JEE Main Previous Year Questions (2026): Statistics

 Page 1


JEE Main Previous Year Questions 
(2025): Statistics 
Question1: The variance of the numbers ?? , ???? , ???? , ???? , … , ?????? is ____ . 
JEE Main 2025 (Online) 23rd January Evening Shift 
 
Answer: 8788 
Solution: 
Var( 8,21,34,47, … … ,320) 
Var( 0,13,26,39, … … ,312) 
13
2
· Var( 0,1,2, … … ,24) 
13
2
· Var( 1,2,3, … … ,25) 
So, ?? 2
= 13
2
× (
25
2
-1
12
)= 8788 
Alternate solution 
8 + ( ?? - 1) 13 = 320 
13?? = 325 
?? = 25 
no. of terms = 25 
mean =
? x
i
n
=
8+21+?+320
25
=
25
2
( 8+320)
25
 
variance ?? 2
=
? x
i
2
n
- ( mean )
2
 
=
8
2
+ 21
2
+ ? + 320
2
13
- ( 164)
2
 
= 8788 
Question2: Marks obtains by all the students of class ???? are presented in a freqency 
distribution with classes of equal width. Let the median of this grouped data be 14 
with median class interval 12-18 and median class frequency 12. If the number of 
students whose marks are less than ???? is ???? , then the total number of students is : 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 52 
B. 44 
C. 40 
D. 48 
Answer: B 
Page 2


JEE Main Previous Year Questions 
(2025): Statistics 
Question1: The variance of the numbers ?? , ???? , ???? , ???? , … , ?????? is ____ . 
JEE Main 2025 (Online) 23rd January Evening Shift 
 
Answer: 8788 
Solution: 
Var( 8,21,34,47, … … ,320) 
Var( 0,13,26,39, … … ,312) 
13
2
· Var( 0,1,2, … … ,24) 
13
2
· Var( 1,2,3, … … ,25) 
So, ?? 2
= 13
2
× (
25
2
-1
12
)= 8788 
Alternate solution 
8 + ( ?? - 1) 13 = 320 
13?? = 325 
?? = 25 
no. of terms = 25 
mean =
? x
i
n
=
8+21+?+320
25
=
25
2
( 8+320)
25
 
variance ?? 2
=
? x
i
2
n
- ( mean )
2
 
=
8
2
+ 21
2
+ ? + 320
2
13
- ( 164)
2
 
= 8788 
Question2: Marks obtains by all the students of class ???? are presented in a freqency 
distribution with classes of equal width. Let the median of this grouped data be 14 
with median class interval 12-18 and median class frequency 12. If the number of 
students whose marks are less than ???? is ???? , then the total number of students is : 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 52 
B. 44 
C. 40 
D. 48 
Answer: B 
Solution: 
The median for grouped data is given by: 
Median = ?? + (
?? 2
-????
?? )× h 
where 
?? is the lower limit (or boundary) of the median class. 
???? is the cumulative frequency of all classes preceding the median class. 
?? is the frequency of the median class. 
h is the class width. 
?? is the total number of students. 
Given: 
Median = 14 
Median class interval is 12 - 18, so ?? = 12 and the class width h = 18 - 12 = 6. 
Frequency of median class ?? = 12. 
Cumulative frequency below the median class = 18 (i.e., ???? = 18 ). 
Plugging these into the formula: 
14 = 12 + (
?? 2
- 18
12
) × 6 
Step 1: Subtract 12 from both sides: 
2 = (
?? 2
- 18
12
) × 6 
Step 2: Simplify the multiplication factor: 
(
6
12
)=
1
2
 
So the equation becomes: 
2 =
1
2
(
?? 2
- 18) 
Step 3: Multiply both sides by 2 to remove the fraction: 
4 =
?? 2
- 18 
Step 4: Solve for 
?? 2
 : 
?? 2
= 4 + 18 = 22 
Step 5: Multiply both sides by 2 to find ?? : 
?? = 44 
Thus, the total number of students is 44 . 
Question3: For a statistical data ?? ?? , ?? ?? , … , ?? ????
 of ???? values, a student obtained the 
mean as 5.5 and ?
?? =?? ????
??? ?? ?? = ?????? . He later found that he had noted two values in the 
data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The 
variance of the corrected data is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
Page 3


JEE Main Previous Year Questions 
(2025): Statistics 
Question1: The variance of the numbers ?? , ???? , ???? , ???? , … , ?????? is ____ . 
JEE Main 2025 (Online) 23rd January Evening Shift 
 
