Statistics Class 10 MCQ with Answers – Maths Class 10 MCQ Online Test are covered in this Article. Statistics Class 10 MCQ Test contains 25 questions. Answers to MCQ on Statistics Class 10 are available after clicking on the answer. MCQ Questions for Class 10 with Answers have been made for Class 10 students to help check the concept you have learnt from detailed classroom sessions and application of your knowledge.
| Board | CBSE |
| Textbook | Maths (NCERT) |
| Class | Class 10 |
| Chapter | Chapter 14 Statistics |
| Category | MCQ Questions for Class 10 Maths with Answers |
Statistics Class 10 MCQ with Answers
1.Mean is
(a) (sum of all observations)/(total number of observations)
(b) (sum of all observations + 1)/(total number of observations)
(c) (sum of all observations-total number of observations)/(total number of observations)
(d)None of these
Answer
Answer: (a) (sum of all observations)/(total number of observations)
2.Formula for mode is
(a) l + ((f1+f0)/2f1-f0-f2)×h
(b) l + ((f1-f0)/2f1+f0-f2)×h
(c) l + ((f1-f0)/2f1-f0-f2)×h
(d) l + ((f1+f0)/2f1+f0+f2)×h
Answer
Answer: (c) l + ((f1-f0)/2f1-f0-f2)×h
3. A class with maximum frequency is called
(a) mode
(b) modal class
(c) median class
(d) mean
Answer
Answer: (b) modal class
4.Which of the following is a measure of central tendency?
(a) Mean
(b) Mode
(c) Median
(d) All of these
Answer
Answer: (d) All of these
5. Mean of the data : 4, 5, 6, 1, 3, 7, 9 is
(a) 5
(b) 6
(c) 7
(d) 8
Answer
Answer: (a) 5
Explanation: Mean = (sum of all observations)/(total number of observations)
=(4+5+6+1+3+7+9)/7
= 35/7 = 5
6. The empirical relationship between the three measures of central tendency is :
(a) Mode = 3 Median – 2 Mean
(b) Mode – 3 Median = 2 Mean
(c) 2 Mean – Mode = 3 Median
(d) Mode = 3 Median + 2 Mean
Answer
Answer: (a) Mode = 3 Median – 2 Mean
Explanation: 3 Median = Mode + 2 Mean
Mode = 3 Median – 2 Mean
7. The mean and mode are 7 and 1 respectively, then median is
(a) 3
(b) 4
(c) 5
(d) 6
Answer
Answer: (c) 5
Explanation: 3 Median = Mode + 2 Mean
3 Median = 1 + 2(7)
3 Median = 1 + 14
Median = 15/3 = 5
Statistics Class 10 MCQ with Answers
8. If AM of x, x+1, x+3 and x+4 is 32, then value of x can be
(a) 20
(b) 30
(c) 35
(d) None of these
Answer
Answer: (b) 30
Explanation: (x+x+1+x+3+x+4)/4 = 32
4x + 8 = 128
4x = 120
x = 30
9. Consider the following distribution of marks of 20 students of a class. Find the mean of the marks of the students
| Marks obtained(xi) | 20 | 33 | 35 | 39 |
| Number of students (fi) | 10 | 5 | 4 | 1 |
(a) 27
(b) 27.2
(c) 27.3
(d) 27.4
Answer
Answer: (b) 27.2
Explanation:
| Marks obtained(xi) | Number of students (fi) | xi fi |
| 20 | 10 | 200 |
| 33 | 5 | 165 |
| 35 | 4 | 140 |
| 39 | 1 | 39 |
| ∑fi=20 | ∑xi fi = 544 |
Mean = (∑xi fi)/(∑fi) = 544/20 = 27.2
10. The method used for finding the mean of a given data is
(a) Direct method
(b) Assumed mean method
(c) Step deviation method
(d) All of these
Answer
Answer: (d) All of these
11.If the mean of the frequency distribution is 6 and ∑xi fi = 90, then ∑fi =
(a) 12
(b) 13
(c) 15
(d) None of these
Answer
Answer: (c) 15
Explanation: Mean = (∑xi fi)/(∑fi)
6 = 90/(∑fi)
∑ fi = 90 /6 = 15
12. The daily wages of 15 workers are as follows:
| 100 | 70 | 50 | 100 | 200 | 200 | 200 | 50 | 50 | 60 | 65 | 115 | 200 | 222 | 240 |
Find the mode of the data
(a) 100
(b) 200
(c) 222
(d) 240
Answer
Answer: (b) 200
Explanation:
| Daily wages (in Rs) | 50 | 60 | 65 | 70 | 100 | 115 | 200 | 222 | 240 |
| Number of workers | 3 | 1 | 1 | 1 | 2 | 1 | 4 | 1 | 1 |
200 has the maximum frequency. So, mode of this data is Rs. 200.
