NCERT Solutions for Miscellaneous Exercise Chapter 15 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, $\bar{X}=\frac{6+7+10+12+12+13+x+y}{8}=9$

60 + x + y = 72

x + y = 12 -(i)

Now, Variance

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

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

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

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

$⟹x^2+y^2=80$ -(ii)

Squaring (i), we get

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

(iii) - (ii) :

2xy = 64 (iv)

Now, (ii) - (iv):

$$\begin{gathered} {x}^2+y^2-2xy=80-64 \\ ⟹(x-y{)}^2=16 \\ ⟹x-y=\pm 4 \end{gathered}$$ -(v)

Hence, From (i) and (v):

x – y = 4 $⟹$ x = 8 and y = 4

x – y = -4 $⟹$ 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 $16$ , respectively. If five of the observations are $2,4,10,12,14$ . 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, $\bar{X}=\frac{2+4+10+12+14+x+y}{7}=8$

42 + x + y = 56

x + y = 14 -(i)

Now, Variance

$=\frac{1}{n}\sum _{i=1}^8(x_i-\bar{x}{)}^2=16$

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

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

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

$⟹x^2+y^2=112-12=100$ -(ii)

Squaring (i), we get

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

(iii) - (ii) :

2xy = 96 (iv)

Now, (ii) - (iv):

$$\begin{gathered} {x}^2+y^2-2xy=100-96 \\ ⟹(x-y{)}^2=4 \\ ⟹x-y=\pm 2 \end{gathered}$$ -(v)

Hence, From (i) and (v):

x – y = 2 $⟹$ x = 8 and y = 6

x – y = -2 $⟹$ 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 $8$ and $4$ , respectively. If each observation is multiplied by $3$ , 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, $\bar{x}=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}=8$

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

$$\begin{gathered} {\bar{y}}_i=3{\bar{x}}_i \\ ⟹{\bar{x}}_i=\frac{{\bar{y}}_i}{3} \end{gathered}$$

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

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

= 24

We know that,

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

$\pi 100=\sqrt{\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^2}$

$$\begin{gathered} \pi 100 \\ ⟹4^2=\frac{1}{6}\sum _{i=1}^6(x_i-\bar{x}{)}^2 \\ ⟹\sum _{i=1}^6(x_i-\bar{x}{)}^2=6\times 16=96 \end{gathered}$$ -(i)

Now, Substituting the values of $x_{i}and\bar{x}$ in (i):

$$\begin{gathered} \pi 100 \\ ⟹\sum _{i=1}^6(\frac{y_i}{3}-\frac{\bar{y}}{3}{)}^2=96 \\ ⟹\sum _{i=1}^6(y_i-\bar{y}{)}^2=96\times 9=864 \end{gathered}$$

Hence, the variance of the new observations = $\frac{1}{6}\times 864=144$

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

Question 4: Given that $\overline{x}$ is the mean and ${\sigma }^2$ is the variance of $n$ observations .Prove that the mean and variance of the observations $a{x}_1,a{x}_2,a{x}_3,....,a{x}_n$ , are $a\overline{x}$ and $a^2{\sigma }^2$ respectively, $(a\ne 0)$ .

Answer:

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

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

$$\begin{gathered} {\bar{y}}_i=a{\bar{x}}_i \\ ⟹{\bar{x}}_i=\frac{{\bar{y}}_i}{a} \end{gathered}$$

$\bar{y}=\frac{1}{n}\sum _{i=1}^n{y}_i=\frac{1}{n}\sum _{i=1}^{n}a{x}_i$

$\bar{y}=a\left[\frac{1}{n}\sum _{i=1}^n{x}_i\right]=a\bar{x}$

Hence the mean of the new observations $a{x}_1,a{x}_2,a{x}_3,....,a{x}_n$ is $a\overline{x}$

We know,

$\pi 100{\sigma }^2=\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

Now, Substituting the values of $x_{i}and\bar{x}$ :

$$\begin{gathered} ⟹{\sigma }^2=\frac{1}{n}\sum _{i=1}^n(\frac{y_i}{a}-\frac{\bar{y}}{a}{)}^2 \\ ⟹a^2{\sigma }^2=\frac{1}{n}\sum _{i=1}^n(y_i-\bar{y}{)}^2 \end{gathered}$$

Hence the variance of the new observations $a{x}_1,a{x}_2,a{x}_3,....,a{x}_n$ is $a^2{\sigma }^2$

Hence proved.

