datavorous/entrance-exam-dataset huggingface dataset:
The mean and variance of a binomial distribution are α and α3 respectively. If PX=1=4243, then PX=4 or 5 is equal to:

Options:
A 59
B 6481
C 1627
D 145243

The correct answer is:
1627

Solution:
Given mean np=α ⋯1

Also given variance npq=α3 ⋯2

From 1 and 2

q=13 and p=23

Now PX=1=4243

⇒C1nqn-1p1=4243

⇒n3n=2243

⇒n=6

So, P4 or 5=C46234132+C56235·130

=1627

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The mean and variance of a data set comprising 15 observations are 15 and 5 respectively. If one of the observation 15 is deleted and two new observations 6 and 8 are added to the data, then the new variance of resulting data is

Options:
A 10.3715
B 11.8125
C 13.25
D 5.7516

The correct answer is:
11.8125

Solution:
Let 15 observations are x1,x2……x15
x1+x2+x3……x1515=15
If x15=15
x̄new=x1+x2+x3…x15-15+6+816=22416=14
∑xi215-x̄2=5 ⇒x12+x22……x142+(15)2=230×15
⇒x12+x22……x142=3225
New variance = x12+x22……x142+62+8216-x̄new2
=332516-196
= 11.8125

Topic: Mathematics, Statistics, JEE Main
The mean and variance of a random variable X having a binomial probability distribution are 6 and 3 respectively, then the probability PXâ¥2 is

Options:
A 134096
B 40834096
C 31024
D 132048

The correct answer is:
40834096

Solution:
np=6, npq=3
q=12, p=12, n=12
PX≥2=1-PX=0-PX=1
=1-12C01201212-12C11211211
=1-13212=40834096

Topic: Mathematics, Probability, JEE Main
The mean and variance of a random variable $\mathrm{X}$ having binomial distribution are 4 and 2 respectively, then $\mathrm{P}$ $(\mathrm{X}=1)$ is

Options:
A
$\frac{1}{4}$

B
$\frac{1}{32}$

C
$\frac{1}{16}$

D
$\frac{1}{8}$

The correct answer is:
$\frac{1}{32}$

Solution:
$\left.\begin{array}{c}\mathrm{np}=4 \\ \mathrm{npq}=2\end{array}\right\} \Rightarrow \mathrm{q}=\frac{1}{2} \cdot \mathrm{p}=\frac{1}{2}, \mathrm{n}=8$
$\mathrm{p}(\mathrm{X}=1)={ }^8 \mathrm{C}_1\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^7=8 \cdot \frac{1}{2^8}=\frac{1}{2^5}=\frac{1}{32}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2003
The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and σ2 respectively. If the variance of all the 30 numbers in the two sets is 13, then σ2 is equal to

Options:
A 10
B 11
C 9
D 12

The correct answer is:
10

Solution:
Given that,

For set 1:

Mean =12 and Variance =14

For set 2:

Mean =14 and Variance =σ2

Now, For set 1 we have,

⇒∑i=1i=15xi15=12

⇒∑i=1i=15xi=15×12

Also, ∑i=1i=15xi215-∑i=1i=15xi152=14

⇒∑i=1i=15xi215-122=14

⇒∑i=1i=15xi2=14+144×15

Now for set 2:

⇒∑i=1i=15yi15=14

and ⇒∑i=1i=15yi=15×14

Also, ∑i=1i=15yi215-∑i=1i=15yi152=σ2

⇒∑i=1i=15yi215-142=σ2

⇒∑i=1i=15yi2=σ2+196×15

Now,

The combined variance will be

ââââââ⇒13=∑i=1i=15xi2×15+∑i=1i=15yi2×1515+15-∑i=1i=15xi+∑i=1i=15yi15+152

⇒13=14+144×15+σ2+196×1530-132

∴σ2=10

Hence, σ2=10

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The mean and variance of eight observations are 9 and $9.25$ respectively. If six of the observations are $6,7,10,12,12$ and 13 , find the remaining two observations.

