datavorous/entrance-exam-dataset huggingface dataset:
The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is

Options:

The correct answer is:
35

Solution:
∑xi25=40 and ∑xi-60+N25=39

Let age of newly appointed teacher is N

⇒1000-60+N=975

⇒N=35 year

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is

Options:
A 35
B 40
C 25
D 30

The correct answer is:
35

Solution:
Mean age can be calculated as
x-=x1+x2+…x2525=40
⇒ x1+x2+…+x25=1000
Let age of new teacher be A, then
1000-60+A=39×25=975
⇒ A=975-940=35

Topic: Mathematics, Statistics, JEE Main, JEE Main 2017 (08 Apr Online)
The mean age of a combined group of men and women is 30 years. If the means of the age of men and women are respectively 32 and 27 , then the percentage of women in the group is

Options:
A $30$
B $40$
C $50$
D $60$

The correct answer is:
$40$

Solution:
The formula for combined mean is $\bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}$
Given, $\bar{x}=30, \bar{x}_1=32, \bar{x}_2=27$
Let $n_1+n_2=100$ and $n_1$ denotes men, $n_2$ denotes women for this $n_2=100-n_1$
$30=\frac{32 n_1+\left(100-n_1\right) 27}{100} \Rightarrow 30=\frac{32 n_1+2700-27 n_1}{100}$
$\Rightarrow 3000-2700=32 n_1-27 n_1 \Rightarrow 300=5 n_1 \Rightarrow n_1=60$
So, $n_2=40$
Hence, the percentage of women in the group is $40$ .

Topic: Mathematics, Statistics, JEE Main
The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is

Options:
A 11
B 13
C 12
D 14

The correct answer is:
13

Solution:
Incorrect mean is

∑i=110xi10=20

⇒∑i=110xi=200 ....i

So, corrected mean is

∑i=110xi-50+4010=200-1010=19

Incorrect variance is

∑i=110xi210-400=64

⇒∑i=110xi2=4640

Now, corrected variance is

=∑i=19xi2+402-50210-192

=4640-90010-192=374-361=13

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The mean and standard deviation of 10 observations x1, x2, x3.....x10 are xÂ¯ and Ï respectively. Let 10 is added to x1, x2.....x9 and 90 is subtracted from x10. If still, the standard deviation is the same, then x10-xÂ¯ is equal to

Options:
A 35
B 45
C 55
D 50

The correct answer is:
45

Solution:
∑i=110xi-x¯210=σ2 … 1
∑i=19xi+10-x¯2+x10-90-x¯210=σ2 … 2
From 1 and (2)
∑i=110xi-x¯210=∑i=110xi-x¯2+20∑i=19xi-x¯-180x10-x¯+100×9+810010
2×∑i=19xi-x¯+2x10-x¯-20x10-x¯+90+810=0
⇒900=20x10-x¯⇒x10-x¯=45

Topic: Mathematics, Statistics, JEE Main
The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to _____.

Options:

The correct answer is:
17

Solution:
Old mean=x1+x2+...x14+515=8 (given)

⇒x1+x2+...x14=115

Correct mean=x1+x2+...x14+2015=x¯new⇒x¯new=9

Given, old variance=9⇒x12+x22+...x142+5215-82=9⇒x12+x22+...x142=1070

Now varnew=x12+x22+...x142+20215-92=1070+40015-81=17

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The mean and standard deviation of 20 observations are found to be 10 and 2 . respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

Options:
A 1.8
B 1.94
C $\sqrt{3.96}$
D $\sqrt{3.86}$

The correct answer is:
$\sqrt{3.96}$

Solution:
$\begin{aligned}
& \text { Mean }(\overline{\mathrm{x}})=10 \\
& \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}=10 \\
& \Sigma \mathrm{x}_{\mathrm{i}}=10 \times 20=200
\end{aligned}$
If 8 is replaced by 12, then $\Sigma x_i=200-8+12=204$
$\therefore$ Correct mean $(\overline{\mathrm{x}})=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}$
$=\frac{204}{20}=10.2$
$\because$ Standard deviation $=2$
$\therefore$ Variance $=(\text { S.D. })^2=2^2=4$
$\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^2=4$
$\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-(10)^2=4$
$\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}=104$
$\Rightarrow \Sigma x_i^2=2080$
Now, replaced ' 8 ' observations by ' 12 '
Then, $\Sigma \mathrm{x}_{\mathrm{i}}^2=2080-8^2+12^2=2160$
$\therefore$ Variance of removing observations
$\begin{aligned}
& \Rightarrow \frac{\Sigma x_i^2}{20}-\left(\frac{\Sigma x_i}{20}\right)^2 \\
& \Rightarrow \frac{2160}{20}-(10.2)^2 \\
& \Rightarrow 108-104.04 \\
& \Rightarrow 3.96
\end{aligned}$
Correct standard deviation
$=\sqrt{3.96}$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (06 Apr Shift 1)
The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If α and β are the mean and standard deviation respectively for correct data, then (α,β) is:

Options:
A (10.5,26)
B (10.5,25)
C (11,25)
D (11,26)

The correct answer is:
(10.5,26)

Solution:
Mean =x1+x2+x3+……+x19+2520=10⇒x1+x2+x3+………+x19+25=200

⇒x1+x2+x3+………+x19=175

New mean =x1+x2+x3+……+x19+3520=175+3520=10.5

S.D. =x12+x22+x32+……+x192+(25)220-(10)2

⇒2.5=x12+x22+x32+……+x192+(25)220-(10)2

⇒x12+x22+x32+……+x192=1500

New S.D. =x12+x22+x32+……+x192+(35)220-(10.5)2

=1500+(35)220-(110.25)

=26

(α,β)=(10.5,26)

Topic: Mathematics, Statistics, JEE Main
The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If α and β are the mean and standard deviation respectively for correct data, then (α,β) is:

Options:
A (10.5,26)
B (10.5,25)
C (11,25)
D (11,26)

The correct answer is:
(10.5,26)

Solution:
Mean =x1+x2+x3+……+x19+2520=10⇒x1+x2+x3+………+x19+25=200

⇒x1+x2+x3+………+x19=175

New mean =x1+x2+x3+……+x19+3520=175+3520=10.5

S.D. =x12+x22+x32+……+x192+(25)220-(10)2

⇒2.5=x12+x22+x32+……+x192+(25)220-(10)2

⇒x12+x22+x32+……+x192=1500

New S.D. =x12+x22+x32+……+x192+(35)220-(10.5)2

=1500+(35)220-(110.25)

=26

(α,β)=(10.5,26)

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (26 Aug Shift 1)
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 is λ49, then the value of λ is equal to

Options:

The correct answer is:
2550

Solution:
x-=âx150=50
ââxi=2500
Correct âxi=2500-99=2401
Correct x-=240149=49
Ï2=âxi2n-x-2
â100=âxi250-2500
ââxi2=130000

Correct âxi2=130000-992=120199
Correct Ï2=âxi2n-x-2
=12019949-492
=120199-11764949=255049=Î»49

Hence, Î»=2550

Topic: Mathematics, Statistics, JEE Main
The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation, then the correct standard deviation is

Options:
A 4
B 6
C 3
D 5

The correct answer is:
5

Solution:
Standard deviation σ=1n∑i=1nxi2-1n2∑i=1nxi2
=1n∑i=1nxi2-x-2

⇒5.1=1100×Incorrect∑i=1nxi2-402
or, 26.01=1100× Incorrect ∑i=1nxi2-1600
Therefore,
Incorrect ∑i=1nxi2=100(26.01+1600)=162601

Correct mean, x ¯=3990100=39.9
Now, Correct ∑i=1nxi2= Incorrect ∑i=1nxi2-502+402
=162601-2500+1600=161701
Therefore correct standard deviation
=Correct∑xi2n-Correctmean2
=161701100-39.92=1617.01-1592.01=25=5

Topic: Mathematics, Statistics, JEE Main
The mean and standard deviation of a distribution of weights of a group of 20 boys are $40 \mathrm{~kg}$ and $5 \mathrm{~kg}$ respectively.
If two boys of weights $43 \mathrm{~kg}$ and $37 \mathrm{~kg}$ are excluded from this group, then the variance of the distribution of weights of the remaining group of boys is

