Explore Exams
Login Signup
Sign in Open App
All Exams  >   Class 10  >   Online MCQ Tests for Class 10  >   All Questions

All questions of Statistics for Class 10 Exam

The wickets taken by a bowler in 10 cricket matches are 2, 6, 4, 5, 0, 3, 1, 3, 2, 3. The mode of the data is
  • a)
    1
  • b)
    2
  • c)
    3
  • d)
    4
Correct answer is option 'C'. Can you explain this answer?

Ananya Das answered
Mode is the most frequently occurring observation. Since it's not a grouped data the frequency is the occurrence of only one observation , So the highest no. of frequency is 3 . So the modal age is 3.
Clear all your doubts with EduRev

Find out the mode of the following data:
  • a)
    450
  • b)
    400
  • c)
    600
  • d)
    478
Correct answer is option 'A'. Can you explain this answer?

Rohini Seth answered
The mode of a set of data values is the value that appears most often .So mode is the most occurrence of observation. 
Converting the distribution in continuous grouped and applying the formula
Mode= 
Modal class is 400-500
Where L=lower limit of the modal class =400
f1= modal class frequency=32
f2 = just after the modal class frequency=20
f0 = just previous the modal class frequency=20
C = Class size=100
Mode= 
=400+50=450

The mean of all the factors of 24 is
  • a)
    7.5
  • b)
    24
  • c)
    7
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

EduRev Class 10 answered
To find the mean of all the factors of 24:

- List the factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
- Sum the factors: (1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60).
- Count the factors: There are 8 factors.
- Calculate the mean: Sum of the factors/ number of the factors = 60/8 = 7.5
Thus, the mean of all the factors is 7.5. The correct answer is A: 7.5.

The marks obtained by 9 students in Mathematics are 59, 46, 30, 23, 27, 44, 52, 40 and 29. The median of the data is
  • a)
    30
  • b)
    29
  • c)
    35
  • d)
    40
Correct answer is option 'D'. Can you explain this answer?

Kiran Mehta answered
Median is the middle value of the observations. So Median cannot be obtained when the observations are not in order. So first we have to arrange in ascending order. 
59, 46, 30, 23, 27, 44, 52, 40 and 29
Gives- 23,27,29,30,40,44,46,52,59
Since the no. of observations is odd 

The measure of central tendency that can be obtained graphically is
  • a)
    mean
  • b)
    median
  • c)
    mode
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Smriti Tripathi answered
I think its Median option(b) 
The 
median
 is a 
simple
 measure of central tendency. To find the 
median
, we arrange the observations in order from smallest to largest value. If there is an odd number of observations, the 
median
 is the middle value. If there is an even number of observations, the 
median
 is the average of the two middle value

The mean of 1, 2, 3, 4 …n is given by
  • a)
  • b)
    n/2
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Krishna Iyer answered
⇒   We have the sequence, 1,2,3.......n
⇒   This is an AP, with the initial term a=1 and the common difference d=1.
∴   The sum of n terms of an AP is given by,


→   Arithmetic mean of n numbers a1​,a2​,a3​,a4​,...an​ is given by the formula

The mean of the first 10 natural numbers is
  • a)
    5
  • b)
    6
  • c)
    4.5
  • d)
    5.5
Correct answer is option 'D'. Can you explain this answer?

Pooja Shah answered
The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

Mean = Sum of Observations/Total No of Observations

Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 ÷ 10

Mean = 55/10

Mean = 5.5

What is the mode of the given data?​
  • a)
    24
  • b)
    21
  • c)
    22
  • d)
    25
Correct answer is option 'A'. Can you explain this answer?

Skyhigh Academy answered
Mode= 
Where L=lower limit of the modal class =20
f1= modal class frequency.=27
f2 = just after the modal class frequency.=21 
f0 = just previous the modal class frequency=23
C = Class size=10
Mode= 
 =20+4=24

The age of 18 students of a class is reported below. Their modal age is 10, 17, 14, 10, 11, 12, 12, 13, 17, 13, 14, 14, 15, 16, 17, 15, 17, 16​
  • a)
    22 years
  • b)
    17 years
  • c)
    14 years
  • d)
    16 years
Correct answer is option 'B'. Can you explain this answer?

Gaurav Kumar answered
Mode is the highest number of times an observation occurs, which means the number that appears most frequently in a set. So modal age is the age which occurs maximum number of times. It is evident that 17 occurs most of the times. So, modal age is 17 years.

