Miscellaneous Exercise (Revised) - Chapter 15 - Statistics - Ncert Solutions class 11 - Maths

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Chapter 13 - Statistics | NCERT Solutions for Class 11 Maths

Miscellaneous Exercise Question 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.

Answer.

Let two required observations be $x$ and $y$. Then,
According to question, $\frac{6+7+10+12+12+13+x+y}{8}=9$
$
\begin{aligned}
& \Rightarrow 60+x+y=72 \\
& \Rightarrow x+y=12 \ldots \ldots
\end{aligned}
$
$
\begin{aligned}
& \text { Also } \frac{1}{8}\left[6^2+7^2+10^2+12^2+12^2+13^2+x^2+y^2\right]-9^2=9.25 \\
& \Rightarrow \frac{1}{8}\left[36+49+100+144+144+169+x^2+y^2\right]-9^2=9.25 \\
& \Rightarrow 642+x^2+y^2=722 \\
& \Rightarrow x^2+y^2=80 \text {.......(ii) } \\
& \because(x+y)^2+(x-y)^2=2\left(x^2+y^2\right) \\
& \Rightarrow(12)^2+(x-y)^2=2 \times 80
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow(x-y)^2=16 \\
& \Rightarrow x-y= \pm 4
\end{aligned}
$

When $x-y=4$, then on solving $x+y=12$ and $x-y=4$, we get $x=8$ and $y=4$
When $x-y=-4$, then on solving $x+y=12$ and $x-y=-4$, we get $x=4$ and $y=8$

Miscellaneous Exercise Question 2.

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 observations.

Answer.

Let two required observations be $x$ and $y$. Then,
According to question, $\frac{2+4+10+12+14+x+y}{7}=8$
$
\begin{aligned}
& \Rightarrow 42+x+y=56 \\
& \Rightarrow x+y=14 \ldots \ldots \ldots(\mathrm{i})
\end{aligned}
$

Also $\frac{1}{7}\left[2^2+4^2+10^2+12^2+14^2+x^2+y^2\right]-8^2=16$
$\Rightarrow \frac{1}{7}\left[4+16+100+144+196+x^2+y^2\right]-64=16$
$\Rightarrow 460+x^2+y^2=560$
$\Rightarrow x^2+y^2=100$
$\because(x+y)^2+(x-y)^2=2\left(x^2+y^2\right)$
$\Rightarrow(14)^2+(x-y)^2=2 \times 100$

$
\begin{aligned}
& \Rightarrow(x-y)^2=4 \\
& \Rightarrow x-y= \pm 2
\end{aligned}
$

When $x-y=2$, then on solving $x+y=14$ and $x-y=2$, we get $x=8$ and $y=6$
When $x-y=-2$, then on solving $x+y=14$ and $x-y=-2$, we get $x=6$ and $y=8$
Miscellaneous Exercise Question 3.

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.

Answer.

Let six observations be $x_1, x_2, x_3, x_4, x_5, x_5$, then

According to questions, $\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}=8$
$
\Rightarrow x_1+x_2+x_3+x_4+x_5+x_5=48
$

Now, if each observation is multiplied by 3 , then
new observations are $3 x_1, 3 x_2, 3 x_3, 3 x_4, 3 x_5$ and $3 x_6$
New mean $=\frac{3\left(x_1+x_2+x_3+x_4+x_5+x_6\right)}{6}=\frac{1}{2} \times 48=24$
Also $\frac{1}{6}\left(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2\right)-(8)^2=16$
$
\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2=480
$

Now, if each observation is multiplied by 3 , then
$
\begin{aligned}
& \text { New Variance }=\frac{1}{6} \times(3)^2\left(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2\right)-(24)^2=\frac{9}{6} \times 480-576 \\
& =720-576=144
\end{aligned}
$

Therefore, New S.D. $=\sqrt{144}=12$

Miscellaneous Exercise Question 4.

Given that $\bar{x}$ is the mean and $\sigma^2$ is the variance of $n$ observations $x_1, x_2, \ldots \ldots ., x_n$. Prove that the mean and variance of the observations $a x_1, a x_2, a x_3, \ldots \ldots, \ldots x_n$ are $a \bar{x}$ and $a^2 \sigma^2$ respectively $(a \neq 0)$.

Answer.

