datavorous/entrance-exam-dataset huggingface dataset:
The mean and standard deviation of $\mathbf{2 0}$ observations are found to be 10 and 2 respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases :
(i) If wrong item is omitted
(ii) If it is replaced by 12

Options:

None

Solution:
$\begin{aligned}
&\text { Mean }=\frac{\sum x_i}{n}=\frac{\sum_{i=1}^{20} x_i}{20}=10 \\
&\Rightarrow \sum_{i=1}^{20} x_i=20 \times 10=200
\end{aligned}$
(i) Since, the wrong item whose observation 8 has been omitted, the correct mean $=\frac{\sum_{i=1}^{20} x-8}{19}$ because on omitting the wrong item, the total number of observations left are 19.
Now, $\frac{\sum_{i=1}^{20}\left(x_i-\bar{x}\right)^2}{n}=\frac{\sum_{i=1}^{20} x_i^2}{n}-\bar{x}^2$
Since, incorrect mean and variance are given as 10 and 4 ,
so $\frac{\sum_{i=1}^{20} x_i^2}{20}-10^2=4$
i.e., $\sum_{i=1}^{20} x_i^2=(4+100) 20=104 \times 20=2080$
Correct variance $=\frac{\sum_{i=1}^{19}\left(x_i-10.11\right)^2}{19}$
i.e., $\frac{\sum_{i=1}^{20} x_i^2-8^2}{19}-(10.11)^2$
$\frac{2080-64}{19}-(10.11)^2=\frac{2016}{19}-(10.11)^2$
$=106.105-102.212=3.893$
Correct S.D $=\sqrt{3.893}=1.97$
(ii) The wrong item which was 8 had been replaced by 12 , as such the correct mean
$=\frac{\sum_{i=1}^{20} x_i-8+12}{20}=\frac{\sum_{i=1}^{20} x_i}{20}+\frac{4}{20}$
$=\frac{200}{20}+\frac{1}{5}=10+0.2=10.2$
Since, incorrect mean and variance are given as 10 and 4 ,
So
$\begin{aligned}
&\frac{\sum_{i=1}^{20} x_i^2}{20}-10^2=4 \Rightarrow \frac{\sum_{i=1}^{20} x_i^2}{20}=4+100=104 \\
&\Rightarrow \sum_{i=1}^{20} x_i^2=20 \times 104 \quad \text { i.e } \sum_{i=1}^{20} x_i^2=2080 \\
&\Rightarrow \text { Correct value of } \sum_i x_i^2=2080-8^2+12^2 \\
&=2080-64+144=2160
\end{aligned}$
Correct variance $=\frac{\sum_{i=1}^{20} x_i^2}{20}-10.2^2$
$=\frac{2160}{20}-(10.2)^2=108-104.04=3.96$
$\therefore \quad$ Correct $\mathrm{SD}=\sqrt{\text { variance }}$
or $\quad \sqrt{3.96}=1.99(\mathrm{appx})$

Topic: Mathematics, Statistics
The mean and standard deviation of 100 items are 50,5 and that of $\begin{array}{ll}150 \text { items are } 40,6 \text { respectively. } & [20] 4-1]\end{array}$
What is the combined mean of all 250 items?

Options:
A 43
B 44
C 45
D 46

The correct answer is:
44

Solution:
Mean of 100 items $=\overline{\mathrm{x}}_{100}=50$
Mean of 150 items $=\overline{\mathrm{x}}_{150}=40$
Standard deviation of 100 items $=\sigma_{100}=5$ Standard deviation of 150 items $=\sigma_{150}=6$
$\begin{aligned} \bar{x}_{250}=\frac{\mathrm{n}_{1} \bar{x}_{100}+\mathrm{n}_{2} \cdot \bar{x}_{150}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{(100 \times 50)+(150 \times 40)}{100+150} \\=& \frac{11000}{250}=44 \end{aligned}$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
The mean and standard deviation of 100 items are 50,5 and that of $\begin{array}{ll}150 \text { items are } 40,6 \text { respectively. } & [20] 4-1]\end{array}$
What is the combined standard deviation of all 250 items?