13. If the mean of first n natural numbers is 2n, then value of n is
(a) 1/3
(b) ½
(c) ¼
(d) 1
Answer
Answer: (a) 1/3
Explanation: We know that sum of first n natural numbers is (n(n+1))/2.
Mean = 2n
Mean = sum of natural numbers/n
2n = (n(n+1))/2n
4n = (n+1)
3n = 1
n = 1/3
14. The cumulative frequency of a class is the frequency obtained by
(a) by adding the frequencies of all the classes preceding the given class.
(b) by subtracting the frequencies of all the classes preceding the given class.
(c) by multiplying the frequencies of all the classes preceding the given class.
(d) by dividing the frequencies of all the classes preceding the given class.
Answer
Answer: (a) by adding the frequencies of all the classes preceding the given class.
Explanation: The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.
Statistics Class 10 MCQ with Answers
15. Median of grouped data can be obtained graphically
(a) True
(b) False
(c) Can’t say
(d) None of these
Answer
Answer: (a) True
Explanation: The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for this data.
16. Consider the following distribution:
| Marks | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 |
| Number of students | 2 | 5 | 6 | 1 | 1 |
Find the sum of the lower modal class and upper modal class of this data.
(a) 10
(b) 5
(c) 7
(d) 12
Answer
Answer: (d) 12
Explanation: The maximum class frequency is 6. The class corresponding to this frequency is 5-7.
So, 5-7 is modal class.
Lower limit of modal class is 5 and upper limit of modal class is 7.
Sum of lower and upper modal class = 5 + 7 = 12
17. Mode is
(a) Mean
(b) Maximum value
(c) Maximum frequent value
(d) Least frequent value
Answer
Answer: (c) Maximum frequent value
18. If 40 is removed from the data: 20, 25, 30, 40, 45 ,50, then median is decreased by
(a) 0
(b) 5
(c) 6
(d) All of these
Answer
Answer: (b) 5
Explanation: n = 6 which is even term.
The median will be the average of the (n/2)th and (n/2 + 1)th observations.
(n/2)th and (n/2 + 1)th observations are 30 and 40 respectively.
Median = (30+40)/2= 70/2 = 35
When 40 is removed from the data. The data becomes 20, 25, 30, 45, 50
n = 5
So, now the median is 3rd term which is 30.
Median is decreased by 35-30 = 5.
