Class 11 Maths NCERT Solutions Chapter 13 Ex 13.2

Step 1: Find the mean

\bar{x}=\dfrac{6+7+10+12+13+4+8+12}{8}=\dfrac{72}{8}=9

Step 2: Tabulate the squared deviations

Step 3: Compute the variance

\sigma^2=\dfrac{1}{n}\sum(x_i-\bar{x})^2=\dfrac{74}{8}=\dfrac{37}{4}=9.25

Step 1: Find the mean

The first n natural numbers are 1,2,3,\ldots,n. Using the sum formula 1+2+\cdots+n=\dfrac{n(n+1)}{2}:

\bar{x}=\dfrac{1}{n}\times\dfrac{n(n+1)}{2}=\dfrac{n+1}{2}

Step 2: Find the variance

Using the shortcut formula \sigma^2=\dfrac{1}{n}\sum x_i^2-\bar{x}^2, and the standard sum \displaystyle\sum_{i=1}^n i^2=\dfrac{n(n+1)(2n+1)}{6}:

\dfrac{1}{n}\sum x_i^2=\dfrac{1}{n}\times\dfrac{n(n+1)(2n+1)}{6}=\dfrac{(n+1)(2n+1)}{6}

So:

\sigma^2=\dfrac{(n+1)(2n+1)}{6}-\left(\dfrac{n+1}{2}\right)^2

Taking (n+1) common:

\sigma^2=(n+1)\left[\dfrac{2n+1}{6}-\dfrac{n+1}{4}\right]=(n+1)\times\dfrac{2(2n+1)-3(n+1)}{12}

=(n+1)\times\dfrac{4n+2-3n-3}{12}=(n+1)\times\dfrac{n-1}{12}=\dfrac{n^2-1}{12}

The first 10 multiples of 3 are 3,6,9,12,15,18,21,24,27,30 — that is, 3 times the first 10 natural numbers 1,2,\ldots,10.

Step 1: Use the mean and variance of the first 10 natural numbers

Using the result of Question 2 with n=10:

\text{Mean of }1,\ldots,10=\dfrac{10+1}{2}=5.5,\qquad\text{Variance of }1,\ldots,10=\dfrac{10^2-1}{12}=\dfrac{99}{12}=8.25

Step 2: Scale by the common factor 3

When every observation is multiplied by a constant k, the new mean is k times the old mean, and the new variance is k^2 times the old variance. Here k=3:

\bar{x}=3\times5.5=16.5=\dfrac{33}{2}

\sigma^2=3^2\times8.25=9\times8.25=74.25=\dfrac{297}{4}

Verification by direct calculation

\sigma^2=\dfrac{742.5}{10}=74.25 — matching the scaling result above.

Step 1: Set up the working table and find the mean

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{760}{40}=19

(The deviation column above uses this mean of 19 — e.g. for x_i=6: 6-19=-13.)

Step 2: Compute the variance

\sigma^2=\dfrac{1}{\text{N}}\sum f_i(x_i-\bar{x})^2=\dfrac{1736}{40}=\dfrac{217}{5}=43.4

Step 1: Set up the working table and find the mean

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{2200}{22}=100

Step 2: Compute the variance

\sigma^2=\dfrac{1}{\text{N}}\sum f_i(x_i-\bar{x})^2=\dfrac{640}{22}=\dfrac{320}{11}\approx29.09

Take the assumed mean \text{A}=64 and h=1, and let y_i=\dfrac{x_i-\text{A}}{h}=x_i-64.

Step 1: Set up the step-deviation table

Step 2: Compute the mean

\bar{x}=\text{A}+h\times\dfrac{\sum f_iy_i}{\text{N}}=64+1\times\dfrac{0}{100}=64

Step 3: Compute the variance and standard deviation

\sigma^2=\dfrac{h^2}{\text{N}^2}\left[\text{N}\sum f_iy_i^2-\left(\sum f_iy_i\right)^2\right]=\dfrac{1}{10000}\left[100\times286-0^2\right]=\dfrac{28600}{10000}=2.86

\sigma=\sqrt{2.86}\approx1.69

Step 1: Find the mid-point of each class and set up the working table

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{3210}{30}=107

Step 2: Compute the variance

\sigma^2=\dfrac{1}{\text{N}}\sum f_i(x_i-\bar{x})^2=\dfrac{68280}{30}=2276

Step 1: Find the mid-point of each class and set up the working table

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{1350}{50}=27

Step 2: Compute the variance

\sigma^2=\dfrac{1}{\text{N}}\sum f_i(x_i-\bar{x})^2=\dfrac{6600}{50}=132

Take the assumed mean \text{A}=92.5 (the mid-point of the class 90-95) and h=5, and let y_i=\dfrac{x_i-\text{A}}{h}.

Step 1: Set up the step-deviation table

Step 2: Compute the mean

\bar{x}=\text{A}+h\times\dfrac{\sum f_iy_i}{\text{N}}=92.5+5\times\dfrac{6}{60}=92.5+0.5=93

Step 3: Compute the variance and standard deviation

\sigma^2=\dfrac{h^2}{\text{N}^2}\left[\text{N}\sum f_iy_i^2-\left(\sum f_iy_i\right)^2\right]=\dfrac{25}{3600}\left[60\times254-6^2\right]

=\dfrac{25}{3600}\left[15240-36\right]=\dfrac{25\times15204}{3600}=\dfrac{1267}{12}\approx105.58

\sigma=\sqrt{\dfrac{1267}{12}}\approx10.28

Step 1: Convert the data into a continuous frequency distribution

Following the hint, 0.5 is subtracted from every lower limit and added to every upper limit:

Step 2: Set up the working table and find the mean

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{4350}{100}=43.5

Step 3: Compute the variance and standard deviation

\sigma^2=\dfrac{1}{\text{N}}\sum f_i(x_i-\bar{x})^2=\dfrac{3084}{100}=30.84

\sigma=\sqrt{30.84}\approx5.55