Answer: 8788 
Solution: 
Var( 8,21,34,47, … … ,320) 
Var( 0,13,26,39, … … ,312) 
13
2
· Var( 0,1,2, … … ,24) 
13
2
· Var( 1,2,3, … … ,25) 
So, ?? 2
= 13
2
× (
25
2
-1
12
)= 8788 
Alternate solution 
8 + ( ?? - 1) 13 = 320 
13?? = 325 
?? = 25 
no. of terms = 25 
mean =
? x
i
n
=
8+21+?+320
25
=
25
2
( 8+320)
25
 
variance ?? 2
=
? x
i
2
n
- ( mean )
2
 
=
8
2
+ 21
2
+ ? + 320
2
13
- ( 164)
2
 
= 8788 
Question2: Marks obtains by all the students of class ???? are presented in a freqency 
distribution with classes of equal width. Let the median of this grouped data be 14 
with median class interval 12-18 and median class frequency 12. If the number of 
students whose marks are less than ???? is ???? , then the total number of students is : 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 52 
B. 44 
C. 40 
D. 48 
Answer: B 
Solution: 
The median for grouped data is given by: 
Median = ?? + (
?? 2
-????
?? )× h 
where 
?? is the lower limit (or boundary) of the median class. 
???? is the cumulative frequency of all classes preceding the median class. 
?? is the frequency of the median class. 
h is the class width. 
?? is the total number of students. 
Given: 
Median = 14 
Median class interval is 12 - 18, so ?? = 12 and the class width h = 18 - 12 = 6. 
Frequency of median class ?? = 12. 
Cumulative frequency below the median class = 18 (i.e., ???? = 18 ). 
Plugging these into the formula: 
14 = 12 + (
?? 2
- 18
12
) × 6 
Step 1: Subtract 12 from both sides: 
2 = (
?? 2
- 18
12
) × 6 
Step 2: Simplify the multiplication factor: 
(
6
12
)=
1
2
 
So the equation becomes: 
2 =
1
2
(
?? 2
- 18) 
Step 3: Multiply both sides by 2 to remove the fraction: 
4 =
?? 2
- 18 
Step 4: Solve for 
?? 2
 : 
?? 2
= 4 + 18 = 22 
Step 5: Multiply both sides by 2 to find ?? : 
?? = 44 
Thus, the total number of students is 44 . 
Question3: For a statistical data ?? ?? , ?? ?? , … , ?? ????
 of ???? values, a student obtained the 
mean as 5.5 and ?
?? =?? ????
??? ?? ?? = ?????? . He later found that he had noted two values in the 
data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The 
variance of the corrected data is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 5 
B. 7 
C. 9 
D. 4 
Answer: B 
Solution: 
Mean x = 5.5 
= ?
i=1
10
?x
i
= 5.5 × 10 = 55 
= ?
i=1
10
?x
i
2
= 371 
( ? x
i
)
new 
= 55 - ( 4 + 5)+ ( 6 + 8)= 60 
( ? x
i
)
new 
= 371 - ( 4
2
+ 5
2
)+ ( 6
2
+ 8
2
)= 430 
Variance ?? 2
=
? x
i
2
10
- (
? x
i
10
)
2
 
?? 2
=
430
10
- (
60
10
)
2
 
?? 2
= 43 - 36 
?? 2
= 7 
Question4: Let ?? ?? , ?? ?? , … , ?? ????
 be ten observations such that ?
?? =?? ????
?( ?? ?? - ?? )= ???? , 
?
?? =?? ????
?( ?? ?? - ?? )
?? = ???? , ?? > ?? , and their variance is 
?? ?? . If ?? and ?? ?? are respectively the 
mean and the variance of ?? ( ?? ?? - ?? )+ ?? ?? , ?? ( ?? ?? - ?? )+ ?? ?? , … , ?? ( ?? ????
- ?? )+ ?? ?? , then 
????
?? ?? is equal to : 
JEE Main 2025 (Online) 29th January Morning Shift  
Options: 
A. 100 
B. 90 
C. 120 
D. 110 
Answer: A 
Solution: 
?
i=1
10
?x
i
= 50, ? mean = 5 
Variance =
4
5
=
? x
i
2
10
- (
? x
i
10
)
2
 
4
5
=
? x
i
2
10
- 25 
Page 4


JEE Main Previous Year Questions 
(2025): Statistics 
Question1: The variance of the numbers ?? , ???? , ???? , ???? , … , ?????? is ____ . 
JEE Main 2025 (Online) 23rd January Evening Shift 
 