Options:
A 26.18
B 5.27
C 26.78
D 5.17

The correct answer is:
26.78

Solution:
Here, $n=20$
Mean $=40$
Mean $=\frac{\text { Sum of observation }}{n}$
$\Rightarrow 40=\frac{\text { Sum of weights }}{20}$
Sum of weights $=20 \times 40=800$
Now, standard deviation, $\sigma=5$
$\because$ variance $=\sigma^2=5^2=25$
Now, two boys are excluded of weight $43 \mathrm{~kg}$ and $73 \mathrm{~kg}$.
New mean $=\frac{\text { Remaining weight }}{20-2}=\frac{800-43-73}{18}$
$$
\bar{X}_{\text {new }}=\text { new mean }=38
$$
Now, variance $\sigma^2=25$
$$
\begin{aligned}
& \frac{\Sigma x^2}{n}-(\bar{X})^2=25 \Rightarrow \frac{\Sigma x^2}{20}-40^2=25 \\
\Rightarrow \quad & \frac{\Sigma x^2}{20}=1600+25 \Rightarrow \Sigma x^2=1625 \times 20 \\
\Rightarrow \quad & \Sigma x^2=32500
\end{aligned}
$$
When two boys are removed
$$
\Sigma x_{\text {new }}^2=32500-(43)^2-(73)^2=25322 .
$$
So, new variance
$$
\begin{aligned}
& \sigma_{\text {new }}^2=\frac{\sum x_{\text {new }}^2}{18}-\left(\bar{X}_{\text {new }}\right)^2=\frac{25322}{18}-(38)^2 \\
& \sigma_{\text {new }}^2=26.78
\end{aligned}
$$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2018 (05 May Shift 1)
The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is

Options:

The correct answer is:
269

Solution:
Let the observations be x1, x2, x3, ..., x8, 45, 50,

Mean=50

⇒x1+x2+x3+...+x8+45+5010=50

⇒x1+x2+x3+...+x8=405 ....i

Hence, new mean

X¯new=405+20+2510=45

Now,

S.D=∑i=18xi2+452+50210-502

⇒12=∑i=18xi2+452510-2500

⇒∑i=18xi2=21915

Now,

Variancenew=∑i=18xi2+202+25210-452

⇒Variancenew=21915+202+25210-452

⇒Variancenew=2294-2025

⇒Variancenew=269

Hence this is the correct answer.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then yx is equal to:

Options:
A 94
B 73
C 72
D 83

The correct answer is:
73

Solution:
Since, number of observations are even n=10

Hence, the median is =n2th term+n2+1th term2=5th term+6th term2

Thus, median =34+x2

⇒34+x2=35

⇒x=36 ...i

And, mean=sum of termsnumber of terms

⇒10+22+26+29+34+x+42+67+70+y10=42

⇒ x+y+300=420

⇒ x+y=120 ...ii

Put the value of x from i in ii, to get

36+y=120

⇒y=84

⇒yx=8436=73.

Topic: Mathematics, Statistics, JEE Main
The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then yx is equal to:

Options:
A 94
B 73
C 72
D 83

The correct answer is:
73

Solution:
Since, number of observations are even n=10

Hence, the median is =n2th term+n2+1th term2=5th term+6th term2

Thus, median =34+x2

⇒34+x2=35

⇒x=36 ...i

And, mean=sum of termsnumber of terms

⇒10+22+26+29+34+x+42+67+70+y10=42

⇒ x+y+300=420

⇒ x+y=120 ...ii

Put the value of x from i in ii, to get

36+y=120

⇒y=84

⇒yx=8436=73.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The mean and the standard deviation of a data of 8 items are 25 and 5 respectively. If two items 15 and 25 are added to this data, then the variance of the new data is

Options:
A 29
B 24
C 26
D $\sqrt{29}$

The correct answer is:
29

Solution:
We have, $n=8, \bar{x}=25$ and $\sigma=5$
$$
\begin{aligned}
& \bar{x}=\frac{\Sigma x_i}{n} \\
& \Rightarrow \quad \Sigma x_i=n \bar{x}=8 \times 25=200 \\
& \Rightarrow \text { In corrected } \Sigma x_i=200 \\
& \text { and } \quad \sigma=5 \\
& \Rightarrow \quad \sigma^2=25 \\
& \Rightarrow \frac{1}{n} \Sigma x_i^2-(\text { mean })^2=25 \\
& \Rightarrow \quad \frac{\Sigma x_i^2}{8}-625=25 \\
& \Rightarrow \quad \Sigma x_i^2=5200 \\
& \text { Corrected } \Sigma x_i^2=5200+225+625=6050 \\
& \text { and corrected mean }=200+15+25=240 \\
& \therefore \text { Corrected variance }=\frac{6050}{10}-\left(\frac{240}{10}\right)^2 \\
& =605-(24)^2=605-576=29 \\
&
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
The mean and the standard deviation S.D. of five observations are 9 and 0, respectively. If one of the observation is increased such that the mean of the new set of five observations becomes 10, then their S.D. is

