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FOUNDATION FOR 11th/ 12th /JEE
BEGINNER’s Course
Statistics Test Solutions
Q1.
If for some 𝑥 ∈ 𝑅, the frequency distribution of the marks obtained by 20 students in a test is
Marks
2
3
5
Frequency
(𝑥 + 1)2
2x-5
𝑥 2 − 3𝑥
7
x
Then the mean of the marks is
a.
b.
c.
d.
3.2
3.0
2.5
2.8
Solution:
Number of students are
(𝑥 + 1)2 + (2𝑥 − 5) + (𝑥 2 − 3𝑥) + 𝑥 = 20
⇒ 2𝑥 2 + 2𝑥 − 4 = 20
⇒ 𝑥 2 + 𝑥 − 12 = 0
⇒ (𝑥 + 4)(𝑥 − 3) = 0 ⇒ 𝑥 = 3
Marks
2
6
No. Of
16
1
Students
32+3+21 56
Average Marks = 20 = 20 = 2.8
Q2.
5
7
0
3
The mean and variance of 10 observation are found to be 10 and 4 respectively. On
rechecking it was found that an observation 8 was incorrect. If it is replaced by 18, then the
correct variance is
a. 7
b. 8
c. 9
56
d. 6
Solution:
𝑥 (𝑜𝑙𝑑) = 10 =
∑ 𝑥𝑖
⇒ ∑ 𝑥𝑖(𝑜𝑙𝑑) = 100
10
∑ 𝑥𝑖(𝑁𝑒𝑤) = 100 − 8 + 18 = 110
𝑥̅(𝑛𝑒𝑤) =
110
= 11
10
𝑉𝑎𝑟(𝑜𝑙𝑑) = 4 =
2
∑ 𝑥𝑖(𝑜𝑙𝑑)
2
∑𝑥
𝑖(𝑜𝑙𝑑)
= 1040
10
− (𝑥̅𝑜𝑙𝑑 )2
Statistics Test Solutions
2
∑𝑥
= 1040 − 64 + 324
𝑖(𝑁𝑒𝑤)
= 1300
𝑉𝑎𝑟𝑒(𝑁𝑒𝑤) =
Q3.
1300
− (11)2 = 130 − 121 = 9
10
If 2 data sets having 10 and 20 observations have coefficients of variation 50 and 60 respectively
and arithmetic means 30 and 25 respectively, then the combined variance of those 30 observations
is
a.
b.
c.
d.
2075
3
2075
9
1000
9
1075
3
Solution:
Given,
𝑛1 = 10, 𝑥̅1 = 30, 𝐶𝑉1 = 50
𝑛2 = 20, 𝑥̅1 = 25, 𝐶𝑉2 = 60
𝜎1
𝜎1
𝐶𝑉1 = × 100 ⇒ 50 =
× 100 ⇒ 𝜎1 = 15
𝑥1
30
𝜎2
𝜎1
𝐶𝑉2 = × 100 ⇒ 60 =
× 100 ⇒ 𝜎2 = 15
𝑥2
25
Combined mean =
𝑑1 = 𝑥̅1 − 𝑥̅ =
10×30+20×25
30
10
5
, 𝑑2 = 𝑥̅ − 𝑥̅2 =
3
3
Combined variance =
=
=
Q4.
80
= 3
2250 + 4500 +
𝑛1 𝜎12 +𝑛2 𝜎22 +𝑛1 𝑑12 +𝑛2 𝑑12
𝑛1 +𝑛2
1000 500
9 + 9
30
6225 2075
=
27
9
If the data 𝑥1 , 𝑥2 … . . , 𝑥10 is such that the mean of first four of these is 11, the mean of the remaining
six is 16 and the sum of squares of all of these is 2000, then the standard deviation of this data is
a.
b.
c.
d.
2√2
2
4
√2
Solution:
According to the question
Statistics Test Solutions
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4
= 11 ⇒ 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 44
4
𝑥5 + 𝑥6 + ⋯ . +𝑥10
= 16 ⇒ 𝑥5 + 𝑥6 + ⋯ + 𝑥10 = 96
6
2
And 𝑥12 + 𝑥22 + ⋯ + 𝑥10
= 2000
Variance = 𝜎 2 =
∑ 𝑥𝑖2
𝑁
− (𝑥̅ )2
2000
44 + 96 2
) =4
=
−(
10
10
⇒𝜎=2
Q5.
The standard deviation of a distribution is 30. I each observation is increased by 5, then the new
standard deviation will be
a.
b.
c.
d.
32
28
27
30
Solution:
S.D of a series is unaltered if each observation is increased (or reduced) by the same scalar quantity,
S.D is independent of change of origin.
Hence, S.D will be the same.
∴ 𝑆. 𝐷 = 30
Q6.