Answer: 8788 
Solution: 
Var( 8,21,34,47, … … ,320) 
Var( 0,13,26,39, … … ,312) 
13
2
· Var( 0,1,2, … … ,24) 
13
2
· Var( 1,2,3, … … ,25) 
So, ?? 2
= 13
2
× (
25
2
-1
12
)= 8788 
Alternate solution 
8 + ( ?? - 1) 13 = 320 
13?? = 325 
?? = 25 
no. of terms = 25 
mean =
? x
i
n
=
8+21+?+320
25
=
25
2
( 8+320)
25
 
variance ?? 2
=
? x
i
2
n
- ( mean )
2
 
=
8
2
+ 21
2
+ ? + 320
2
13
- ( 164)
2
 
= 8788 
Question2: Marks obtains by all the students of class ???? are presented in a freqency 
distribution with classes of equal width. Let the median of this grouped data be 14 
with median class interval 12-18 and median class frequency 12. If the number of 
students whose marks are less than ???? is ???? , then the total number of students is : 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 52 
B. 44 
C. 40 
D. 48 
Answer: B 
Solution: 
The median for grouped data is given by: 
Median = ?? + (
?? 2
-????
?? )× h 
where 
?? is the lower limit (or boundary) of the median class. 
???? is the cumulative frequency of all classes preceding the median class. 
?? is the frequency of the median class. 
h is the class width. 
?? is the total number of students. 
Given: 
Median = 14 
Median class interval is 12 - 18, so ?? = 12 and the class width h = 18 - 12 = 6. 
Frequency of median class ?? = 12. 
Cumulative frequency below the median class = 18 (i.e., ???? = 18 ). 
Plugging these into the formula: 
14 = 12 + (
?? 2
- 18
12
) × 6 
Step 1: Subtract 12 from both sides: 
2 = (
?? 2
- 18
12
) × 6 
Step 2: Simplify the multiplication factor: 
(
6
12
)=
1
2
 
So the equation becomes: 
2 =
1
2
(
?? 2
- 18) 
Step 3: Multiply both sides by 2 to remove the fraction: 
4 =
?? 2
- 18 
Step 4: Solve for 
?? 2
 : 
?? 2
= 4 + 18 = 22 
Step 5: Multiply both sides by 2 to find ?? : 
?? = 44 
Thus, the total number of students is 44 . 
Question3: For a statistical data ?? ?? , ?? ?? , … , ?? ????
 of ???? values, a student obtained the 
mean as 5.5 and ?
?? =?? ????
??? ?? ?? = ?????? . He later found that he had noted two values in the 
data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The 
variance of the corrected data is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 5 
B. 7 
C. 9 
D. 4 
Answer: B 
Solution: 
Mean x = 5.5 
= ?
i=1
10
?x
i
= 5.5 × 10 = 55 
= ?
i=1
10
?x
i
2
= 371 
( ? x
i
)
new 
= 55 - ( 4 + 5)+ ( 6 + 8)= 60 
( ? x
i
)
new 
= 371 - ( 4
2
+ 5
2
)+ ( 6
2
+ 8
2
)= 430 
Variance ?? 2
=
? x
i
2
10
- (
? x
i
10
)
2
 
?? 2
=
430
10
- (
60
10
)
2
 
?? 2
= 43 - 36 
?? 2
= 7 
Question4: Let ?? ?? , ?? ?? , … , ?? ????
 be ten observations such that ?
?? =?? ????
?( ?? ?? - ?? )= ???? , 
?
?? =?? ????
?( ?? ?? - ?? )
?? = ???? , ?? > ?? , and their variance is 
?? ?? . If ?? and ?? ?? are respectively the 
mean and the variance of ?? ( ?? ?? - ?? )+ ?? ?? , ?? ( ?? ?? - ?? )+ ?? ?? , … , ?? ( ?? ????
- ?? )+ ?? ?? , then 
????
?? ?? is equal to : 
JEE Main 2025 (Online) 29th January Morning Shift  
Options: 
A. 100 
B. 90 
C. 120 
D. 110 
Answer: A 
Solution: 
?
i=1
10
?x
i
= 50, ? mean = 5 
Variance =
4
5
=
? x
i
2
10
- (
? x
i
10
)
2
 
4
5
=
? x
i
2
10
- 25 
? ? x
i
2
= 258 ( 1) 
Now ? ?
10
i=1
( x
i
- ?? )
2
= 98 
?
i=1
10
?( x
i
2
- 2?? · x
i
+ ?? 2
)= 98 
258 - 2?? ( 50)+ 10?? 2
= 98 
( ?? - 8) ( ?? - 2)= 0 
?? = or ?? = 2 ( as ?? > 2) 
? ?? = 8 ( 2) 
Now as per the question 
2( x
1
- 1)+ 4?? , 2( x
2
- 1)+ 4?? , … .2( x
10
- 1)+ 4?? 
can be simplified to 
2x
1
+ 30,2x
2
+ 30, … .2x
10
+ 30 using eq. (2) 
?? = 2( 5)+ 30 = 40 ( 3) 
?? 2
= 2
2
(
4
5
)=
16
5
 