Given: $\bar{x}=\frac{x_1+x_2+x_3+x_4+\ldots \ldots \ldots . .+x_n}{n}=\frac{\sum x}{n}$
Also $\frac{x_1^2+x_2^2+x_3^2+x_4^2+\ldots \ldots \ldots . .+x_n^2}{n}=\frac{\sum x^2}{n}$
New Mean =

$
\bar{x}=\frac{a x_1+a x_2+a x_3+a x_4+\ldots \ldots+a x_n}{n}=\frac{a\left(x_1+x_2+x_3+x_4+\ldots \ldots \ldots+x_n\right)}{n}=a \bar{x}
$

Also $\sigma^2=\frac{\sum_{i=1}^n x_i^2}{n}-\left(\frac{\sum_{i=1}^n x_i}{n}\right)^2$
New Variance $=\sigma^{\prime 2}=\frac{\sum_{i=1}^n\left(a x_i\right)^2}{n}-\left(\frac{\sum_{i=1}^n a x_i}{n}\right)^2$
$
\begin{aligned}
& \sigma^{\prime 2}=a^2\left[\frac{\sum_{i=1}^n x_i{ }^2}{n}-\left(\frac{\sum_{i=1}^n x_i}{n}\right)^2\right] \\
& =a^2 \sigma^2
\end{aligned}
$
Miscellaneous Exercise Question 5.

The mean and standard deviation of 20 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 .

Answer.

Given: $n=20, \bar{x}=10$ and $\sigma=2$
Now, Mean $=\frac{\sum x_i}{20}=10$
$
\begin{aligned}
& \Rightarrow \sum x_i=20 \times 10 \\
& \Rightarrow \sum x_i=200
\end{aligned}
$
$\therefore$ Incorrect $\sum x_i=200$
Now $\frac{1}{n} \sum x_i^2-(x)^2=4$

$
\begin{aligned}
& \Rightarrow \frac{1}{20} \sum x_i^2-(10)^2=4 \\
& \Rightarrow \sum x_i^2=2080
\end{aligned}
$
(i) If wrong item is omitted: then we have 19 observations.
$
\begin{aligned}
& \therefore \text { Correct } \sum x_i=\text { Incorrect } \sum x_i-8 \\
& \Rightarrow \text { Correct } \sum x_i=200-8=192 \\
& \therefore \text { Correct Mean }=\frac{192}{19}=10.11
\end{aligned}
$

Also Correct $\sum x_i^2=$ Incorrect $\sum x_i^2-(8)^2=2080-64=2016$
$
\begin{aligned}
& \therefore \text { Correct Variance }=\frac{1}{19}\left(\text { Correct } \sum x_1^2\right)-(\text { Correct mean })^2 \\
& =\frac{1}{19} \times 2016-\left(\frac{192}{19}\right)^2=\frac{2016}{19}-\frac{36864}{361}=\frac{1440}{361} \\
& \text { Correct S.D. }=\sqrt{\frac{1440}{361}}=1.997
\end{aligned}
$
(ii) If it is replaced by 12 : then

$
\begin{aligned}
& \therefore \text { Correct } \sum x_i=\text { Incorrect } \sum x_i-8+12 \\
& \Rightarrow \text { Correct } \sum x_i=200-8+12=204 \\
& \therefore \text { Correct Mean }=\frac{204}{20}=10.2
\end{aligned}
$

Also Correct $\sum x_i^2=$ Incorrect $\sum x_i^2-(8)^2+(12)^2=2080-64+144=2160$

$
\begin{aligned}
& \therefore \text { Correct Variance }=\frac{1}{20}\left(\text { Correct } \sum x_i^2\right)-(\text { Correct mean })^2 \\
& =\frac{2160}{20}-(10.2)^2=108-104.04=3.96 \\
& \text { Correct S.D. }=\sqrt{3.96}=1.989
\end{aligned}
$

Miscellaneous Exercise Question 6.

The mean and standard deviation of a 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.

Answer.

Given: $n=100, \bar{x}=20$ and $\sigma=3$
Now, $\bar{x}=\frac{1}{n} \sum x_i=20$
$
\begin{aligned}
& \Rightarrow \frac{1}{100} \sum x_i=20 \\
& \Rightarrow \sum x_i=2000
\end{aligned}
$
$\therefore$ Incorrect $\sum x_i=2000$
Now $\frac{1}{n} \sum x_i^2-(x)^2=9$
$
\begin{aligned}
& \Rightarrow \frac{1}{100} \sum x_i^2-(20)^2=9 \\
& \Rightarrow \sum x_i^2=40900
\end{aligned}
$

When wrong items 21, 21 and 18 are omitted from the data, then there will be 97observations.
$\therefore$ Correct $\sum x_i=$ Incorrect $\sum x_i-21-21-18$
$\Rightarrow$ Correct $\sum x_i=2000-21-21-18=1940$

$\begin{aligned}
& \therefore \text { Correct Mean }=\frac{1940}{97}=20 \\
& \text { Also Correct } \sum x_i^2=\text { Incorrect } \sum x_i^2-(21)^2-(21)^2-(18)^2 \\
& =40900-441-441-324=39694 \\
& \therefore \text { Correct Variance }=\frac{1}{97}\left(\text { Correct } \sum x_i^2\right)-(\text { Correct mean })^2 \\
& =\frac{1}{97} \times 39694-(20)^2=409.22-400=9.22 \\
& \text { Correct S.D. }=\sqrt{9.22}=3.036
\end{aligned}$