The mean of five numbers is 28. If one of the numbers is excluded, the mean gets reduced by 2. The excluded number is
  • a)
    45
  • b)
    36
  • c)
    30
  • d)
    25
Correct answer is option 'B'. Can you explain this answer?

Vikas Kumar answered
Mean of 5 nos.= 28

Therefore, sum of the 5 nos. = 28 * 5 = 140 

The the no. exclude by 'x'.

New sum of 4 nos.= 140 - x

New mean = 28- 2 = 26

Therefore, (140 - x ) / 4 = 26

=> 140 - x = 104

=> x = 36

ANSWER= 36

In a data, if l = 40, h = 15, f1=7, f0=3, f2=6, then the mode is
  • a)
    52
  • b)
    62
  • c)
    72
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Vivek Rana answered
Mode=L+ 
Where L=lower limit of the modal class =40
f1= modal class frequency.=7
f2 = just after the modal class frequency.=6
f0 = just previous the modal class frequency=3
C = Class size=15
Mode=40+ * 15 = 40 + 12 = 52

The median of the given data is 46 and the total number of items is 230.

Using the formula Mode = 3 Median – 2 Mean, the mean will be
  • a)
    44
  • b)
    43
  • c)
    44.5
  • d)
    45.9
Correct answer is option 'D'. Can you explain this answer?

Ananya Das answered
To find Mode
Mode=L+ 
Where L=lower limit of the modal class  = 40
f1 = modal class frequency.= 65
f2 = just after the modal class frequency.=46 
f0 = just previous the modal class frequency=34
C = Class size=10
Mode = 
=46.2
Mode = 3 Median – 2 Mean
46.2 =3*46-2Mean
2Mean=138-46.2=91.8
Mean=91.8/2=45.9

If the median of the following data is 166.79, then the mean and mode are
  • a)
    Mode = 167.3, Mean = 168.03
  • b)
    Mode = 160.9 Mean = 167
  • c)
    Mode = 161.9 Mean = 168
  • d)
    Mode = 152.9 Mean = 166.73
Correct answer is option 'A'. Can you explain this answer?

Amit Kumar answered
We have
Mode=L+ 
Where L=lower limit of the modal class =160
f1= modal class frequency.=28
f2 = just after the modal class frequency.=24
f0 = just previous the modal class frequency=17
C = Class size=10
Mode= 
=167.3
Mode = 3 Median – 2 Mean
Mean=(3*166.79-167.3)/2=168.03

The mean of the following data 12, 22, 32,.......n2 is –
  • a)
    (n+1)(2n+1)/6
  • b)
    n(n – 1)(2n+1)/6
  • c)
    n(n+1)(2n – 1)/6
  • d)
    n(n – 1)(2n – 1)/6
Correct answer is option 'A'. Can you explain this answer?

Lighτ Yαmɪ answered
Sum of Series 1²+2²+3²+….+n² = n(n+1)(2n+1)/6
[as, (1+2+3+4+…..n) = n(n+1) / 2
Using the identity: a^3 -b^3 = (a-b)(a² +ab +b²)

We get, x^3 -(x-1)^3 = 3x² -3x +1 , by putting x = 1,2,3…n

We get,

1^3 - 0^3 = 3×1² - 3×1 + 1

2^3 -1^3 = 3×2² - 3×2 +1

3^3 - 2^3 = 3×3² - 3×3 +1

…..

n^3 - (n-1)^3 = 3*n² - 3n +1

By adding LHS up to x=3

1^3–0^3 + 2^3–1^3 + 3^3–2^3 = 3^3 -0^3

So by adding LHS up to x= n

We get, n^3 - 0^3 …….. LHS

Now by adding RHS =

3( 1²+2²+3² +…n²) - 3( 1+2+3+….n) + n

=3(1²+2²+3²+..n²) = n^3 +3(1+2+3+…n) -n

=(1² +2²+3²+..n²) = 1/3 [n^3 +(3 n(n+1)/2) - n]

=(1²+2² +3² +..n²) = 1/3 [ (2n^3 +3n²+3n - 2n) /2 ]

= (1²+2²+3²+..n²) = ( 2n^3 +3n² +n) /6

=n (n+1)(2n+1) /6]

mean=sum/number of terms
=n(n+1)(2n+1)/6×n
=(n+1) (2n+1) /6

Hence, option A is correct

In a data, if l = 60, h = 15, f1 = 16, f0 = 6, f2 = 6, then the mode is
  • a)
    60
  • b)
    72
  • c)
    67.5
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Shreya Shajilal answered
Mode=l+(f1-f0/2f1-f0-f2)×h
=60+(16-6/2×16-6-6)×15
=60+(10/20)×15
=60+(1/2)×15
=60+15/2
=60+7.5
=67.5.
Therefore the mode is 67.5.