Options:

None

Solution:
Let the remaining two observation be $x$ and $y$.
Then Mean, $\bar{x}=9=\frac{\sum x_i}{N}$
$9=\frac{6+7+10+12+12+13+x+y}{8}$
$\Rightarrow 72=60+x+y$ or $x+y=12 \quad \ldots(i)$
Now, Variance $=9.25=\frac{\sum\left(x_i-\bar{x}\right)^2}{8}=\frac{\sum\left(x_i-9\right)^2}{8}$
$=\frac{\begin{array}{r}(6-9)^2+(7-9)^2+(10-9)^2+(12-9)^2 \\ +(12-9)^2+(13-9)^2+(x-9)^2+(y-9)^2\end{array}}{8}$
$=\frac{3^2+2^2+1^2+3^2+3^2+4^2+x^2+y^2-18 x-18 y+162}{8}$
$=9+4+1+9+9+16+x^2+y^2-18(x+y)+162$
i.e. $\frac{37}{4}=\frac{48+x^2+y^2-18 \times 12+162}{8}$
$\Rightarrow 74=210+x^2+y^2-216$
$\Rightarrow 74+216-210=x^2+y^2=80 \Rightarrow x^2+y^2=80 \quad \ldots(ii)$
$\begin{aligned}
&\text { from (i), } \begin{array}{l}
(x+y)^2=144 \\
80+2 x y=144 \quad \ldots \text { (from ii) } \\
2 x y=144-80=64
\end{array} \\
&\text { So, }(x-y)^2=x^2+y^2-2 x y=80-64=16 \\
&\Rightarrow x-\mathrm{y}=\pm 4
\end{aligned}$
On solving equations $x+y=12$ and $x-y=\pm 4$, we have,
We get
and $y=8$ Thus, the remaining terms are 4 and 8

Topic: Mathematics, Statistics
The mean and variance of seven observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12 and 14, then the remaining two observations are

Options:
A 5,7
B 3,5
C 6,8
D 4,2

The correct answer is:
6,8

Solution:
Let x and y be remaining 2 observations
Mean =8=2+4+10+12+14+x+y7
â x+y=14 ...(i)
Variance =16=1722+42+102+122+x2+y2-82
â x2+y2=100 ...(ii)
now solving equation (i) and (ii), we get,

x=8, y=6 or x=6, y=8

Topic: Mathematics, Statistics, JEE Main
The mean and variance of the data 4,5,6,6,7,8,x,y where x<y are 6 and 94 respectively. Then x4+y2 is equal to

Options:
A 320
B 420
C 162
D 674

The correct answer is:
320

Solution:
Mean =4+5+6+6+7+8+x+y8=6

⇒x+y=12 …1

Variance =∑xi2n-x¯2

⇒ 94=42+52+62+62+72+82+x2+y28-62

⇒ x2+y2=80 …2

On solving equations 1 and 2, we get x2+12-x2=80

⇒ x2+144+x2-24x=80

⇒ 2x2-24x+64=0

⇒ x2-12x+32=0

⇒ x-4x-8=0

⇒ x=4 or 8

From equation 2, when x=4, y=8 and when x=8, y=4.

∴ x4+y2=44+82=256+64=320

(or) x4+y2=84+42=4096+16=4112

Topic: Mathematics, Statistics, JEE Main
The mean and variance of the data 4,5,6,6,7,8,x,y where x<y are 6 and 94 respectively. Then x4+y2 is equal to

Options:
A 320
B 420
C 162
D 674

The correct answer is:
320

Solution:
Mean =4+5+6+6+7+8+x+y8=6

⇒x+y=12 …1

Variance =∑xi2n-x¯2

⇒ 94=42+52+62+62+72+82+x2+y28-62

⇒ x2+y2=80 …2

On solving equations 1 and 2, we get x2+12-x2=80

⇒ x2+144+x2-24x=80

⇒ 2x2-24x+64=0

⇒ x2-12x+32=0

⇒ x-4x-8=0

⇒ x=4 or 8

From equation 2, when x=4, y=8 and when x=8, y=4.