Options:
A 0
B 2
C 4
D 1

The correct answer is:
2

Solution:
x¯=∑xin=9

s.d=∑xi-x¯2n=0

⇒xi=x¯∀i

∴ each term in the original observation is 9

⇒ observation 1={9,9,9,9,9}

x¯new =∑xin=9+9+9+9+x55

⇒10=36+x55

⇒x5=14

⇒ new set of observations ={9,9,9,9,14}

s.dnew =∑xi-x¯new2n

=(9-10)2+(9-10)2+(9-10)2+(14-10)25

=4+425

=205=2

Topic: Mathematics, Statistics, JEE Main, JEE Main 2018 (16 Apr Online)
The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by p and then reduced by q, where p≠0 and q≠0. If the new mean and new s.d. become half of their original values, then q is equal to

Options:
A -5
B 10
C -20
D -10

The correct answer is:
-20

Solution:
If each observation is multiplied with p & then q is subtracted, then
New mean x-1=px--q
⇒ 10=p20-q .....1
and new standard deviation
σ2=pσ1 ⇒ 1=p2   ⇒  p=12  ⇒ p=±12
If p=12
then q = 0 (from equation (1))
If p=-12
q=-20

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

Options:
A
$\frac{37}{256}$

B
$\frac{219}{256}$

C
$\frac{128}{256}$

D
$\frac{28}{256}$

The correct answer is:
$\frac{28}{256}$

Solution:
Given that $n \mathrm{p}=4, \mathrm{n} p \mathrm{q}=2 \Rightarrow \mathrm{q}=1 / 2 \Rightarrow \mathrm{p}=1 / 2, \mathrm{n}=8 \Rightarrow \mathrm{p}(\mathrm{x}=2)={ }^8 \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6=\frac{28}{256}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2004
The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3,4 and 4; then the absolute value of the difference of the other two observations, is :

Options:
A 3
B 5
C 7
D 1

The correct answer is:
7

Solution:
Let rest two observations are x and y.

Given mean =4

⇒3+4+4+x+y5=4⇒x+y=9 ...1

Given variance =5.2

⇒9+16+16+x2+y25-42=5.2

⇒x2+y2=65 ...2

Squaring equation 1 and subtracting equation 2, we get

2xy=81-65⇒2xy=16⇒xy=8

∴ x-y2=x2+y2-2xy=65-16=49

∴ x-y=7

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The mean and variance for seven observations are 8 and 16 respectively. If 5 of the observations are 2,4,10,12,14, then the product of the remaining two observations is

Options:
A 48
B 45
C 49
D 40

The correct answer is:
48

Solution:
Let the remaining two observations are x and y.

Mean x¯=∑xiN=2+4+10+12+14+x+y7

⇒ 42+x+y7=8

⇒x+y=14 ...1
Variance,

σ2=∑xi2N-x¯2=4+16+100+144+196+x2+y27-64

⇒σ2=460+x2+y27-64=16

⇒x2+y2=100 …2

From 1 and 2, we get,

x=6, y=8 and xy=48

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1,3,5, then the sum of cubes of the remaining two observations is

Options:
A 1072
B 1792
C 1216
D 1456

The correct answer is:
1072

Solution:
Let the two observations be a and b.

Given:

Mean=5

⇒1+3+5+a+b5=5

⇒a+b=16 …1

And

σ2=∑xi25-∑xi52

⇒8=12+32+52+a2+b25-25

⇒a2+b2=130 …2

Solving 1 & 2, we get

16-b2+b2=130

⇒b=9

So a=7

Hence, a3+b3=343+729=1072

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The mean and variance of 5 observations are 6 and 6.8 respectively. If a number equal to mean is included in the set of observations and the variance of 6 observations is k, then the value of 34k is equal to

Options:

The correct answer is:
6

Solution:
Let 5 observations are x1,x2,x3,x4,x5
x¯old=6=x1+x2+x3+x4+x55
⇒x1+x2+x3+x4+x5=30 ... 1
Var.old=x12+x22+....x525-x¯old2=6.8
=x12+x22+....x52=214 ... 2
An observation 6 is included
x¯new=x1+x +x3+x4+x5+66=6
x12+x22+x32....x52+62=214+36=250
Var.new=2506-62=346

Topic: Mathematics, Statistics, JEE Main
The mean and variance of 7 observations are 7 and 22 respectively. If 5 of the observations are 2,4,10,12,14, then the remaining 2 observations are

