NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 11 - Statistics

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:

Given,

The mean and variance of 8 observations are 9 and 9.25, respectively

Let the remaining two observations be x and y,

Observations: 6, 7, 10, 12, 12, 13, x, y.

∴ Mean, $\overline X = \frac{6+ 7+ 10+ 12+ 12+ 13+ x+ y}{8} = 9$

60 + x + y = 72

x + y = 12 -(i)

Now, Variance

$= \frac{1}{n}\sum_{i=1}^8(x_i - x)^2 = 9.25$

$\implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2 -18(x+y)+ 2.9^2 \right ]$

$\implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2+ -18(12)+ 2.9^2 \right ]$ (Using (i))

$\implies 9.25 = \frac{1}{8}\left[48+x^2+ y^2 -216+ 162 \right ] = \frac{1}{8}\left[x^2+ y^2 - 6 \right ]$

$\implies x^2+ y^2 = 80$ -(ii)

Squaring (i), we get

$x^2+ y^2 +2xy= 144$ (iii)

(iii) - (ii) :

2xy = 64 (iv)

Now, (ii) - (iv):

$\\ x^2+ y^2 -2xy= 80-64 \\ \implies (x-y)^2 = 16 \\ \implies x-y = \pm 4$ -(v)

Hence, From (i) and (v):

x – y = 4 $\implies$ x = 8 and y = 4

x – y = -4 $\implies$ x = 4 and y = 8

Therefore, The remaining observations are 4 and 8. (in no order)

Question 2: The mean and variance of $7$ observations are $8$ and $\small 16$ , respectively. If five of the observations are $\small 2,4,10,12,14$ . Find the remaining two observations.

Answer:

Given,

The mean and variance of 7 observations are 8 and 16, respectively

Let the remaining two observations be x and y,

Observations: 2, 4, 10, 12, 14, x, y

∴ Mean, $\overline X = \frac{2+ 4+ 10+ 12+ 14+ x+ y}{7} = 8$

42 + x + y = 56

x + y = 14 -(i)

Now, Variance

$= \frac{1}{n}\sum_{i=1}^8(x_i - \overline x)^2 = 16$

$\implies 16 = \frac{1}{7}\left[(-6)^2+ (-4)^2+ 2^2+ 4^2+ 6^2+ x^2+ y^2 -16(x+y)+ 2.8^2 \right ]$

$\implies 16 = \frac{1}{7}\left[36+16+4+16+36+ x^2+ y^2 -16(14)+ 2(64) \right ]$ (Using (i))

$\implies 16 = \frac{1}{7}\left[108+x^2+ y^2 -96 \right ] = \frac{1}{7}\left[x^2+ y^2 + 12\right ]$

$\implies x^2+ y^2 = 112- 12 =100$ -(ii)

Squaring (i), we get

$x^2+ y^2 +2xy= 196$ (iii)

(iii) - (ii) :

2xy = 96 (iv)

Now, (ii) - (iv):

$\\ x^2+ y^2 -2xy= 100-96 \\ \implies (x-y)^2 = 4 \\ \implies x-y = \pm 2$ -(v)

Hence, From (i) and (v):

x – y = 2 $\implies$ x = 8 and y = 6

x – y = -2 $\implies$ x = 6 and y = 8

Therefore, The remaining observations are 6 and 8. (in no order)

Question 3: The mean and standard deviation of six observations are $\small 8$ and $\small 4$ , respectively. If each observation is multiplied by $\small 3$ , find the new mean and new standard deviation of the resulting observations.

Answer:

Given,

Mean = 8 and Standard deviation = 4

Let the observations be $x_1, x_2, x_3, x_4, x_5\ and\ x_6$

Mean, $\overline x = \frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} = 8$

Now, Let $y_i$ be the the resulting observations if each observation is multiplied by 3:

$\\ \overline y_i = 3\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{3}$

New mean, $\overline y = \frac{y_1+ y_2+ y_3+ y_4+ y_5+ y_6}{6}$

$= 3\left [\frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} \right] =3\times 8$

= 24

We know that,

Standard Deviation = $\sigma = \sqrt{Variance}$

$\pi{100} =\sqrt{ \frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2}$

$\pi{100}\\ \implies 4^2=\frac{1}{6}\sum_{i=1}^6(x_i - \overline x)^2 \\ \implies \sum_{i=1}^6(x_i - \overline x)^2 = 6\times16 = 96$ -(i)

Now, Substituting the values of $x_i\ and\ \overline x$ in (i):

$\pi{100} \\ \implies \sum_{i=1}^6(\frac{y_i}{3} - \frac{\overline y}{3})^2 = 96 \\ \implies \sum_{i=1}^6(y_i - \overline y)^2 = 96\times9 =864$

Hence, the variance of the new observations = $\frac{1}{6}\times864 = 144$

Therefore, Standard Deviation = $\sigma = \sqrt{Variance}$ = $\sqrt{144}$ = 12

Question 4: Given that $\small \bar {x}$ is the mean and $\small \sigma^2$ is the variance of $\small n$ observations .Prove that the mean and variance of the observations $\small ax_1,ax_2,ax_3,....,ax_n$ , are $\small a\bar{x}$ and $\small a^2 \sigma^2$ respectively, $\small (a \neq 0)$ .