∴ x4+y2=44+82=256+64=320

(or) x4+y2=84+42=4096+16=4112

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12 . If the new mean of the marks is 10.2. then their new variance is equal to:

Options:
A 4.04
B 4.08
C 3.96
D 3.92

The correct answer is:
3.96

Solution:
Let the number of observations of the data set is n.

So,

∑xi=10n ....i

Also,

∑xi-8+12=10.2n

⇒10n+4=10.2n

⇒0.2n=4

⇒n=20

Also,

σ2=4=∑xi220-100

⇒∑xi2=104×20=2080

Now, new variance is

=∑xi2-82+12220-10.22

=2080+4×2020-104.04

=108-104.04

=3.96

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The mean deviation about the mean for the following data:

Options:
A $14.33$
B $15.66$
C $18$
D $23.84$

The correct answer is:
$23.84$

Solution:
Mean $(\bar{x})=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{2560}{50}=51.2$
Mean deviation about mean $=\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}$
$\begin{aligned} & \frac{1}{50}\left\{\begin{array}{l}10 \mid(10-51.2|+8| 30-51.2|+12| 50-51.2 \mid \\ +9|70-51.2|+11|90-51.2|\end{array}\right\} \\ & \frac{1}{50}[412+169.6+14.4+169.2+426.8] \\ & =23.84\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
The mean deviation from the median for the following data is

Options:
A $0.75$
B $7.5$
C $0.65$
D $0.40$

The correct answer is:
$0.75$

Solution:
$\begin{aligned} & N=48 \\ & \frac{N}{2}=24\end{aligned}$
$\therefore$ Median $=24$ th observation $=12$
$\begin{aligned} N & =48 \\ \mathrm{MD} & =\frac{\Sigma f_i\left|x_i-M\right|}{N}=\frac{36}{48}=0.75\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
The mean deviation from the median of the data $16,22,3,14,5,10,8,6,11,4$ is

Options:
A 5
B 5.7
C 4.7
D 4

The correct answer is:
4.7

Solution:
Given data, $16,22,3,14,5,10,8,6,11,4$ Arranging is ascending order $3,4,5,6,8,10,11,14,16,22$ Median $=\frac{\left(5^{\text {th }}+6^{\text {th }}\right) \text { observation }}{2}=\frac{8+10}{2}=9$
Mean deviation from the median
$$
\begin{aligned}
& |3-9|+|4-9|+|5-9|+|6-9|+|8-9| \\
= & \frac{+|10-9|+|11-9|+|14-9|+|16-9|+|22-9|}{10} \\
= & \frac{6+5+4+3+1+1+2+5+7+13}{10} \\
= & \frac{47}{10}=4.7
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
The mean deviation of the numbers $a, a+d, a+2 d, \ldots, a+2 n d$ from their mean is equal to

Options:
A $\frac{(n+1) d}{2 n+1}$
B $\frac{n(n+1) d}{2 n+1}$
C $\frac{(n+1)|d|}{2 n}$
D $\frac{n(n+1)|d|}{2 n+1}$

The correct answer is:
$\frac{n(n+1)|d|}{2 n+1}$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, JEE Main
The mean deviation of the series a2 , a2+d , a2+2d, ....., a2+2nd from its median is

Options:
A n+1d2n+1
B nd2n+1
C nn+1d2n+1
D 2n+1dnn+1

The correct answer is:
nn+1d2n+1

Solution:
Median or mean x¯=2n+12⋅a2+a2+2nd2n+1=a2+nd ∵ Mean=Median
Σxi-x¯=2d1+2+....+n=nn+1d
∴ M.D.=nn+1d2n+1