Options:
A 4,3
B 2,5
C 6,1
D 4,2

The correct answer is:
6,1

Solution:
∴ Let x and y be remaining 2 observation
Mean =7=2+4+10+12+14+x+y7
⇒x+y=7 ...(i)
Variance =22=1722+42+102+122+142+x2+y2-72
⇒x2+y2=37 ...(ii)
Now solving (i) and (ii), we get,

x=1,y=6 or x=6,y=1

Topic: Mathematics, Statistics, JEE Main
The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2,4,10,12,14 then the absolute difference of the remaining two observations is :

Options:
A 1
B 4
C 2
D 3

The correct answer is:
2

Solution:
Assume, the two remaining observations are x and y

x¯=2+4+10+12+14+x+y7=8

⇒ 42+x+y=56

⇒ x+y=14 ……1

σ2=Σx12n-Σx1n2

16=4+16+100+144+196+x2+y27-82

⇒ 16+64=460+x2+y27

⇒ 560=460+x2+y2

⇒ x2+y2=100 ……2

⇒ xy=48, solving equations 1 & 2

x-y2=x+y2-4xy=4

∴ x-y=2

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, $10,12,14$. Find the remaining two observation.

Options:

None

Solution:
Let the required remaining observations be $x$ and $y$, then Mean $\bar{x}=8=\frac{\sum x_i}{7}$
$\begin{aligned}
&\Rightarrow 8=\frac{2+4+10+12+14+x+y}{7} \\
&\Rightarrow 56=42+x+y \text { or } x+y=14 \quad \ldots(i)
\end{aligned}$
i.e. Variance $=16=\frac{\sum\left(x_i-\bar{x}\right)^2}{n}$
$\Rightarrow 112=236+x^2+y^2-16(x+y)$
$\Rightarrow 112=236+x^2+y^2-16 \times 14$
$\Rightarrow x^2+y^2=100$
i.e. $(x+y)^2=x^2+y^2+2 x y \Rightarrow 14^2=100+2 x y$
i.e. $196-100=2 x y \quad \Rightarrow 2 x y=96$
So, $(x-y)^2=x^2+y^2-2 x y \Rightarrow 100-96=4$
i.e. $x-y=\pm 2 \quad \ldots(ii)$
Solving (i) and (ii), we get
are the remaining observations.

Topic: Mathematics, Statistics
The mean and variance of 7 observations are 8 and 16 respetively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :

Options:
A 925
B 1345
C 1125
D 53625

The correct answer is:
53625

Solution:
Let a, b, c, d, e be 5 unknown observations.

Given that No. of observation, n=7, Mean =8, Variance =16

∴ sum of observations =No. of observation ×Mean=7×8=56

⇒ Mean of 5 remaining observations =56-8-65=425

As, we know that,

Variance for 7 observations=Σxi2n-x¯2

⇒16=Σxi27-64

⇒Σxi2=560

⇒a2+b2+c2+d2+e2+64+36=560

⇒a2+b2+c2+d2+e2=460

Variance of remaining 5 observations =Σxi2n-x¯2=4605-4252=53625

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5,7,10,12,14,15, then the absolute difference of the remaining two observations is :

Options:
A 9
B 5
C 3
D 7

The correct answer is:
7

Solution:
Let remaining two observations are a and b

Σxin=5+7+12+10+15+14+a+b8=10

⇒ 63+a+b=80

⇒ a+b=17

σ2=Σxi2n−Σxin2

⇒13.5=25+49+144+100+225+196+a2+b28−100

⇒908=a2+b2+739

⇒a2+b2=169

⇒a+b2−2ab=169

⇒289−169=2ab

⇒ ab=60

∴ a−b2=a+b2−4ab=289−240=49

∴ a−b=7.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The mean and variance of 10 observations 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

Options:
A 7
B 8
C 9
D 556

The correct answer is:
9

Solution:
x¯old=10=Σxi10⇒Σxiold=100
Σxinew=100-8+18=110
x¯new=11010=11
Varold=4=Σxiold210-x¯old2
Σxiold2=1040
Σxinew2=1040-64+324
=1300
Varnew=130010-112=130-121=9

Topic: Mathematics, Statistics, JEE Main
The mean and variance of 10 observations are found to be 10 and 5 respectively. On rechecking it is found that an observation 5 is incorrect. If the incorrect observation is replaced by 15, then the correct variance is

Options:
A 7
B 8
C 9
D 4

The correct answer is:
4

Solution:
x¯old=10=Σxi10⇒Σxiold=100
Σxinew=100-5+15=110
x¯new=11010=11
Varold=5=Σxiold210-x¯old2
Σxiold2=1050
Σxinew2=1050-25+225
=1250
Varnew=125010-112=125-121=4

Topic: Mathematics, Statistics, JEE Main
The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is _______.