The mean and the median of a distribution are 45.9 and 46 respectively. The mode will be
  • a)
    45
  • b)
    47
  • c)
    48
  • d)
    46.2
Correct answer is option 'D'. Can you explain this answer?

Bye Bye answered
Given, mean=45.9 and median=46
to find:- mode

as we know,
3median= 2mean+mode

3(46)=2(45.9)+mode

mode=138-91.8

mode=46.2
so correct answer is
(d)46.2

 The following table gives the marks obtained by 80 students in a selection test:

The median marks is​
  • a)
    34.3
  • b)
    45.5
  • c)
    20
  • d)
    28.5
Correct answer is option 'A'. Can you explain this answer?

Subset Academy answered
Median = 
First change the table into continuous data.
Middle of the frequencies=40 which will lie in 30-40. So 
L=30
N/2=80/2=40
F=27
f=30
c=10
Median = 

The median of first 10 prime numbers is
  • a)
    11
  • b)
    12
  • c)
    13
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Gaurav Kumar answered
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. So 10 prime numbers are: 2,3,5,7,11,13,17,19,23,29
Since the number of terms are even ,
Median = 
 10/2=5th term=11
10/2 +1=6th term=13
Median = 

The mean of the following data is: 45, 35, 20, 15, 25, 40​
  • a)
    15
  • b)
    25
  • c)
    30
  • d)
    35
Correct answer is option 'C'. Can you explain this answer?

Pankaj verma answered
Mean of Data Set

To find the mean of a data set, we add up all the numbers in the set and then divide by the total number of values in the set.

Calculation

To find the mean of the given data set, we need to add up all the numbers in the set and then divide by the total number of values in the set:

Mean = (45 + 35 + 20 + 15 + 25 + 40) / 6

Mean = 180 / 6

Mean = 30

Therefore, the mean of the given data set is 30. The correct answer is option C.

The mean of following distribution is –
  • a)
    18.50
  • b)
    18.05
  • c)
    18.15
  • d)
    18.25
Correct answer is option 'B'. Can you explain this answer?

Ræjû Bhæï answered
B option sahi hi n.. Bss isme khuss raho.. Aur explain chaiye toh subh ko fir milenge abhi sone ja rha hu.

The mean of the first 10 multiples of 6 is
  • a)
    3.3
  • b)
    33
  • c)
    34
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Komal nayar answered
Mean of First 10 Multiples of 6

To find the mean of the first 10 multiples of 6, we need to add the first 10 multiples of 6 and divide by 10.

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60

Addition: 6 + 12 + 18 + 24 + 30 + 36 + 42 + 48 + 54 + 60 = 330

Mean: 330/10 = 33

Therefore, the correct answer is option B, 33.

If median of 20 observations is 50 and mode is also 50, then the mean is
  • a)
    45
  • b)
    55
  • c)
    50
  • d)
    49
Correct answer is option 'C'. Can you explain this answer?

Ria Madan answered
3Median =2Mean+Mode
3(50) = 2Mean+ 50
150-50=2Mean
100=2Mean
Mean=100/2
=50
Gende the mean is also 50
Hope it helps

Mode is not affected by
  • a)
    Maximum value
  • b)
    Minimum value
  • c)
    Extreme values
  • d)
    All of the above
Correct answer is option 'D'. Can you explain this answer?

Anika Singh answered
Mode is the measure which counts for the maximum no. Of frequency of a number. So for eg. I have 2,36,3,4,3.
So the mode is 3..the maximum value or minimum value doesn't matter

The mean of a data set with 12 observations is calculated as 19.25. If one more value is included in the data, then for the new data with 13 observations, mean becomes 20. Value of this 13th observation is​
  • a)
    28
  • b)
    29
  • c)
    31
  • d)
    30
Correct answer is option 'B'. Can you explain this answer?