Question 5:(i) 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:

If wrong item is omitted.

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

$\bar{x}=\frac{1}{n}\sum _{i=1}^n{x}_i$

$⟹10=\frac{1}{20}\sum _{i=1}^{20}{x}_i$ $⟹\sum _{i=1}^{20}{x}_i=200$

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,

$\sigma =\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-(\frac{1}{n}\sum _{i=1}^{n}x{)}^2}$

$$\begin{gathered} ⟹2^2=\frac{1}{n}\sum _{i=1}^n{x}_i^2-(\bar{x}{)}^2 \\ ⟹\frac{1}{n}\sum _{i=1}^n{x}_i^2=4+(\bar{x}{)}^2 \\ ⟹\frac{1}{20}\sum _{i=1}^n{x}_i^2=4+100=104 \\ ⟹\sum _{i=1}^n{x}_i^2=2080 \end{gathered}$$ ,which is the incorrect sum.

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

= 2080 – 64

= 2016

Hence, Correct Standard Deviation =

${\sigma }^{\prime }=\sqrt{\frac{1}{n^{\prime }}{\left(\sum _{i=1}^n{x}_i^2\right)}^{\prime }-({\bar{x}}^{\prime }{)}^2}=\sqrt{\frac{2016}{19}-(10.1{)}^2}$

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

=2.02

Question 5:(ii) The mean and standard deviation of $20$ 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:

If it is replaced by $12.$

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

$\bar{x}=\frac{1}{n}\sum _{i=1}^n{x}_i$

$⟹10=\frac{1}{20}\sum _{i=1}^{20}{x}_i$ $⟹\sum _{i=1}^{20}{x}_i=200$

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,

$\sigma =\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-(\frac{1}{n}\sum _{i=1}^{n}x{)}^2}$

$$\begin{gathered} ⟹2^2=\frac{1}{n}\sum _{i=1}^n{x}_i^2-(\bar{x}{)}^2 \\ ⟹\frac{1}{n}\sum _{i=1}^n{x}_i^2=4+(\bar{x}{)}^2 \\ ⟹\frac{1}{20}\sum _{i=1}^n{x}_i^2=4+100=104 \\ ⟹\sum _{i=1}^n{x}_i^2=2080 \end{gathered}$$ ,which is the incorrect sum.

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

= 2080 – 64 + 144

= 2160

Hence, Correct Standard Deviation =

${\sigma }^{\prime }=\sqrt{\frac{1}{n}{\left(\sum _{i=1}^n{x}_i^2\right)}^{\prime }-({\bar{x}}^{\prime }{)}^2}=\sqrt{\frac{2160}{20}-(10.2{)}^2}$

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

= 1.98

Question 6: The mean and standard deviation of a group of $100$ 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,

Initial Number of observations, n = 100

$\bar{x}=\frac{1}{n}\sum _{i=1}^n{x}_i$

$\pi 100⟹20=\frac{1}{100}\sum _{i=1}^{100}{x}_i$ $⟹\sum _{i=1}^{100}{x}_i=2000$

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,

$\sigma =\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-(\frac{1}{n}\sum _{i=1}^{n}x{)}^2}$

$$\begin{gathered} ⟹3^2=\frac{1}{n}\sum _{i=1}^n{x}_i^2-(\bar{x}{)}^2 \\ ⟹\frac{1}{n}\sum _{i=1}^n{x}_i^2=9+(\bar{x}{)}^2 \\ ⟹\frac{1}{100}\sum _{i=1}^n{x}_i^2=9+400=409 \\ ⟹\sum _{i=1}^n{x}_i^2=40900 \end{gathered}$$ ,which is the incorrect sum.

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

= 40900 - 441 - 441 - 324

= 39694

Hence, Correct Standard Deviation =

${\sigma }^{\prime }=\sqrt{\frac{1}{n^{\prime }}{\left(\sum _{i=1}^n{x}_i^2\right)}^{\prime }-({\bar{x}}^{\prime }{)}^2}=\sqrt{\frac{39694}{97}-(20{)}^2}$

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

= 3.036