19. For one term, absentee record of 20 employees of a company is given below. If mean is 15, then the missing frequencies x and y are
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| frequency | x | 4 | y | 2 | 1 |
Find value of x and y
(a) X =3 and y =10
(b) X = 10 and y =3
(c) X = 10 and y = 10
(d) X = 3 and y =3
Answer
Answer: (b) X = 10 and y =3
Explanation:
| Class interval | Frequency(fi) | Class mark (xi) | fi xi |
| 0-10 | x | 5 | 5x |
| 10-20 | 4 | 15 | 60 |
| 20-30 | y | 25 | 25y |
| 30-40 | 2 | 35 | 70 |
| 40-50 | 1 | 45 | 45 |
| 20 | 5x +25y+175 |
Mean = (∑xi fi)/(∑fi)
15 = (5x+25y+175)/20
300 = 5x + 25y + 175
5x + 25y = 300-175
5 (x+5y) = 125
x + 5y = 25 ………………………..(a)
And
x + y + 7 =20
x +y = 13 …………….. (b)
Subtracting (b) from (a)
X + 5y -x-y = 25-13
4y = 12
Y = 3
From (b)
X + 3 = 13
X = 10
20. Mean of first 10 natural number is
(a) 3.5
(b) 4.5
(c) 5.5
(d) 6.5
Answer
Answer: (c) 5.5
Explanation: Mean = (sum of all observations)/(total number of observations)
= (1+2+3+4+5+6+7+8+9+10)/10 =55/10 = 5.5
21.In a data, if l =20, f1= 10, f2 = 2, f0 = 3 and h = 15. Then mode of this data will be
(a) 20
(b) 23
(c) 27
(d) 28
Answer
Answer: (c) 27
Explanation: Mode = l + ((f1– f0)/(2f1-f0-f2))×h
Mode = 20 + ((10-3)/(2×10-3-2))×15
Mode = 20 + (7/(20-5)) ×15
Mode = 20 + (7/15) ×15
Mode = 20 + 7 = 27
22. The distribution given below gives the weight of 30 students of a class.
| Weight(in kg) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |
| Number of students | 11 | 5 | 7 | 6 | 1 |
Find the sum of upper limit of the modal class and upper limit of the median class is
(a) 110
(b) 115
(c) 120
(d) 125
Answer
Answer: (b) 115
Explanation:
| Weight (in kg) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |
| Number of students | 11 | 5 | 7 | 6 | 1 |
| Cumulative frequency | 11 | 16 | 23 | 29 | 30 |
n/ 2 = 30/2 = 15
Cumulative frequency nearer and greater than 15 is 16 which corresponds to the class interval 55-60.
Median class is 55-60.
Lower limit of the median class is 55 and upper limit of the median class is 60.
11 is the maximum frequency.
So, the modal class is 50 -55.
Lower limit of the modal class 50 and upper limit of the modal class is 55.
sum of upper limit of the modal class and upper limit of the median class is = 55 + 60
= 115.
23. A survey regarding the heights (in cm) of 15 workers of LIFE FOUNDATION was conducted and the following data was obtained:
| Height (in cm) | Number of workers |
| Less than 135 | 3 |
| Less than 140 | 4 |
| Less than 145 | 11 |
| Less than 150 | 15 |
The frequency of class 140-145 is
(a) 3
(b) 4
(c) 7
(d) 15
Answer
Answer: (c) 7
Explanation:
| Class interval | Frequency | Cumulative frequency |
| Below 135 | 3 | 3 |
| 135-140 | 1 | 4 |
| 140-145 | 7 | 11 |
| 145-150 | 4 | 15 |
The frequency of class 140-145 is 11.
24. The median of the data: 23, 25, 13, 15, 18, 20, 11, 12 is
(a) 15
(b) 16.5
(c) 15.5
(d) 16
Answer
Answer: (b) 16.5
Explanation: n = 8
Median is the average of 4th and 5th observation i.e., 15 and 18
Median = (15+18)/2 = 33/2 = 16.5
25.If mode and median of a data is 5 and 6 respectively, then mean is
(a) 6.5
(b) 3
(c) 2
(d) None of these
Answer
Answer: (a) 6.5
Explanation: We know that
3 Median = Mode + 2Mean
3 (6) = 5 + 2Mean
2Mean = 18 – 5
Mean =13/2 = 6.5
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Frequently Asked Questions on Statistics Class 10 MCQ with Answers
1. Are these MCQ on Statistics Class 10 are based on 2021-22 CBSE Syllabus?
Yes . There are 25 MCQ’s on this Chapter in this blog.
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