Topic: Mathematics, Statistics, JEE Main
The mean electric energy density between the plates of a charged capacitor is (here $q=$ charge on the capacitor and $A=$ area of the capacitor plate)

Options:
A $\frac{q^2}{2 \varepsilon_0 A^2}$
B $\frac{q}{2 s_0 A^2}$
C $\frac{q^2}{2 \varepsilon_0 A}$
D None of the above

The correct answer is:
$\frac{q^2}{2 \varepsilon_0 A^2}$

Solution:
Energy density
$=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \varepsilon_0\left(\frac{\sigma}{\varepsilon_0}\right)^2=\frac{\sigma^2}{2 \varepsilon_0}=\frac{q^2}{2 \varepsilon_0 A^2}$

Topic: Physics, Capacitance, JEE Main
The mean free path of electrons in a metal is $4 \times 10^{-8} \mathrm{~m}$. The electric field which can give on an average $2 \mathrm{eV}$ energy to an electron in the metal will be in units of $\mathrm{V} / \mathrm{m}$ :

Options:
A $5 \times 10^7$
B $8 \times 10^7$
C $5 \times 10^{-11}$
D $8 \times 10^{-11}$

The correct answer is:
$5 \times 10^7$

Solution:
$$
\begin{aligned}
\mathrm{qV} & =2 \mathrm{eV} \\
\Rightarrow \quad 1.6 \times 10^{-19} \mathrm{~V} & =2 \times 1.6 \times 10^{-19} \mathrm{~V} \\
\Rightarrow \quad \mathrm{V} & =2 \mathrm{~V} \\
\mathrm{E} & =\frac{\mathrm{V}}{\mathrm{d}} \\
\Rightarrow \quad \mathrm{E} & =\frac{2 \mathrm{~V}}{4 \times 10^{-8}}=5 \times 10^7
\end{aligned}
$$

Topic: Chemistry, States of Matter, NEET, NEET 2009 (Mains)
The mean intensity of radiation on the surface of the Sun is about 108W/m2. The rms value of the corresponding magnetic field is closest to:

Options:
A 1T
B 10-2T
C 102T
D 10-4T

The correct answer is:
10-4T

Solution:
I=B0 22μ0×C
B0=2μ0IC
Brms=μ0IC=4π×10-7×1083×108
≈10-4 T

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The mean kinetic energy of a gas at $300 \mathrm{~K}$ is $100 \mathrm{~J}$. The mean energy of the gas at $450 \mathrm{~K}$ is equal to

Options:
A $100 \mathrm{~J}$
B $3000 \mathrm{~J}$
C $450 \mathrm{~J}$
D $150 \mathrm{~J}$

The correct answer is:
$150 \mathrm{~J}$

Solution:
$\frac{E_1}{E_2}=\frac{T_1}{T_2} \Rightarrow E_2=E_1 \frac{T_2}{T_1}=100 \times \frac{450}{300}=150 \mathrm{~J}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The mean life of a radioactive substance is

Options:
A Equal to disintegration constant
B Reciprocal of disintegration constant
C Directly proportional to disintegration constant
D More than one of the above

The correct answer is:
Reciprocal of disintegration constant

Solution:
The mean life $(\tau)$ of a radioactive substance is reciprocal of disintegration constant $\lambda$.
$\tau=\frac{1}{\lambda}$

Topic: Physics, Nuclear Physics, JEE Main
The mean lives of a radioactive substance are 1620 years and 405 years for Î± and Î² emission respectively. If it is decaying by both Î± and Î² emission simultaneously, then the time during which three-fourths of the sample will decay is

Options:
A 643 years
B 449 years
C 528 years
D 279 years

The correct answer is:
449 years

Solution:
Î»=Î»Î±+Î»Î²
=1TÎ±+1TÎ²âµÎ»=1T
=11620+1405 [given, TÎ±=1620 yr and TÎ²=405 yr ]
=51620yr-1
34th sample will decay, i.e. remaining 14th,
N=N012n
N04=N012n
â n=2
â´t=nT12=nlnâ¡2Î»
=2Ã0.69351620=449 yr

Topic: Physics, Nuclear Physics, JEE Main
The mean momentum of a nucleon inside a nucleus with mass number A varies as

Options:
A A3
B A-23
C A-13
D A13

The correct answer is:
A-13

Solution:
pÎ»=constant.
pââ1Î»â1râ1A13

Topic: Physics, Dual Nature of Matter, JEE Main
The mean of \( 5 \) numbers is \( 18 \). If one number is excluded, their mean becomes \( 16 \). Then find the excluded number.