?
????
?? 2
=
8 × 40
16/5
= 100 
Question5: If the mean and the variance of ?? , ?? , ?? , ?? , ?? , ???? , ???? , ???? are 9 and 9.25 
respectively, then ?? + ?? + ???? is equal to : 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. 103 
B. 106 
C. 100 
D. 105 
Answer: A 
Solution: 
Let's set up two equations from the given mean and variance: 
Mean = 9 ? total sum = 8 · 9 = 72 
Known values sum to 6 + 4 + 8 + 12 + 10 + 13 = 53, so 
?? + ?? = 72 - 53 = 19. 
Population variance = 9.25 ? 
? ?? ?? 2
8
- 9
2
= 9.25 ? ? ?? ?? 2
= 8( 81 + 9.25)= 722. 
Known squares sum to 6
2
+ 4
2
+ 8
2
+ 12
2
+ 10
2
+ 13
2
= 529 , so 
?? 2
+ ?? 2
= 722 - 529 = 193 . 
Page 5


JEE Main Previous Year Questions 
(2025): Statistics 
Question1: The variance of the numbers ?? , ???? , ???? , ???? , … , ?????? is ____ . 
JEE Main 2025 (Online) 23rd January Evening Shift 
 
Answer: 8788 
Solution: 
Var( 8,21,34,47, … … ,320) 
Var( 0,13,26,39, … … ,312) 
13
2
· Var( 0,1,2, … … ,24) 
13
2
· Var( 1,2,3, … … ,25) 
So, ?? 2
= 13
2
× (
25
2
-1
12
)= 8788 
Alternate solution 
8 + ( ?? - 1) 13 = 320 
13?? = 325 
?? = 25 
no. of terms = 25 
mean =
? x
i
n
=
8+21+?+320
25
=
25
2
( 8+320)
25
 
variance ?? 2
=
? x
i
2
n
- ( mean )
2
 
=
8
2
+ 21
2
+ ? + 320
2
13
- ( 164)
2
 
= 8788 
Question2: Marks obtains by all the students of class ???? are presented in a freqency 
distribution with classes of equal width. Let the median of this grouped data be 14 
with median class interval 12-18 and median class frequency 12. If the number of 
students whose marks are less than ???? is ???? , then the total number of students is : 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 52 
B. 44 
C. 40 
D. 48 
Answer: B 
Solution: 
The median for grouped data is given by: 
Median = ?? + (
?? 2
-????
?? )× h 
where 
?? is the lower limit (or boundary) of the median class. 
???? is the cumulative frequency of all classes preceding the median class. 
?? is the frequency of the median class. 
h is the class width. 
?? is the total number of students. 
Given: 
Median = 14 
Median class interval is 12 - 18, so ?? = 12 and the class width h = 18 - 12 = 6. 
Frequency of median class ?? = 12. 
Cumulative frequency below the median class = 18 (i.e., ???? = 18 ). 
Plugging these into the formula: 
14 = 12 + (
?? 2
- 18
12
) × 6 
Step 1: Subtract 12 from both sides: 
2 = (
?? 2
- 18
12
) × 6 
Step 2: Simplify the multiplication factor: 
(
6
12
)=
1
2
 
So the equation becomes: 
2 =
1
2
(
?? 2
- 18) 
Step 3: Multiply both sides by 2 to remove the fraction: 
4 =
?? 2
- 18 
Step 4: Solve for 
?? 2
 : 
?? 2
= 4 + 18 = 22 
Step 5: Multiply both sides by 2 to find ?? : 
?? = 44 
Thus, the total number of students is 44 . 
Question3: For a statistical data ?? ?? , ?? ?? , … , ?? ????
 of ???? values, a student obtained the 
mean as 5.5 and ?
?? =?? ????
??? ?? ?? = ?????? . He later found that he had noted two values in the 
data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The 
variance of the corrected data is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 5 
B. 7 
C. 9 
D. 4 
Answer: B 
Solution: 
Mean x = 5.5 
= ?
i=1
10
?x
i
= 5.5 × 10 = 55 
= ?
i=1
10
?x
i
2
= 371 
( ? x
i
)
new 
= 55 - ( 4 + 5)+ ( 6 + 8)= 60 
( ? x
i
)
new 
= 371 - ( 4
2
+ 5
2
)+ ( 6
2
+ 8
2
)= 430 
Variance ?? 2
=
? x
i
2
10
- (
? x
i
10
)
2
 