Options:
A $\quad 7.1$
B 73
C 75
D $7.7$

The correct answer is:
75

Solution:
Mean of 100 items $=\overline{\mathrm{x}}_{100}=50$
Mean of 150 items $=\overline{\mathrm{x}}_{150}=40$
Standard deviation of 100 items $=\sigma_{100}=5$ Standard deviation of 150 items $=\sigma_{150}=6$
$\begin{array}{ll}\mathrm{d}_{1}=50-44=6 & \mathrm{~d}_{1}^{2}=36 \\ \mathrm{~d}_{2}=40-44=-4 & \mathrm{~d}_{2}^{2}=16\end{array}$
$\sigma_{250}=\frac{\sqrt{\mathrm{n}_{1}\left(\sigma_{100}^{2}+\mathrm{d}_{1}^{2}\right)+\mathrm{n}_{2}\left(\sigma_{150}^{2}+\mathrm{d}_{2}^{2}\right)}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
$=\frac{\sqrt{390}}{5}=\frac{37.28}{5}=7.456=7.5$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
The mean and standard deviation of 100 items are 50,5 and that of $\begin{array}{ll}150 \text { items are } 40,6 \text { respectively. } & [20] 4-1]\end{array}$
What is the variance of all 250 items?

Options:
A $50.6$
B $53.3$
C $55.6$
D $59.6$

The correct answer is:
$55.6$

Solution:
Mean of 100 items $=\overline{\mathrm{x}}_{100}=50$
Mean of 150 items $=\overline{\mathrm{x}}_{150}=40$
Standard deviation of 100 items $=\sigma_{100}=5$ Standard deviation of 150 items $=\sigma_{150}=6$
Variance of all 250 items $=\left(\sigma_{250}\right)^{2}=(7.456)^{2}=55.6$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively. Later on it was found that one of the observations was taken as 50 in the place of 40 . If the wrong entry is replaced by the correct one, then the sum of the squares of all the observations is

Options:
A 162701
B 163501
C 162601
D 161701

The correct answer is:
161701

Solution:
$\bar{x}=40$
$\begin{aligned} & \sigma=5.1 \\ & n=100\end{aligned}$
Mean $=\frac{\sum x_i}{n}=40$
$\begin{aligned} & \therefore \sum x_i=4000 \\ & \sigma=5.1 \\ & \operatorname{Var}(x)=(5.1)^2=26.01 \\ & \frac{\sum x_i^2}{100}-(40)^2=26.01 \\ & \therefore \sum x_i^2=162601\end{aligned}$
Incorrect $\sum x_i^2=162601$
Correct $\sum x_i^2=162601-50^2+40^2=161701$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2023 (14 May Shift 2)
The mean and standard deviation of 100 observations $x_1, x_2, \ldots, x_{100}$ 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 value of $\sum_{i=1}^{100} x_i^2=$

Options:
A 3990
B 161701
C 162601
D 4000

The correct answer is:
161701

Solution:
$\begin{aligned} & \text { Given, } x=40, \sigma=5 \cdot 1, n=100 \\ & \qquad \sigma^2=\frac{\Sigma x_i^2}{100}-(\bar{x})^2 \Rightarrow(5 \cdot 1)^2=\frac{\Sigma x_1^2}{100}-(40)^2\end{aligned}$
$\Sigma x_1^2=(26.01+1600) \times 100=2601+1600 \times 100$
The correct value of $\Sigma x_1^2=2601+160000-(50)^2$
$+(40)^2$
$=2601+160000-2500+1600=161701$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 1)
The mean and standard deviation of a binomial distribution are 12 and 2 respectively. What is the number of trials?

Options:
A 2
B 12
C 18
D 24

The correct answer is:
18

Solution:
Mean, $x=n p=12$ ... (i)
Standard deviation
$\Rightarrow \sqrt{\mathrm{npq}}=2$
$\Rightarrow n p q=4$ ... (ii)
$\frac{(\mathrm{ii})}{(\mathrm{i})} \Rightarrow \frac{\not \mathrm{h} \mathrm{pq}}{\not \mathrm{h} \not \mathrm{p}}=\frac{4}{12} \Rightarrow \mathrm{q}=\frac{1}{3}$
$\mathrm{p}=1-\frac{1}{3}=\frac{2}{3}$
Now, (i) $\Rightarrow \mathrm{np}=12$
$\Rightarrow n\left(\frac{2}{3}\right)=12 \Rightarrow 2 n=36 \Rightarrow n=18$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 1)
The mean and standard deviation of a binomial variate $X$ are 4 and $\sqrt{3}$ respectively. Then $P(X \geq 1)$ is equal to