Options:

The correct answer is:
26.00

Solution:
We have, mean of 5 numbers is 18.

Therefore, sum of observations = 5×18=90

Let excluded number be x

Now, 16=90-x4

⇒48=90-x

⇒x=90-64=26

Topic: Mathematics, Statistics, JEE Main
The mean of 5 observations is 5 and their variance is 12.4. If three of the observations are 1,2 & 6; then the value of the remaining two is :

Options:
A 1,11
B 5,5
C 5,11
D None of these

The correct answer is:
5,11

Solution:
x+y+1+2+65=5

⇒x+y=16

Now 12.4=16+9+1+x-52+y-525

⇒62 =26+x-52+y-52

⇒36=x-52+y-52

Shifting of the origin to 5,5 the equations will become

62=x2+y2 and x+y=6

Solving them and again will get the x=0 or y=0.

If x=0⇒y=6 or

If y=0⇒x=6.

But by shifting the origin back to original position, we get x=0+5=5 & y=6+5=11

(or)

x=6+5=11 & y=0+5=5

Hence, the values of the remaining two are 5, 11.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2016 (10 Apr Online)
The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2,4,5 and 7, then the remaining two observations are:

Options:
A 10,11
B 3,18
C 8,13
D 1,20

The correct answer is:
10,11

Solution:
Let the other two observations be a and b, then

a+b+2+4+5+76=6.5

⇒a+b=21

⇒b=21-a

Then, other two numbers are a and 21-a

Now, variance is 10.25, therefore

10.25=4+16+25+49+a2+21-a26-6.52

⇒610.25+66.52=94+a2+21-a2

⇒a2+21-a2=221

∴ a=10 and 21-a=21-10=11

So, remaining two observations are 10,11.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The mean of 10 numbers
7×8,10×10,13×12,16×14,… is

Options:

The correct answer is:
398

Solution:
Given that

7,10,13-- & 8,10,12,--- ∈ A.P

rth term of an A.P ⇒ tr = a +r-1d

Here a = 7, d = 3 & a = 8, d =2

tr =3r +4 & tr =2r +6

∑r=110(3r+4)(2r+6)
=∑r=1106r2+26r+24=6∑r=110r2+26∑r=110r+∑r=11024

∑r=1n1 = n, ∑r=1nr = nn+12, ∑r=1nr2 = nn+12n+16, ∑r=1nr3 = nn+122

=6 10×11×216+2610×112+2410
=2310+1430+240=3980
mean =398

Topic: Mathematics, Statistics, JEE Main
The mean of 10 numbers
7×8,10×10,13×12,16×14,… is

Options:

The correct answer is:
398

Solution:
Given that

7,10,13-- & 8,10,12,--- ∈ A.P

rth term of an A.P ⇒ tr = a +r-1d

Here a = 7, d = 3 & a = 8, d =2

tr =3r +4 & tr =2r +6

∑r=110(3r+4)(2r+6)
=∑r=1106r2+26r+24=6∑r=110r2+26∑r=110r+∑r=11024

∑r=1n1 = n, ∑r=1nr = nn+12, ∑r=1nr2 = nn+12n+16, ∑r=1nr3 = nn+122

=6 10×11×216+2610×112+2410
=2310+1430+240=3980
mean =398

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The mean of 40 observations is 20 and their standard deviation is 5. If the sum of the squares of the observations is k, then the value of k1000 is

