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

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Chapter 13 - Statistics | NCERT Solutions for Class 11 Maths

Example 1

 The variance of 20 observations is 5 . If each observation is multiplied by 2 , find the new variance of the resulting observations.

Solution

Let the observations be $x_1, x_2, \ldots, x_{20}$ and $\bar{x}$ be their mean. Given that variance $=5$ and $n=20$. We know that
Variance $\left(\sigma^2\right)=\frac{1}{n} \sum_{i=1}^{20}\left(x_i-\bar{x}\right)^2$, i.e., $5=\frac{1}{20} \sum_{i=1}^{20}\left(x_i-\bar{x}\right)^2$
or
$
\sum_{i=1}^{20}\left(x_i-\bar{x}\right)^2=100
$

If each observation is multiplied by 2 , and the new resulting observations are $y_{\mathrm{i}}$, then
$
y_{\mathrm{i}}=2 x_{\mathrm{i}} \text { i.e., } x_{\mathrm{i}}=\frac{1}{2} y_i
$

Therefore
$
\bar{y}=\frac{1}{n} \sum_{i=1}^{20} y_i=\frac{1}{20} \sum_{i=1}^{20} 2 x_i=2 \cdot \frac{1}{20} \sum_{i=1}^{20} x_i
$
i.e. $\bar{y}=2 \bar{x} \quad$ or $\quad \bar{x}=\frac{1}{2} \bar{y}$

Substituting the values of $x_i$ and $\bar{x}$ in (1), we get
$
\sum_{i=1}^{20}\left(\frac{1}{2} y_i-\frac{1}{2} \bar{y}\right)^2=100 \text {, i.e., } \sum_{i=1}^{20}\left(y_i-\bar{y}\right)^2=400
$

Thus the variance of new observations $=\frac{1}{20} \times 400=20=2^2 \times 5$

Example 2

The mean of 5 observations is 4.4 and their variance is 8.24 . If three of the observations are 1,2 and 6 , find the other two observations.

Solution

Let the other two observations be $x$ and $y$.
Therefore, the series is $1,2,6, x, y$.
Now
$
\text { Mean } \bar{x}=4.4=\frac{1+2+6+x+y}{5}
$
or
$
22=9+x+y
$

Therefore
$
x+y=13
$

Also
$
\text { variance }=8.24=\frac{1}{n} \sum_{i=1}^5\left(x_i-\bar{x}\right)^2
$
i.e. $8.24=\frac{1}{5}\left[(3.4)^2+(2.4)^2+(1.6)^2+x^2+y^2-2 \times 4.4(x+y)+2 \times(4.4)^2\right]$
or $41.20=11.56+5.76+2.56+x^2+y^2-8.8 \times 13+38.72$
Therefore
$
x^2+y^2=97
$

But from (1), we have
$
x^2+y^2+2 x y=169
$

From (2) and (3), we have
$
2 x y=72
$

Subtracting (4) from (2), we get

or
$
\begin{aligned}
& x^2+y^2-2 x y=97-72 \text { i.e. }(x-y)^2=25 \\
& x-y= \pm 5
\end{aligned}
$

So, from (1) and (5), we get
$
x=9, y=4 \text { when } x-y=5
$
or $\quad x=4, y=9$ when $x-y=-5$
Thus, the remaining observations are 4 and 9 .
Example 3

 If each of the observation $x_1, x_2, \ldots, x_n$ is increased by ' $a$ ', where $a$ is a negative or positive number, show that the variance remains unchanged.
Solution

Let $\bar{x}$ be the mean of $x_1, x_2, \ldots, x_n$. Then the variance is given by
$
\sigma_1^2=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2
$

If ' $a$ is added to each observation, the new observations will be
$
y_i=x_{\mathrm{i}}+a
$

Let the mean of the new observations be $\bar{y}$. Then
$
\begin{aligned}
\bar{y} & =\frac{1}{n} \sum_{i=1}^n y_i=\frac{1}{n} \sum_{i=1}^n\left(x_i+a\right) \\
& =\frac{1}{n}\left[\sum_{i=1}^n x_i+\sum_{i=1}^n a\right]=\frac{1}{n} \sum_{i=1}^n x_i+\frac{n a}{n}=\bar{x}+a \\
\bar{y} & =\bar{x}+a
\end{aligned}
$
i.e.

Thus, the variance of the new observations
$
\sigma_2^2=\frac{1}{n} \sum_{i=1}^n\left(y_i-\bar{y}\right)^2=\frac{1}{n} \sum_{i=1}^n\left(x_i+a-\bar{x}-a\right)^2 \quad \text { [Using (1) and (2)] }
$

$
=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2=\sigma_1{ }^2
$

Thus, the variance of the new observations is same as that of the original observations.

Example 4

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. What are the correct mean and standard deviation?

Solution

Given that number of observations $(n)=100$
Incorrect mean $(\bar{x})=40$,
Incorrect standard deviation $(\sigma)=5.1$

We know that $\quad \bar{x}=\frac{1}{n} \sum_{i=1}^n x_i$
i.e.
$
40=\frac{1}{100} \sum_{i=1}^{100} x_i \text { or } \quad \sum_{i=1}^{100} x_i=4000
$
i.e. $\quad$ Incorrect sum of observations $=4000$

Thus the correct sum of observations $=$ Incorrect sum $-50+40$
$
=4000-50+40=3990
$

Hence
$
\text { Correct mean }=\frac{\text { correct sum }}{100}=\frac{3990}{100}=39.9
$

Also
Standard deviation $\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-\frac{1}{n^2}\left(\sum_{i=1}^n x_i\right)^2}$
$
=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-(\bar{x})^2}
$

i.e.
$
5.1=\sqrt{\frac{1}{100} \times \text { Incorrect } \sum_{i=1}^n x_i^2-(40)^2}
$
or
$
26.01=\frac{1}{100} \times \text { Incorrect } \sum_{i=1}^n x_i^2-1600
$

Therefore $\quad$ Incorrect $\sum_{i=1}^n x_i^2=100(26.01+1600)=162601$

Now
$
\text { Correct } \begin{aligned}
\sum_{i=1}^n x_i^2 & =\text { Incorrect } \sum_{i=1}^n x_i^2-(50)^2+(40)^2 \\
& =162601-2500+1600=161701
\end{aligned}
$

Therefore
Correct standard deviation

$\begin{aligned}
& =\sqrt{\frac{\text { Correct } \sum x_i^2}{n}-(\text { Correct mean })^2} \\
& =\sqrt{\frac{161701}{100}-(39.9)^2} \\
& =\sqrt{1617.01-1592.01}=\sqrt{25}=5
\end{aligned}$