Options:
A $1-\left(\frac{1}{4}\right)^{16}$
B $1-\left(\frac{3}{4}\right)^{16}$
C $1-\left(\frac{2}{3}\right)^{16}$
D $1-\left(\frac{1}{3}\right)^{16}$

The correct answer is:
$1-\left(\frac{3}{4}\right)^{16}$

Solution:
Mean $=n p=4$, variance $=n q p=3$
On solving, we get $q=\frac{3}{4}, n=16, p=\frac{1}{4}$
Now, $\begin{aligned} p(X \geq 1) & =1-p(X=0)=1-{ }^n C_0 p^0 q^{n-0} \\ & =1-\left(\frac{3}{4}\right)^{16}\end{aligned}$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2007
The mean and standard deviation of a binomial variate $X$ are 4 and $\sqrt{3}$ respectively. Then $P(X \geq 1)$ is equal to

Options:
A $1-\left(\frac{1}{4}\right)^{16}$
B $1-\left(\frac{3}{4}\right)^{16}$
C $1-\left(\frac{2}{3}\right)^{16}$
D $1-\left(\frac{1}{3}\right)^{16}$

The correct answer is:
$1-\left(\frac{3}{4}\right)^{16}$

Solution:
Mean $=n p=4$, variance $=n q p=3$
On solving, we get $q=\frac{3}{4}, n=16, p=\frac{1}{4}$
Now, $\begin{aligned} p(X \geq 1) & =1-p(X=0)=1-{ }^n C_0 p^0 q^{n-0} \\ & =1-\left(\frac{3}{4}\right)^{16}\end{aligned}$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2007
The mean and standard deviation of a set $n_1$ observations are $\bar{x}_1$ and $s_1$, respectively while the mean and standard deviation of another set of $n_2$ observations are $\overline{\mathrm{x}}_2$ and $s_2$, respectively. Show that the standard deviation of the
combined set of $\left(n_1+n_2\right)$ observations is given by
$$
\text { S.D. }=\sqrt{\frac{n_1\left(s_1\right)^2+n_2\left(s_2\right)^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}-\bar{x}_2\right)^2}{\left(n_1+n_2\right)}}
$$

Options:

None

Solution:
Let $d_1=\left(\bar{x}_1-\bar{x}_{12}\right), d_2=\left(\bar{x}_2-\bar{x}_{12}\right)$
$\therefore$ Standard deviation (combined set) is given as
$\mathrm{SD}=\sqrt{\frac{n_1\left(s_1^2+d_1^2\right)+n_2\left(s_2^2+d_2^2\right)}{n_1+n_2}}$
$\Rightarrow \quad \mathrm{SD}=\sqrt{\frac{\mathrm{n}_1 \mathrm{~s}_1^2+\mathrm{n}_2 \mathrm{~s}_2^2}{\mathrm{n}_1+\mathrm{n}_2}+\frac{\mathrm{n}_1 \mathrm{~d}_1^2+\mathrm{n}_2 \mathrm{~d}_2^2}{\mathrm{n}_1+\mathrm{n}_2}}$
Now $\frac{n_1 d_1^2+n_2 d_2^2}{n_1+n_2}=\frac{n_1\left(\bar{x}_1-\bar{x}_{12}\right)^2+n_2\left(\bar{x}_2-\bar{x}_{12}\right)^2}{n_1+n_2}$
$=\left(\frac{1}{n_1+n_2}\right)\left[n_1\left\{\bar{x}_1-\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\right\}^2+n_2\left\{\bar{x}_2-\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\right\}^2\right]$
$=\left(\frac{1}{n_1+n_2}\right)\left[n_1\left\{\frac{n_2\left(\bar{x}_1-\bar{x}_2\right)}{n_1+n_2}\right\}^2+n_2\left\{\frac{n_1\left(\bar{x}_2-\bar{x}_1\right)}{n_1+n_2}\right\}^2\right]$
$=\left(\frac{1}{n_1+n_2}\right)\left(\frac{\bar{x}_1-\bar{x}_2}{n_1+n_2}\right)^2\left(n_1 n_2^2+n_2 n_1^2\right)$
$=\left(\frac{1}{n_1+n_2}\right)\left(\frac{\bar{x}_1-\bar{x}_2}{n_1+n_2}\right)^2\left(n_1 n_2\right)\left(n_2+n_1\right)$
$=\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}$, put in (i) we have :
$$
\mathrm{SD}=\sqrt{\frac{n_1 s_1^2+n_2 s_2^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)}}
$$