Options:

The correct answer is:
17

Solution:
n=40
x¯=20
σ=05
σ2=∑xi240-x¯2
25=∑xi240-400⇒∑xi2=17000

Topic: Mathematics, Statistics, JEE Main
The mean of a data set comprising of 16 observations is 16. If one of the observation value 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data is

Options:
A 14.0
B 16.8
C 16.0
D 15.8

The correct answer is:
14.0

Solution:
Given, ( ∑ i=1 15 x i )+16 16 =16
⇒ ∑i=115xi +16= 256
⇒∑i=115xi=240
Required mean =∑i = 115xi+3+4+518=240+3+4+518
=25218=14

Topic: Mathematics, Statistics, JEE Main, JEE Main 2015 (04 Apr)
The mean of a data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data is

Options:
A 14.0
B 16.8
C 16.0
D 15.8

The correct answer is:
14.0

Solution:
Given, ( â i=1 15 x i )+16 16 =16
â â i=1 15 x i +16= 256
â i=1 15 x i =240
Required mean = â iâ=â1 15 x i +3+4+5 18 = 240+3+4+5 18
= 252 18 = 14

Topic: Mathematics, Statistics, JEE Main
The mean of a data set consisting of 20 observations is 40 . If one observation 53 was wrongly recorded as 33 , then the correct mean will be:

Options:
A
41

B
49

C
$40.5$

D
$42.5$

The correct answer is:
41

Solution:
$$
\text { Correct mean }=\frac{20 \times 40-33+55}{20}=41.1
$$
Nearest option : (a) 41

Topic: Mathematics, Statistics, JEE Main, JEE Main 2013 (09 Apr Online)
The mean of a set of 30 observation is 75. If each observations is multiplied by non-zero number λ and then each of them is decreased by 25, their mean remains the same. Then, λ is equal to :

Options:
A 43
B 13
C 103
D 23

The correct answer is:
43

Solution:
Let 30 observations be x1, x2, .....…, x30.

Given mean =75

⇒x1+ x2+…+x30=75×30

Now according to question,

λ(x1+ x2+…+x30)-25×3030=75×3030

⇒λ75×30=100×30

⇒λ=43

Topic: Mathematics, Statistics, JEE Main, JEE Main 2018 (15 Apr)
The mean of a set of 30 observations is 75 . If each other observation is multiplied by a nonzero number $\lambda$ and then each of them is decreased by 25 , their mean remains the same. The $\lambda$ is equal to
equal to $\{0\}$

Options:
A
$\frac{10}{3}$

B
$\frac{4}{3}$

C
$\frac{1}{3}$

D
$\frac{2}{3}$

The correct answer is:
$\frac{4}{3}$

Solution:
As mean is a linear operation, so if each observation is multiplied by $\lambda$ and decreased by 25 then the mean becomes 75 $\lambda-25$.
According to the question,
$$
75 \lambda-25=75 \Rightarrow \lambda=\frac{4}{3}
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The mean of five numbers is 0 and their variance is 2 . If three of those numbers are -1,1 and 2, then the other two numbers are

Options:
A -5 and 3
B -4 and 2
C -3 and 1
D -2 and 0

The correct answer is:
-2 and 0

Solution:
Let the other two numbers be x and y.
According to the question,
Mean=-1+1+2+x+y5=0⇒x+y=-2…i
Also, σ2=2
⇒-1-02+1-02+2-02+x-02+y-025=2
⇒1+1+4+x2+y2=10⇒x2+y2=4…ii
⇒x+y2-2xy=4
⇒4-2xy=4⇒xy=0…iii
Now, x-y2=x2+y2-2xy=4-0=4 {using ii and iii } …iv
⇒x-y=±2
Solving i and iv , we get,
If x-y=2,x=0,y=-2
If x-y=-2,x=-2,y=0
So, the other two numbers are -2,0