Options:

The correct answer is:
2

Solution:
Given,

n=10 and mean x¯=∑xi10=15

⇒∑i=110xi=150

⇒∑i=19xi+25=150

⇒∑i=19xi=125

Now new sum will be ∑i=19xi+15=140

So, actual mean =14010=14=x¯new

Now finding variance

∑xi210-x¯2=15

⇒∑xi210-152=15

⇒∑i=110xi2=2400

Removing observation of 25 we get,

⇒∑i=19xi2+625=2400

⇒∑i=19xi2=1775

Now adding observation 15 we get,

∑i=19xi2+152=2000=∑xi2actual

Now calculating actual standard deviation we get,

σactual 2=∑xi2actual 10-x¯new 2

=200010-142

=200-196=4

Calculating standard deviation, S.Dactual =4=2

Topic: Mathematics, Statistics, JEE Main
The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is _______.

Options:

The correct answer is:
2

Solution:
Given,

n=10 and mean x¯=∑xi10=15

⇒∑i=110xi=150

⇒∑i=19xi+25=150

⇒∑i=19xi=125

Now new sum will be ∑i=19xi+15=140

So, actual mean =14010=14=x¯new

Now finding variance

∑xi210-x¯2=15

⇒∑xi210-152=15

⇒∑i=110xi2=2400

Removing observation of 25 we get,

⇒∑i=19xi2+625=2400

⇒∑i=19xi2=1775

Now adding observation 15 we get,

∑i=19xi2+152=2000=∑xi2actual

Now calculating actual standard deviation we get,

σactual 2=∑xi2actual 10-x¯new 2

=200010-142

=200-196=4

Calculating standard deviation, S.Dactual =4=2

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (27 Jul Shift 1)
The mean and variance of 20 observations are found to be 10 and 4 respectively. On rechecking, it was found that an observation 8 is incorrect. If the wrong observation is omitted, then the correct variance is

Options:
A 7
B 10019
C 1400361
D 1440361

The correct answer is:
1440361

Solution:
We have n=20, xÂ¯old=10, Varold=4
âxÂ¯=Î£xi20âÎ£xi=200
âÎ£xinew=192, XÂ¯new=1929
Varold=Î£xi220-xÂ¯old2
4=Î£xi220-100
Î£xi2old=2080
Î£xi2new=2080-64=2016
Varnew=201619-192192=1440361

Topic: Mathematics, Statistics, JEE Main
The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11, then the correct variance is

Options:
A 3.99
B 4.01
C 4.02
D 3.98

The correct answer is:
3.99

Solution:
∑xi20=10 … (i)
∑xi220-100=4 … (ii)
∑xi2=104×20=2080
Actual mean =200-9+1120=20220
Variance =2080-81+12120-202202
=212020-10.12=106-102.01=3.99

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11, then the correct variance is

Options:
A 3.99
B 4.01
C 4.02
D 3.98

The correct answer is:
3.99

Solution:
∑xi20=10 … (i)
∑xi220-100=4 … (ii)
∑xi2=104×20=2080
Actual mean =200-9+1120=20220
Variance =2080-81+12120-202202
=212020-10.12=106-102.01=3.99

Topic: Mathematics, Statistics, JEE Main
The mean and variance of a binomial distribution are 8 and 4 respectively. What is $P(X=1)$ equal to?

Options:
A $\frac{1}{2^{12}}$
B $\frac{1}{2^{8}}$
C $\frac{1}{2^{6}}$
D $\frac{1}{2^{4}}$

The correct answer is:
$\frac{1}{2^{12}}$

Solution:
We know that mean and variance of Binomial distribution are $n p$ and $n p q$ respectively therefore $n p=$ 8 and $n p q=4$ On dividing we get
$q=\frac{n p q}{n p}=\frac{4}{8}=\frac{1}{2}$ and $p+q=1 \Rightarrow p=1-\frac{1}{2}=\frac{1}{2}$
$\Rightarrow n\left(\frac{1}{2}\right)=8 \Rightarrow n=16$
We know that $P(X=r)=n_{C_{r}} p^{r} q^{n-r}$
therefore
$P(X=1)={ }^{16} C_{1}\left(\frac{1}{2}\right)^{16-1}\left(\frac{1}{2}\right)^{1}$
$=\frac{16}{2^{15} \cdot 2}=\frac{1}{2^{12}}$

Topic: Mathematics, Probability, JEE Main