Topic: Mathematics, Statistics
The mean and standard deviation of a set of values are 5 and 2 respectively. If 5 is added to each value, then what is the coefficient of variation for the new set of values

Options:
A 10
B 20
C 70
D 40

The correct answer is:
20

Solution:
Given, Mean $=5$ Standard deviation $=2$ If 5 is added to each value, mean $=5+5=10$. Standard deviation will not change.
$=\frac{2}{10} \times 100=2 \times 10$
$=20$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
The mean and standard deviation of group of 100 observations were found to be 20 and 3 , respectively, Later on it was found that three observations were incorrect, which were recorded as 21,21 and 18 . Find the mean and standard deviation if the incorrect observations are omitted.

Options:

None

Solution:
$\operatorname{Mean} \bar{x}=\frac{\sum x_i}{n}, \bar{x}=20$
$\therefore \sum x_i=n \times \bar{x}=100 \times 20=2000$
Incorrect value of $\sum x_i=2000$
Correct value of $\sum x_i=2000-21-21-18$
$=1940$
Correct value of mean $=\frac{1940}{97}=20$
$\sigma^2($ incorrect $)=\frac{\sum x_i^2}{n}-(\bar{x})^2$
$\begin{aligned}
&9=\frac{\sum x_i}{100}-(20)^2 \\
&\therefore \sum x_i=409 \times 100=40900
\end{aligned}$
Incorrect value of $\sum x_i=40900$
Correct value of $\sum x_i=40900-(21)^2-(21)^2$
$-(18)^2=40900-441-441-324=39694$
$\begin{aligned}
\therefore \text { Correct } & \sigma^2=\frac{39694}{97}-(20)^2 \\
&=409.22-400=9.22
\end{aligned}$
Correct value of standard deviation $\sigma=3.04$, Thus correct mean $=20$
Correct value of $\sigma=3.04$

Topic: Mathematics, Statistics
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below :
Which of the three subjects shows the height variability in marks and which shows the lowest ? Subject Mathematics

Options:

None

Solution:
Subject Mathematics
Mean $\bar{x}=42$, standard deviation $\sigma=12$
$\therefore$ Coefficient of variaion $=\frac{\sigma}{\bar{x}} \times 100$ or $\frac{12}{42} \times 100=28.56$
Subject Physics
Mean $\bar{x}=32$
Standard deviation, $\sigma=15$
$\therefore$ Coefficient of variation $=\frac{\sigma}{\bar{x}} \times 100$
$=\frac{15}{32} \times 100=46.875$
Subject Chemistry
Mean $\bar{x}=40.9$, Standard deviation $\sigma=20$,
$\therefore \quad$ Coefficient of variation
$=\frac{\sigma}{\bar{x}} \times 100=\frac{20}{40.9} \times 100=48.9$
Highest variability of marks is in Chemistry and it is least in Mathematics.

Topic: Mathematics, Statistics
The mean and standard deviation of six observations are 8 and 4 , respectively. If each observation is multiplied by 3 , find the new mean and new standard deviation of the resulting observations.

Options:

None

Solution:
Let the observation be $x_1, x_2, x_3, x_4, x_5$ and $x_6$
So, their mean $\bar{x}=\frac{\sum_{i=1}^6 x_i}{6}=8 \Rightarrow \sum_{i=1}^6 x_i=48$.
and variance $=\frac{\sum_{i=1}^6\left(x_i-8\right)^2}{6}=(4)^2=16$
$\frac{\sum_{i=1}^6\left(x_i-8\right)^2}{6}=96$
On multiplying each observation by 3 , we get the new observations as $3 x_1, 3 x_2, 3 x_3, 3 x_4, 3 x_5$, and $3 x_6$, Now, their mean
$\bar{x}=\frac{\sum_{i=1}^6 3 x_i}{6}=\frac{3 \sum_{i=1}^6 x_i}{6} \text { i.e. } \bar{x}=\frac{3 \times 48}{6}=24$
Variance of new observations $=\frac{\sum_{i=1}^6\left(3 x_i-24\right)^2}{6}$
$=\frac{3^2 \sum_{i=1}^6\left(x_i-8\right)^2}{6}=\frac{9 \times 96}{6}=144$
Hence, Standard deviation of new observations $=\sqrt{\text { Variance }}=\sqrt{144}=12$

