165

Highest frequency = 15

So, the modal class = 150 – 155

Lower limit of modal class = 150

Therefore, the sum of the lower limit of the modal class and upper limit of the median class = 150 + 165 = 315

**15. 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) 17

(b) 17.5

(c) 18

(d) 18.5

Answer:** (b) 17.5**

Explanation:

Let us write the continuous classes for the given frequency distribution.

Class | -0.5 – 5.5 | 5.5 – 11.5 | 11.5 – 17.5 | 17.5 – 23.5 | 23.5 – 29.5 |

Frequency | 13 | 10 | 15 | 8 | 11 |

Cumulative frequency | 13 | 23 | 38 | 46 | 57 |

N/2 = 57/2 = 28.5

28.5 lies in between the interval 11.5 – 17.5.

Thus, the median class is 11.5 – 17.5.

Therefore, the upper limit of the median class is 17.5.

**16. The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below: **

Class |
13.8-14 | 14-14.2 | 14.2-14.4 | 14.4-14.6 | 14.6-14.8 | 14.8-15 |

Frequency |
2 | 4 | 5 | 71 | 48 | 20 |

**The number of athletes who completed the race in less then 14.6 seconds is**

(a) 11

(b) 71

(c) 82

(d) 130

Answer: **(c) 82**

Explanation:

The number of athletes who completed the race in less than 14.6 seconds = 2 + 4 + 5 + 71 = 82

**17. Consider the following distribution: **

Marks obtained |
Number 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 **

(a) 3

(b) 4

(c) 48

(d) 51

Answer: **(a) 3**

Explanation:

The given data can be written as:

Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |

cf | 63 | 58 | 55 | 51 | 48 | 42 |

f | 5 | 3 | 4 | 3 | 6 | 42 |

Therefore, the frequency of the class 30-40 is 3.

**18. The empirical relationship between the three measures of central tendency is**

(a) 3 Median = Mode + 2 Mean

(b) 2 Median = Mode + 2 Mean

(c) 3 Median = Mode + Mean

(d) 3 Median = Mode – 2 Mean

Answer: **(a) 3 Median = Mode + 2 Mean**

The empirical relationship between the three measures of central tendency is 3 Median = Mode + 2 Mean.

**19. The ________ of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.**

(a) Class mark

(b) Class height

(c) Average frequency

(d) Cumulative frequency

Answer:** (d) Cumulative frequency**

The cumulative frequency of a class is the frequency obtained by adding the frequencies

of all the classes preceding the given class.

**20. The method used to find the mean of a given data is(are):**

(a) direct method

(b) assumed mean method

(c) step deviation method

(d) all the above

Answer:** (d) all the above**

The mean for a given data can be calculated using either of the following methods.

Direct method

Assumed mean method

Step deviation method

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