Answer:

Given, Mean = $\small \bar {x}$ and variance = $\small \sigma^2$

Now, Let $y_i$ be the the resulting observations if each observation is multiplied by a:

$\\ \overline y_i = a\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{a}$

$\overline y =\frac{1}{n}\sum_{i=1}^ny_i = \frac{1}{n}\sum_{i=1}^nax_i$

$\overline y = a\left [\frac{1}{n}\sum_{i=1}^nx_i \right] = a\overline x$

Hence the mean of the new observations $\small ax_1,ax_2,ax_3,....,ax_n$ is $\small a\bar{x}$

We know,

$\pi{100} \sigma^2=\frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2$

Now, Substituting the values of $x_i\ and\ \overline x$ :

$\\ \implies \sigma^2= \frac{1}{n}\sum_{i=1}^n(\frac{y_i}{a} - \frac{\overline y}{a})^2 \\ \implies a^2\sigma^2= \frac{1}{n}\sum_{i=1}^n(y_i - \overline y)^2$

Hence the variance of the new observations $\small ax_1,ax_2,ax_3,....,ax_n$ is $a^2\sigma^2$

Hence proved.

Question 5:(i) The mean and standard deviation of $\small 20$ observations are found to be $\small 10$ and $\small 2$ , respectively. On rechecking, it was found that an observation $\small 8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

If wrong item is omitted.

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

$\overline x =\frac{1}{n}\sum_{i=1}^nx_i$

$\implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i$ $\implies \sum_{i=1}^{20}x_i = 200$

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 = 192

Therefore, Correct Mean = (Correct Sum)/19

$=\frac{192}{19}$

= 10.1

Now, Standard Deviation,

$\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}$

$\\ \implies 2^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 4 + (\overline x)^2 \\ \implies \frac{1}{20}\sum_{i=1}^nx_i ^2 = 4 + 100 = 104 \\ \implies \sum_{i=1}^nx_i ^2 = 2080$ ,which is the incorrect sum.

Thus, New sum = Old sum - (8x8)

= 2080 – 64

= 2016

Hence, Correct Standard Deviation =

$\sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2016}{19} - (10.1)^2}$

$\pi{100} = \sqrt{106.1 - 102.01} = \sqrt{4.09}$

=2.02

Question 5:(ii) The mean and standard deviation of $\small 20$ observations are found to be $\small 10$ and $\small 2$ , respectively. On rechecking, it was found that an observation $\small 8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

If it is replaced by $\small 12.$

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

$\overline x =\frac{1}{n}\sum_{i=1}^nx_i$

$\implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i$ $\implies \sum_{i=1}^{20}x_i = 200$

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 + 12 = 204

Therefore, Correct Mean = (Correct Sum)/20

$\pi{100} =\frac{204}{20}$

= 10.2

Now, Standard Deviation,

$\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}$

$\\ \implies 2^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 4 + (\overline x)^2 \\ \implies \frac{1}{20}\sum_{i=1}^nx_i ^2 = 4 + 100 = 104 \\ \implies \sum_{i=1}^nx_i ^2 = 2080$ ,which is the incorrect sum.

Thus, New sum = Old sum - (8x8) + (12x12)

= 2080 – 64 + 144

= 2160

Hence, Correct Standard Deviation =

$\sigma' =\sqrt{\frac{1}{n}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2160}{20} - (10.2)^2}$

$= \sqrt{108 - 104.04} = \sqrt{3.96}$

= 1.98

Question 6: The mean and standard deviation of a group of $\small 100$ observations were found to be $\small 20$ and $\small 3$ , respectively. Later on it was found that three observations were incorrect, which were recorded as $\small 21,21$ and $\small 18$ . Find the mean and standard deviation if the incorrect observations are omitted.

Answer:

Given,

Initial Number of observations, n = 100

$\overline x =\frac{1}{n}\sum_{i=1}^nx_i$

$\pi{100} \implies 20 =\frac{1}{100}\sum_{i=1}^{100}x_i$ $\implies \sum_{i=1}^{100}x_i = 2000$

Thus, incorrect sum = 2000

Hence, New sum of observations = 2000 - 21-21-18 = 1940

New number of observation, n' = 100-3 =97

Therefore, New Mean = New Sum)/100

$=\frac{1940}{97}$

= 20

Now, Standard Deviation,

$\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}$

$\\ \implies 3^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 9 + (\overline x)^2 \\ \implies \frac{1}{100}\sum_{i=1}^nx_i ^2 = 9 + 400 = 409 \\ \implies \sum_{i=1}^nx_i ^2 = 40900$ ,which is the incorrect sum.

Thus, New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18)

= 40900 - 441 - 441 - 324

= 39694

Hence, Correct Standard Deviation =

$\sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{39694}{97} - (20)^2}$

$= \sqrt{108 - 104.04} = \sqrt{3.96}$

= 3.036