Topic: Mathematics, Statistics
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results :
Number of observations $=25$, mean $=18.2$ seconds, standard deviation $=3.25$ seconds.
Further, another set of 15 observations $\mathrm{x}_1, \mathrm{x}_2, \ldots . ., \mathrm{x}_{15}$, also in seconds, is now available and we have $\sum_{i=1}^{15} x_i=279$ and $\sum_{i=1}^{15} x_i^2=5524$. Calculate the standard deviation based on all 40 observations.

Options:

None

Solution:
$$
\begin{aligned}
&\mathrm{n}_1=25, \overline{\mathrm{x}}_1=18.2, \sigma_1=3.25 \\
&\mathrm{n}_2=15, \overline{\mathrm{x}}_2=\frac{279}{15}=18.6 \\
&\Sigma \mathrm{x}_2=279, \Sigma \mathrm{x}_2^2=5524 \\
&\overline{\mathrm{x}}_{12}=\frac{\mathrm{n}_1 \overline{\mathrm{x}}_1+\mathrm{n}_2 \overline{\mathrm{x}}_2}{\mathrm{n}_1+\mathrm{n}_2} \\
&=\frac{25(18.2)+15(18.6)}{40}=\frac{455+279}{40} \\
&=\frac{734}{40}=18.35 \\
&\sigma_2=\sqrt{\frac{5524}{15}-(18.6)^2}
\end{aligned}
$$
$$
\begin{aligned}
&=\sqrt{368.266-345.960}=\sqrt{22.306}\\
&=4.7229=4.723\\
&\text { Let } \mathrm{d}_1=\overline{\mathrm{x}}_1-\overline{\mathrm{x}}_{12}=18.2-18.35=(-0.15)\\
&\mathrm{d}_2=\overline{\mathrm{x}}_2-\overline{\mathrm{x}}_{12}=18.6-18.35=(0.25)\\
&\therefore \sigma_{12}=\sqrt{\frac{\mathrm{n}_1\left(\sigma_1^2+\mathrm{d}_1^2\right)+\mathrm{n}_2\left(\sigma_2^2+\mathrm{d}_2^2\right)}{\mathrm{n}_1+\mathrm{n}_2}}\\
&=\sqrt{\frac{25\left\{(3.25)^2+(-0.15)^2\right\}+15\left\{(4.723)^2+(0.25)^2\right\}}{40}}\\
&=\sqrt{\frac{25(10.5625+0.0225)+15(22.3067+0.0625)}{40}}\\
&=\sqrt{\frac{264.625+335.538}{40}}=\sqrt{\frac{600.163}{40}}\\
&=\sqrt{15.004}=3.873
\end{aligned}
$$

Topic: Mathematics, Statistics
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, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 1)
The mean and the variance 10 observations are given to be 4 and 2 respectively. If every observation is multiplied by 2 , the mean and the variance of the new series will be respectively

Options:
A 8 and 20
B 8 and 4
C 8 and 8
D 80 and 40

The correct answer is:
8 and 8

Solution:
$\quad$ Mean $(\bar{X})=4$
Variance $(\mathrm{X})=2$
Number of observations $=10$
New average $=\frac{(4 \times 10) \times 2}{10}=8$
New variance $=(2)^{2} \times$ Variance $(\mathrm{X})$ $=4 \times 2=8$
Option (c) is correct.