Topic: Mathematics, Statistics, JEE Main
The mean of five observation is 4 and their variance is 5.2. If three of these observations are 1,2 and 6 , then the other two are

Options:
A 2 and 9
B 3 and 8
C 4 and 7
D 5 and 6

The correct answer is:
4 and 7

Solution:

Topic: Mathematics, Statistics, JEE Main
The mean of five observations is 4 and their variance is 2.8. If three of these observations are 2, 2 and 5, then the other two are

Options:
A 2 and 9
B 3 and 8
C 4 and 7
D 5 and 6

The correct answer is:
5 and 6

Solution:
x¯=4,N=5
and Σx-x¯2N=2.8
⇒Σx-x¯2=2.85
∴Σx-x¯2=14
∴2-42+2-42+5-42+α-42+β-42=14
Where α,β are the other two observations.
∴4+4+1+α-42+β-42=14
∴α-42+β-42=5
Also, 2+2+5+α+β5=4
∴α+β=20-9=11
Clearly 5,6 only satisfy the above equations in α,β.
Hence, the required numbers are 5,6.

Topic: Mathematics, Statistics, JEE Main
The mean of five observations is $4$ and their variance is $5.2$. If three of these observations are $1$,$2$ and $6$ , then the other two are

Options:
A
$2$ and $9$

B
$3$ and $8$

C
$4$ and $7$

D
$5$ and $6$

The correct answer is:

$4$ and $7$

Solution:
Let the two unknown items be $x$ and $y$. Then,
$\operatorname{Mean}=4 \Rightarrow \frac{1+2+6+x+y}{5}=4$
$\Rightarrow \quad x+y=11\dots(i)$
and variance $=5.2$
$\Rightarrow \frac{1^2+2^2+6^2+x^2+y^2}{5}-(\text { mean })^2=5.2$
$\Rightarrow \quad 41+x^2+y^2=5(5.2+16)$
$\begin{array}{ll}\Rightarrow & 41+x^2+y^2=106 \\ \Rightarrow & x^2+y^2=65\end{array}$
On solving Eqs. (i) and (ii), we get
$x=4, y=7 \text { or } x=7, y=4$

Topic: Mathematics, Statistics, JEE Main
The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is

Options:
A 10:3
B 4:9
C 6:7
D 5:8

The correct answer is:
4:9

Solution:
Let, the two unknown terms are x & y.

The mean is defined as μ=∑xiN

⇒5=1+3+8+x+y5

⇒25=12+x+y

⇒x+y=13 ...1

And, the variance is defined as σ2=∑xi-μ2N=∑xi2N-μ2

⇒9.2=1+9+64+x2+y25-25

⇒34.2×5=74+x2+y2

⇒171=74+x2+y2

⇒97=x2+y2 ...2

Put the value of y from equation 1 and put in 2, to get

97=x2+13-x2

⇒97=x2+169-26x+x2

⇒2x2-26x+72=0

⇒x2-13x+36=0

⇒x-4x-9=0

⇒x=4 or x=9

And, y=9 or y=4 respectively.

Hence, the ratio of x : y is 4 : 9 or 9 : 4.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The mean of four observations is 3 . If the sum of the squares of these observations is 48 , then their standard deviation is

Options:
A $\sqrt{7}$
B $\sqrt{2}$
C $\sqrt{3}$
D $\sqrt{5}$

The correct answer is:
$\sqrt{3}$

Solution:
Let four observations are be $x_1, x_2, x_3$ and $x_4$
Given, mean
and
$$
\begin{aligned}
(\bar{x}) & =3 \\
\Sigma x_i^2 & =48
\end{aligned}
$$
$\begin{aligned} \therefore \quad \mathrm{SD} & =\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2} \\ =\sqrt{\frac{48}{4}-(3)^2}=\sqrt{12-9} & =\sqrt{3}\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
The mean of n items is x-. If the first item is increased by n, second by n-1 and so on and last by 1, then the new mean is