?? 2
=
430
10
- (
60
10
)
2
 
?? 2
= 43 - 36 
?? 2
= 7 
Question4: Let ?? ?? , ?? ?? , … , ?? ????
 be ten observations such that ?
?? =?? ????
?( ?? ?? - ?? )= ???? , 
?
?? =?? ????
?( ?? ?? - ?? )
?? = ???? , ?? > ?? , and their variance is 
?? ?? . If ?? and ?? ?? are respectively the 
mean and the variance of ?? ( ?? ?? - ?? )+ ?? ?? , ?? ( ?? ?? - ?? )+ ?? ?? , … , ?? ( ?? ????
- ?? )+ ?? ?? , then 
????
?? ?? is equal to : 
JEE Main 2025 (Online) 29th January Morning Shift  
Options: 
A. 100 
B. 90 
C. 120 
D. 110 
Answer: A 
Solution: 
?
i=1
10
?x
i
= 50, ? mean = 5 
Variance =
4
5
=
? x
i
2
10
- (
? x
i
10
)
2
 
4
5
=
? x
i
2
10
- 25 
? ? x
i
2
= 258 ( 1) 
Now ? ?
10
i=1
( x
i
- ?? )
2
= 98 
?
i=1
10
?( x
i
2
- 2?? · x
i
+ ?? 2
)= 98 
258 - 2?? ( 50)+ 10?? 2
= 98 
( ?? - 8) ( ?? - 2)= 0 
?? = or ?? = 2 ( as ?? > 2) 
? ?? = 8 ( 2) 
Now as per the question 
2( x
1
- 1)+ 4?? , 2( x
2
- 1)+ 4?? , … .2( x
10
- 1)+ 4?? 
can be simplified to 
2x
1
+ 30,2x
2
+ 30, … .2x
10
+ 30 using eq. (2) 
?? = 2( 5)+ 30 = 40 ( 3) 
?? 2
= 2
2
(
4
5
)=
16
5
 
?
????
?? 2
=
8 × 40
16/5
= 100 
Question5: If the mean and the variance of ?? , ?? , ?? , ?? , ?? , ???? , ???? , ???? are 9 and 9.25 
respectively, then ?? + ?? + ???? is equal to : 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. 103 
B. 106 
C. 100 
D. 105 
Answer: A 
Solution: 
Let's set up two equations from the given mean and variance: 
Mean = 9 ? total sum = 8 · 9 = 72 
Known values sum to 6 + 4 + 8 + 12 + 10 + 13 = 53, so 
?? + ?? = 72 - 53 = 19. 
Population variance = 9.25 ? 
? ?? ?? 2
8
- 9
2
= 9.25 ? ? ?? ?? 2
= 8( 81 + 9.25)= 722. 
Known squares sum to 6
2
+ 4
2
+ 8
2
+ 12
2
+ 10
2
+ 13
2
= 529 , so 
?? 2
+ ?? 2
= 722 - 529 = 193 . 
Now use 
( ?? + ?? )
2
= ?? 2
+ 2???? + ?? 2
 ? 19
2
= 193+ 2???? 
so 
361 = 193+ 2???? ? ???? = 84. 
Finally, 
?? + ?? + ???? = 19 + 84 = 103. 
Answer: 103 (Option A). 
Question6: Let the Mean and Variance of five observations 
?? ?? = ?? , ?? ?? = ?? , ?? ?? = ?? , ?? ?? = ?? and ?? ?? = ?? , ?? > ?? , be 5 and 10 
respectively. Then the Variance of the observations 
?? + ?? ?? , ?? = ?? , ?? , … , ?? is 
JEE Main 2025 (Online) 3rd April Evening Shift 
Options: 
A. 17 
B. 16 
C. 16.4 
D. 17.4 
Answer: B 
Solution: 
First, find the values of ?? and ?? using the given conditions: 
Mean Condition: 
5 =
1 + 3 + ?? + 7 + ?? 5
 
This implies: 
?? + ?? = 14 
Variance Condition: 
1 + 9 + ?? 2
+ 49 + ?? 2
5
- ( 5)
2
= 10 
Simplifying, we get: 
?? 2
+ ?? 2
= 116 
Using ?? + ?? = 14, we have: 
?? = 14 - ?? 
Substitute ?? into the equation for ?? 2
+ ?? 2
 : 
?? 2
+ ( 14 - ?? )
2
= 116 
Expand and simplify: 
?? 2
+ 196 - 28?? + ?? 2
= 116 
2?? 2
- 28?? + 80 = 0