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 1)
The mean and the variance in a binomial distribution are found to be 2 and 1 respectively. The probability $P(X=0)$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{8}$
D $\frac{1}{16}$

The correct answer is:
$\frac{1}{16}$

Solution:
$\mathrm{nP}=2$ $\mathrm{n} \mathrm{Pq}=1$
$\mathrm{q}=\frac{1}{2}$
$\mathrm{P}+\mathrm{q}=1$
$\mathrm{P}=1-\frac{1}{2}=\frac{1}{2}$
$\mathrm{n}\left(\frac{1}{2}\right)=2$
$\therefore \quad \mathrm{n}=4$
${ }^{4} \mathrm{C}_{0}=\left(\frac{1}{2}\right)^{4-0}=\frac{1}{16} .$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 1)
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, NDA, NDA 2010 (Phase 1)
The mean deviation about the mean for the following data, is
\begin{array}{c|c|c|c|c|c}
\hline \boldsymbol{x}_i & 2 & 4 & 5 & 7 & 9 \\
\hline \boldsymbol{f}_i & 2 & 4 & 10 & 8 & 6 \\
\hline
\end{array}

Options:
A 6.3
B 1.5
C 2.83
D 1.733

The correct answer is:
1.733

Solution:
We have, the following data.
Here, mean
$$
\text { (र) }=\frac{\sum x_1 f_i}{\sum f_i}=\frac{4+16+50+56+54}{2+4+10+8+6}=\frac{180}{30}=6
$$
$\therefore$ Mean deviation about mean $=\frac{\sum f_i\left|x_{\dot{i}}-\bar{x}\right|}{\sum f_{\dot{i}}}$
$$
\begin{aligned}
& =\frac{2 \cdot(4)+4 \cdot(2)+10 \cdot(1)+8 \cdot(1)+6 \cdot(3)}{30} \\
& =\frac{8+8+10+8+18}{30}=\frac{52}{30} \approx \frac{17.33}{10}=1.733
\end{aligned}
$$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2018 (05 May Shift 2)
The mean deviation of the following data about its mean is
\begin{array}{|l|l|l|l|l|l|}
\hlinex_i & 5 & 15 & 25 & 35 & 45 \\
\hlinef_i & 8 & 48 & 56 & 32 & 16 \\
\hline
\end{array}

Options:
A 25
B 12
C 8
D 16

The correct answer is:
8

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 2)
The mean kinetic energy of monoatomic gas molecules under standard conditions is $\left\langle E_1\right\rangle$. If the gas is compressed adiabatically 8 times to its initial volume, the mean kinetic energy of gas molecules changes to $\left\langle E_2\right\rangle$. The ratio $\frac{\left\langle E_2\right\rangle}{\left\langle E_1\right\rangle}$ is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
In adiabatic process,
$$
T V^{\gamma-1}=\text { constant }
$$
In given conditions,
$$
V_1=V\left(\text { say), so } V_2=\left(\frac{V}{8}\right)\right.
$$
and $\gamma=\frac{5}{3}$ (gas is monoatomic)
So, $T_1 V^{\frac{5}{3}-1}=T_2\left(\frac{V}{8}\right)^{\frac{5}{3}-1} \Rightarrow \frac{T_2}{T_1}=\frac{V^{\frac{2}{3}}}{\left(\frac{V}{8}\right)^{\frac{2}{3}}}=8^{\frac{2}{3}}=4$
So, temperature of gas increases by 4 times.
As mean KE of gas molecule is proportional to its temperature, then $\mathrm{KE}$ also increases by 4 times.

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2018 (05 May Shift 2)
The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.

Options:

None

Solution:
Here, $\mathrm{n}_1=60, \overline{\mathrm{x}}_1=650, \sigma_1=8$
$$
\begin{aligned}
\mathrm{n}_2 &=80, \overline{\mathrm{x}}_2=660, \sigma_2=7 \\
\overline{\mathrm{x}}_{12} &=\frac{\mathrm{n}_1 \overline{\mathrm{x}}_1+\mathrm{n}_2 \overline{\mathrm{x}}_2}{\mathrm{n}_1+\mathrm{n}_2} \\
&=\frac{60(650)+80(660)}{60+80}=\frac{91800}{140}=655.71 \\
\mathrm{~d}_1 &=\overline{\mathrm{x}}_1-\overline{\mathrm{x}}_{12}=650-655.71=-5.714 \\
\mathrm{~d}_2 &=\overline{\mathrm{x}}_2-\overline{\mathrm{x}}_{12}=660-655.71=4.29 \\
\sigma_{12} &=\sqrt{\frac{\mathrm{n}_1\left(\sigma_1^2+\mathrm{d}_1^2\right)+\mathrm{n}_2\left(\sigma_2^2+\mathrm{d}_2^2\right)}{\mathrm{n}_1+\mathrm{n}_2}} \\
&=\sqrt{\frac{60(64+32.65)+80(49+18.40)}{60+80}} \\
&=\sqrt{\frac{11191}{140}=\sqrt{79.9}=8.94}
\end{aligned}
$$

Topic: Mathematics, Statistics
The mean of 7 observations is 10 and that of 3 observations is $5 .$ What is the mean of all the 10 observations?