Options:
A x-+nn+12n+16
B x-+n+12n+16
C x-+n+12
D x-+n+14

The correct answer is:
x-+n+12

Solution:
Let, x1,x2,â¦â¦..xn be n items.
Then, x-=1nâxi .
Let, y1=x1+n,y2=x2+n-1
y3=x3+n-2,â¦â¦.,yn=xn+1
Then, the mean of the new series is
1nâyi=1nâxi+1nn+n-1+n-2+.â¦â¦.1
= x Â¯ + 1 n â n n+1 2 = x Â¯ + n+1 2

Topic: Mathematics, Statistics, JEE Main
The mean of n observations is XÂ¯. If the first observation is increased by 12, second by 22 and so on, then the new mean is

Options:
A X¯+nn+12n+16
B X¯+n+12n+16
C X¯+n+12
D X¯+n+14

The correct answer is:
X¯+n+12n+16

Solution:
Let, x1,x2,.....xn be n observations.
Then, XÂ¯=1nÎ£xi.
Let, y1=x1+12, y2=x2+22,
y3=x3+32,....,yn=xn+n2
Then, the mean of the new series is
1nÎ£yi=1nÎ£xi+i2=1nÎ£xi+1n12+22+32+....+n2
=XÂ¯+1nânn+12n+16=XÂ¯+n+12n+16

Topic: Mathematics, Statistics, JEE Main
The mean of the coefficients of x,x2,……,x7 in the binomial expression of (2+x)9 is _________

Options:

The correct answer is:
2736

Solution:
We know that.

Binomial expansion of 2+x9 is given by,

2+x9=C09x029+C19x128+...........C99x920

Now put x=1 we get,

2+19=C0929+C1928++C2927+...........C9920

⇒39=C0929+C1928++C2927+...........C9920

So, sum of coefficient of x, x2, x3, ........., x7 will be,

⇒C1928+C2927+...........C7922=39-C0929-C8921-C9920

⇒C1928+C2927+...........C7922=39-29-18-1

⇒C1928+C2927+...........C7922=19683-512-19

⇒C1928+C2927+...........C7922=19152

Now mean is given by 191527=2736

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The mean of the numbers $a, b, 8,5,10$ is 6 and the variance is $6.80$. Then which one of the following gives possible values of $a$ and $b$ ?

Options:
A
$a=0, b=7$

B
$a=5, b=2$

C
$a=1, b=6$

D
$a=3, b=4$

The correct answer is:
$a=3, b=4$

Solution:
Mean of $a, b, 8,5,10$ is 6
$$
\begin{aligned}
& \Rightarrow \frac{a+b+8+5+10}{5}=6 \\
& \Rightarrow a+b=7
\end{aligned}
$$
Given that Variance is $6.8$
$$
\begin{aligned}
& \therefore \text { Variance }=\frac{\sum\left(\mathrm{X}_{\mathrm{i}}-\mathrm{A}\right)^2}{\mathrm{n}} \\
& =\frac{(\mathrm{a}-6)^2+(\mathrm{b}-6)^2+4+1+16}{5}=6.8
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow a^2+b^2=25 \\
& a^2+(7-a)^2=25 \\
& \Rightarrow a^2-7 a+12=0 \\
& \therefore a=4,3 \text { and } b=3,4
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2008
The mean of the numbers a,b,8,5,10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25M is equal to

Options:
A 60
B 55
C 50
D 75

The correct answer is:
60

Solution:
Mean x¯=a+6+8+5+105=6 (given)

⇒a+b=7 ...i

Variance =∑xi2n-x¯2=a2+b2+82+52+1025-62=6.8 (given)

⇒a2+b2=25 ...ii

on solving equations i & ii, we get a=3, b=4

Now, Mean deviation about Mean M=∑xi-x¯n

⇒3-6+4-6+8-6+5-6+10-65

⇒ 3+2+2+1+45=125

∴ M=125⇒25M=60.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (26 Jun Shift 1)