Anita Menon answered
Suppose be all twelevth terms = x
Mean = Terms / Observation
19 . 25 = x / 12
x = 19.25 × 12
x = 231 .00

Now let thirteenth term will be y
Mean = 20
So, 20 = x + y / 13
20 = 231 + y / 13
20 × 13 = 231 + x
260 = 231 + x
260 - 231 = x
29 = x

For a symmetrical distribution, which is correct
  • a)
    Mean > Mode > Median
  • b)
    Mode = Mean+ Median/2
  • c)
    Mean < Mode < Median
  • d)
    Mean = Median = Mode
Correct answer is option 'D'. Can you explain this answer?

Anjana Khatri answered
Symmetrical distribution occurs when the values of variables occur at regular frequencies and the mean, median and mode occur at the same point. In graph form, symmetrical distribution often appears as a bell curve. If a line were drawn dissecting the middle of the graph, it would show two sides that mirror each other.

The class mark of the class 15.5 - 20.5 
  • a)
    15.5
  • b)
    20.5
  • c)
    18
  • d)
    5
Correct answer is option 'C'. Can you explain this answer?

Crafty Classes answered
class mark= (upper limit+lower limit)/2
                 = (15.5+20.5)/2
                 =36/2
                 =18
 

The lower limit of the modal class of the following data is:
  • a)
    10
  • b)
    20
  • c)
    30
  • d)
    50
Correct answer is option 'B'. Can you explain this answer?

Raghav Bansal answered
Modal class is the interval which has the highest frequency ie, which occur most of the times.Here highest frequency is 13 in the class interval 20-30. So lower limit of the class interval is 20.

The mean of the marks obtained by 7 students in a group is 226. If the marks obtained by six of them are 340, 180, 260, 56, 275, 307, then the marks obtained by the seventh student are​
  • a)
    200
  • b)
    125
  • c)
    174
  • d)
    164
Correct answer is option 'D'. Can you explain this answer?

Deepak trivedi answered
Let's start by finding the sum of the marks obtained by the six students:

Sum of marks = 340 + 180 + 260 + 56 + 275 + 307 = 1418

We can then use the formula for the mean to find the total marks obtained by all seven students:

Mean = (Total marks) / (Number of students)

226 = (Total marks) / 7

Total marks = 1582

Finally, we can find the marks obtained by the seventh student by subtracting the sum of the marks obtained by the six students from the total marks obtained by all seven students:

Marks obtained by seventh student = Total marks - Sum of marks

Marks obtained by seventh student = 1582 - 1418

Marks obtained by seventh student = 164

Therefore, the answer is (d) 164.

Mode and mean of a data are 12k and 15k. Median of the data is
  • a)
    12k
  • b)
    14k
  • c)
    15k
  • d)
    16k
Correct answer is option 'B'. Can you explain this answer?

Amit Kumar answered
Here's the solution to your question:

If two values are given in the below formula than we can use the formula if one is given than we cannot use the formula
3median = 2mean+mode
3median = 2(15k)+12k
3median = 30k+12k
3median = 42k
median = 42k/3
median = 14k

You can understand all the concepts of Class 10 mathematics through the link:
Mathematics (Maths) Class 10

The class marks of the class 18-22 is
  • a)
    4
  • b)
    18
  • c)
    22
  • d)
    20
Correct answer is option 'D'. Can you explain this answer?

Harshit Agrawal answered
We know that ,
Class Mark = upper limit + lower limit /2
--->. Class Mark = 22 + 18/2
--->. Class Mark = 40/2
--->. Class Mark = 20

Hence , option D is correct.

The mean of the first 10 natural odd numbers is
  • a)
    9
  • b)
    10
  • c)
    11
  • d)
    12
Correct answer is option 'B'. Can you explain this answer?

Ananya Das answered
Natural numbers start from 1
First ten natural odd numbers=1,3,5,7,9,11,13,15,17,19
Mean = 
So mean is 10.

Chapter doubts & questions for Statistics - Online MCQ Tests for Class 10 2025 is part of Class 10 exam preparation. The chapters have been prepared according to the Class 10 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Class 10 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Statistics - Online MCQ Tests for Class 10 in English & Hindi are available as part of Class 10 exam. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free.

Top Courses Class 10

Related Class 10 Content

© EduRev
Education Revolution
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily