Class 11 Maths NCERT Solutions Chapter 13 Miscellaneous Exercise

Let the remaining two observations be x and y.

Step 1: Use the mean to find x + y

Since the mean of all 8 observations is 9:

\dfrac{6+7+10+12+12+13+x+y}{8}=9

60+x+y=72\;\Rightarrow\;x+y=12\quad\ldots(1)

Step 2: Use the variance to find x² + y²

Since \sigma^2=\dfrac{1}{n}\sum x_i^2-\bar{x}^2, rearranging gives \sum x_i^2=n(\sigma^2+\bar{x}^2):

\sum x_i^2=8\times(9.25+81)=8\times90.25=722

The sum of squares of the six known observations:

6^2+7^2+10^2+12^2+12^2+13^2=36+49+100+144+144+169=642

So:

x^2+y^2=722-642=80\quad\ldots(2)

Step 3: Solve for x and y

Squaring (1): (x+y)^2=144\;\Rightarrow\;x^2+y^2+2xy=144. Using (2):

80+2xy=144\;\Rightarrow\;xy=32

So x and y are roots of t^2-12t+32=0:

(t-4)(t-8)=0\;\Rightarrow\;t=4\text{ or }t=8

Let the remaining two observations be x and y.

Step 1: Use the mean to find x + y

Since the mean of all 7 observations is 8:

\dfrac{2+4+10+12+14+x+y}{7}=8

42+x+y=56\;\Rightarrow\;x+y=14\quad\ldots(1)

Step 2: Use the variance to find x² + y²

\sum x_i^2=n(\sigma^2+\bar{x}^2)=7\times(16+64)=7\times80=560

The sum of squares of the five known observations:

2^2+4^2+10^2+12^2+14^2=4+16+100+144+196=460

So:

x^2+y^2=560-460=100\quad\ldots(2)

Step 3: Solve for x and y

Squaring (1): (x+y)^2=196\;\Rightarrow\;x^2+y^2+2xy=196. Using (2):

100+2xy=196\;\Rightarrow\;xy=48

So x and y are roots of t^2-14t+48=0:

(t-6)(t-8)=0\;\Rightarrow\;t=6\text{ or }t=8

When every observation of a data set is multiplied by a constant k, the mean scales by k and the standard deviation scales by |k| (equivalently, the variance scales by k^2). Here k=3, original mean =8, original SD =4:

\text{New mean}=3\times8=24

\text{New SD}=3\times4=12

Let y_i=ax_i for i=1,2,\ldots,n, and let \bar{y} be the mean of the y_i.

Part 1: New mean

\bar{y}=\dfrac{1}{n}\sum_{i=1}^n y_i=\dfrac{1}{n}\sum_{i=1}^n ax_i=\dfrac{a}{n}\sum_{i=1}^n x_i=a\bar{x}

So the mean of the new observations is a\bar{x}, as required.

Part 2: New variance

Let \sigma_y^2 denote the variance of the y_i. By definition:

\sigma_y^2=\dfrac{1}{n}\sum_{i=1}^n(y_i-\bar{y})^2

Substituting y_i=ax_i and \bar{y}=a\bar{x}:

\sigma_y^2=\dfrac{1}{n}\sum_{i=1}^n(ax_i-a\bar{x})^2=\dfrac{1}{n}\sum_{i=1}^n a^2(x_i-\bar{x})^2

=a^2\times\dfrac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2=a^2\sigma^2

Given n=20, incorrect mean =10, incorrect SD =2.

Recover the incorrect sum and sum of squares

\sum x_i=n\bar{x}=20\times10=200

\sum x_i^2=n(\sigma^2+\bar{x}^2)=20\times(4+100)=20\times104=2080

(i) Wrong item omitted

Removing the incorrect value 8, the new number of observations is 19:

\text{New }\sum x_i=200-8=192,\qquad\text{New }\sum x_i^2=2080-8^2=2080-64=2016

\text{New mean}=\dfrac{192}{19}\approx10.11

\text{New variance}=\dfrac{2016}{19}-\left(\dfrac{192}{19}\right)^2=\dfrac{1440}{361}\approx3.99

\text{New SD}=\sqrt{3.99}\approx2.00

(ii) Wrong item replaced by 12

Removing 8 and adding 12 instead, n stays at 20:

\text{New }\sum x_i=200-8+12=204,\qquad\text{New }\sum x_i^2=2080-64+144=2160

\text{New mean}=\dfrac{204}{20}=10.2

\text{New variance}=\dfrac{2160}{20}-(10.2)^2=108-104.04=3.96

\text{New SD}=\sqrt{3.96}\approx1.99

Given n=100, incorrect mean =20, incorrect SD =3.

Recover the incorrect sum and sum of squares

\sum x_i=n\bar{x}=100\times20=2000

\sum x_i^2=n(\sigma^2+\bar{x}^2)=100\times(9+400)=100\times409=40900

Remove the three incorrect observations

Sum and sum of squares of the incorrect values 21,21,18:

21+21+18=60,\qquad21^2+21^2+18^2=441+441+324=1206

With these three observations omitted, the new number of observations is 97:

\text{New }\sum x_i=2000-60=1940,\qquad\text{New }\sum x_i^2=40900-1206=39694

Compute the corrected mean and standard deviation

\text{New mean}=\dfrac{1940}{97}=20

\text{New variance}=\dfrac{39694}{97}-(20)^2=\dfrac{39694}{97}-400=\dfrac{894}{97}\approx9.22

\text{New SD}=\sqrt{9.22}\approx3.04