Options:
A 15
B 10
C $8.5$
D $7.5$

The correct answer is:
$8.5$

Solution:
Given mean of 7 observations is $\underline{10}$.
$$
\frac{x_{1}+x_{2}+\ldots .+x_{7}}{7}=10
$$
$$
\Rightarrow x_{1}+x_{2}+\ldots .+x_{7}=70
$$
$(1)$
Also, mean of 3 observations
$$
\frac{x_{8}+x_{9}+x_{10}}{3}=5
$$
$\Rightarrow x_{8}+x_{9}+x_{10}=15$
(2)
So, from $(1)$ and $(2)$ Required mean
$\frac{x_{1}+x_{2}+\ldots+x_{7}+x_{8}+x_{9}+x_{10}}{10}$
$=\frac{70+15}{10}=\frac{85}{10}=8.5$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
The mean of 10 observations is 5 . If 2 is added to each observation and then multiplied by 3 , then what will be the new mean

Options:
A 5
B 7
C $15 \quad$
D 21

The correct answer is:
21

Solution:
Given : Mean of 10 observations is 5 .
Mean of 10 observations is 5
$\Rightarrow \frac{\sum_{i=1}^{10} x_{i}}{10}=5$
According to the Question
$\frac{\sum_{i=1}^{10} 3\left(x_{i}+2\right)}{10}=$ New mean
$\Rightarrow \frac{3 \sum_{i=1}^{10} x_{i}}{10}+\frac{3 \times 2 \times 10}{10}=$ New mean
$\Rightarrow 3 \times 5+6=$ New mean $\quad \Rightarrow 21=$ Newmean

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 2)
The mean of 20 observations is $15 .$ On checking, it was found that two observations were wrongly copied as 3 and 7 wrong observations are replaced by correct value 8 and 4, then the correct mean is

Options:
A 15
B $15.15$
C $\quad 15.35$
D 16

The correct answer is:
$15.15$

Solution:
Sum of all observations $=20 \times 15=300$ Sum of correct observations $=300-(3+6)+(8+4)=303$
Correct mean $=\frac{303}{20}=15.1$ ?

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 2)
The mean of 100 observations is 50 and the standard deviation is 10 . If 5 is subtracted from each observation and then it is divided by 4 , then what will be the new mean and the new standard deviation respectively ?

Options:
A 45,5
B $11.25,1.25$
C $11.25,2.5$
D $12.5,2.5$

The correct answer is:
$11.25,2.5$

Solution:
Number of observations $(\mathrm{N})=100$
Mean $(\bar{x})=50$
Standard deviation $(\sigma)=10$
New mean, $\bar{x}_{1}=\frac{\Sigma \frac{x-5}{4}}{N}=\frac{1}{4}\left(\frac{\Sigma x}{N}-5\right)=\frac{1}{4}(50-5)$
$=\frac{45}{4}=11.25$
New standard deviation $\left(\sigma_{1}\right)=\sigma\left(\frac{x-5}{4}\right)$
$=\frac{1}{4} \sigma(x-5)=\frac{1}{4} \sigma(x)=\frac{10}{4}=2.5$

Topic: Mathematics, Statistics, NDA, NDA 2019 (Phase 1)
The mean of 100 observations is 50 and their standard deviation is 5 . The sum of squares of all observation is

Options:
A
50000

B
250000

C
252500

D
255000

The correct answer is:
252500

Solution:
We have $\mathrm{n}=100, \overline{\mathrm{x}}=50, \mathrm{~s}=5, \mathrm{~s}^2=25$
We know that
$$
\begin{aligned}
& \sigma^2=\frac{\sum x_i^2}{n}-\left(\frac{1}{n} \sum x_i\right)^2 \Rightarrow 25=\frac{\sum x_i^2}{100}-(50)^2 \\
\Rightarrow & 2500=\sum x_i^2-250000 \\
\Rightarrow & \sum x_i^2=252500
\end{aligned}
$$

Topic: Mathematics, Statistics