If the mean and the standard deviation of the data 3,5,7,𝑎, 𝑏 𝑎𝑟𝑒 5 and 2 respectively, then 𝑎 𝑎𝑛𝑑 𝑏
are the roots of the equation:
a.
b.
c.
d.
𝑥 2 − 10𝑥 + 18 = 0
2𝑥 2 − 20𝑥 + 19 = 0
𝑥 2 − 10𝑥 + 19 = 0
𝑥 2 − 20𝑥 + 18 = 0
Solution:
Mean =
3+5+7+𝑎+𝑏
Variance =
5
= 5 ⇒ 𝑎 + 𝑏 = 10
32 +52 +72 +𝑎2 +𝑏 2
5
− (5)2 = 4
⇒ 𝑎2 + 𝑏 2 = 62
⇒ (𝑎 + 𝑏)2 − 2𝑎𝑏 = 62
⇒ 𝑎𝑏 = 19
Hence, a and b are the roots of the equation,
𝑥 2 − 10𝑥 + 19 = 0
Statistics Test Solutions
Q7.
The mean and variance of a data set comprising 15 observations are 15 and 5 respectively. If one of
the observations 15 is deleted and two new observation 6 and 8 are added to the data, then the
new variance of resulting data is.
a.
b.
c.
d.
10.3715
11.8125
13.25
5.7516
Solution:
Let 15 observations are 𝑥1 , 𝑥2 , … . 𝑥15
𝑥1 + 𝑥2 + 𝑥3 … . . +𝑥15
= 15
15
If 𝑥15 = 15
𝑥̅𝑛𝑒𝑤 =
𝑥1 + 𝑥2 + 𝑥3 … . . 𝑥15 − 15 + 6 + 8 224
=
= 14
16
16
∑ 𝑥𝑖2
− (𝑥̅ )2 = 5
15
2
⇒ 𝑥12 + 𝑥22 + ⋯ + 𝑥14
+ (15)2 = 230 × 15
2
⇒ 𝑥12 + 𝑥22 + ⋯ + 𝑥14
= 3225
New variance =
2
𝑥12 + 𝑥22 + ⋯ + 𝑥14
+ 62 + 82
− (𝑥̅𝑛𝑒𝑤 )2
16
3325
− 196
16
= 11.8125
Q8.
If the mean deviation about median for the numbers 3, 5, 7, 2𝑘, 12, 16, 21, 24 arranged in the
ascending order is 6, then the median is__________.
Solution:
2𝑘+12
Median =
2
=𝑘+6
Mean deviation = ∑
|𝑥𝑖 −𝑀|
𝑛
=6
(𝑘+3)+(𝑘+1)+(𝑘−1)+(6−𝑘)+(6−𝑘)+(10−𝑘)+(15−𝑘)+(18−𝑘)
8
∴
58−2𝑘
=6
8
k=5
Median=
Q9.
2×5+12
2
= 11
For the following observations 2, 3, 5, 6, 8, 10, 12, 17, 20, 25. Find the mean deviation about median.
Solution:
Here, Total number of observations (𝑛) = 10 For 𝑛 𝑒𝑣𝑒𝑛,
Statistics Test Solutions
𝑁 𝑡ℎ
⇒ 𝑀𝑒𝑑𝑖𝑎𝑛 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑓 ( 2 )
𝑁
𝑎𝑛𝑑 ( 2 + 1)
𝑡ℎ
𝑖𝑡𝑒𝑚𝑠.
8 + 10
=9
2
=
⇒ Mean deviation about median
=
=
∑|𝑥−𝑀𝑒𝑑𝑖𝑎𝑛|
𝑛
7+6+4+3+1+1+3+8+11+16
10
60
= 10 = 6
Q10. The mean and standard deviation of 50 observations are 50 and 10. Later on, it was decided to omit
an observation that was incorrectly recorded as 99. If the variance of the remaining 49 observations
𝜆
is 49, then the value of 𝜆 is equal to_______________.
Solution:
∑𝑥
𝑥̅ = 501 = 50
⇒ ∑ 𝑥𝑖 = 2500 − 99 = 2401
𝐶𝑜𝑟𝑟𝑒𝑐𝑡 𝑥̅ =
𝜎2 =
∑ 𝑥𝑖2
2401
= 49
49
− 𝑥̅ 2
𝑛
∑ 𝑥2
⇒ 100 = 50𝑖 − 2500
⇒ ∑ 𝑥𝑖2 = 130000
Correct ∑ 𝑥𝑖2 = 130000 − 992 = 120199
Correct 𝜎 2 =
=
=
120199
49
∑ 𝑥𝑖2
𝑛
− 𝑥̅ 2
− (49)2
120199−117649
49
2550
𝜆
= 49 = 49
Hence, 𝜆 = 2550