NCERT Solutions for Class 11 Maths Chapter 13 Statistics

Question 1: Find the mean deviation about the mean for the data. $\small 4,7,8,9,10,12,13,17$

Answer:

Mean ($\overline{x}$) of the given data:

$\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10$

The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are
6, 3, 2, 1, 0, 2, 3, 7.

$\therefore$ $\sum_{i=1}^{8}|x_i - 10| = 24$

$\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|= \frac{24}{8} = 3$

Hence, the mean deviation from the mean is 3.

Question 2: Find the mean deviation about the mean for the data. $\small 38,70,48,40,42,55,63,46,54,44$

Answer:

Mean ( $\overline{x}$ ) of the given data:

$\\ \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{38+70+48+40+42+55+63+46+54+44}{10} \\ = \frac{500}{10} = 50$

The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6.

$\therefore$ $\sum_{i=1}^{8}|x_i - 50| = 84$

$\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|= \frac{84}{10} = 8.4$

Hence, the mean deviation from the mean is 8.4.

Question 3: Find the mean deviation about the median. $\small 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$

Answer:

Number of observations, n = 12, which is even.

Arranging the values in ascending order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

Now, Median (M)

$\\ = \frac{(\frac{12}{2})^{th} \text{observation} + (\frac{12}{2}+ 1)^{th} \text{observation}}{2} \\ = \frac{13 + 14}{2} = \frac{27}{2}= 13.5$

The respective absolute values of the deviations from the median, $|x_i - M|$ are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

$\therefore$ $\sum_{i=1}^{8}|x_i - 13.5| = 28$

$\therefore$ $M.D.(M) = \frac{1}{12}\sum_{i=1}^{n}|x_i - M|$

$= \frac{28}{12} = 2.33$

Hence, the mean deviation from the median is 2.33.

Question 4: Find the mean deviation about the median. $\small 36, 72, 46, 42, 60, 45, 53, 46, 51, 49$

Answer:

Number of observations, n = 10, which is even.

Arranging the values in ascending order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Now, Median (M)

$\\ = \frac{(\frac{10}{2})^{th} \text{observation} + (\frac{10}{2}+ 1)^{th} \text{observation}}{2} \\ = \frac{46 + 49}{2} = \frac{95}{2}= 47.5$

The respective absolute values of the deviations from the median, $|x_i - M|$ are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

$\therefore$ $\sum_{i=1}^{8}|x_i - 47.5| = 70$

$\therefore$ $M.D.(M) = \frac{1}{10}\sum_{i=1}^{n}|x_i - M|$

$= \frac{70}{10} = 7$

Hence, the mean deviation from the median is 7.

Question 5: Find the mean deviation from the mean.

$\small \\x_i\hspace{1cm}5\hspace{1cm}10\hspace{1cm}15\hspace{1cm}20\hspace{1cm}25$
$ f_i\hspace{1cm}7\hspace{1.1cm}4\hspace{1.1cm}6\hspace{1.1cm}3\hspace{1.2cm}5$

Answer:

$N = \sum_{i=1}^{5}{f_i} = 25 ; \sum_{i=1}^{5}{f_ix_i} = 350$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{350}{25} = 14$

Now, we calculate the absolute values of the deviations from the an, $|x_i - \overline{x}|$, and

$\sum f_i|x_i - \overline{x}|$ = 158

$\therefore$ $M.D.(\overline{x}) = \frac{1}{25}\sum_{i=1}^{n}f_i|x_i - \overline{x}|$

$= \frac{158}{25} = 6.32$

Hence, the mean deviation from the mean is 6.32.

Question 6: Find the mean deviation from the mean.

$\small \\x_i\hspace{1cm}10\hspace{1cm}30\hspace{1cm}50\hspace{1cm}70\hspace{1cm}90$
$\small f_i\hspace{1cm}4\hspace{1.1cm}24\hspace{1.1cm}28\hspace{1.1cm}16\hspace{1.2cm}8$

Answer:

$N = \sum_{i=1}^{5}{f_i} = 80 ; \sum_{i=1}^{5}{f_ix_i} = 4000$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{4000}{80} = 50$

Now, we calculate the absolute values of the deviations from the an, $|x_i - \overline{x}|$, and

$\sum f_i|x_i - \overline{x}|$ = 1280

$\therefore$ $M.D.(\overline{x}) = \frac{1}{80}\sum_{i=1}^{5}f_i|x_i - \overline{x}|$

$= \frac{1280}{80} = 16$

Hence, the mean deviation from the mean is 16.

Question 7: Find the mean deviation about the median.

Answer:

Now, N = 26 which is even.

Median is the mean of $13^{th}$ and $14^{th}$ observations.

Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Therefore, Median, M $= \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7$

Now, we calculate the absolute values of the deviations from the median, $|x_i - M|$, and

$\sum f_i|x_i - M|$ = 84

$\therefore$ $M.D.(M) = \frac{1}{26}\sum_{i=1}^{6}|x_i - M|$

$= \frac{84}{26} = 3.23$

Hence, the mean deviation from the median is 3.23.

Question 8: Find the mean deviation about the median.

Answer:

Now, N = 30, which is even.

Median is the mean of $15^{th}$ and $16^{th}$ observations.

Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore, Median, M $= \frac{15^{th} \text{observation} + 16^{th} \text{observation}}{2} = \frac{30 + 30}{2} = 30$

Now, we calculate the absolute values of the deviations from the median, $|x_i - M|$, and

$\sum f_i|x_i - M|$ = 149.5

$\therefore$ $M.D.(M) = \frac{1}{29}\sum_{i=1}^{5}|x_i - M|$

$= \frac{149.5}{29} = 5.1$

Hence, the mean deviation from the median is 5.1.

Question 9: Find the mean deviation from the mean.

Answer:

$N = \sum_{i=1}^{8}{f_i} = 50 ; \sum_{i=1}^{8}{f_ix_i} = 17900$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{8}f_ix_i = \frac{17900}{50} = 358$

Now, we calculate the absolute values of the deviations from the mean n, $|x_i - \overline{x}|$, and

$\sum f_i|x_i - \overline{x}|$ = 7896

$\therefore$ $M.D.(\overline{x}) = \frac{1}{50}\sum_{i=1}^{8}f_i|x_i - \overline{x}|$

$= \frac{7896}{50} = 157.92$

Hence, the mean deviation from the mean is 157.92.

Question 10: Find the mean deviation from the mean.

Answer:

$N = \sum_{i=1}^{6}{f_i} = 100 ; \sum_{i=1}^{6}{f_ix_i} = 12530$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{6}f_ix_i = \frac{12530}{100} = 125.3$

Now, we calculate the absolute values of the deviations from the mean, $|x_i - \overline{x}|$, and

$\sum f_i|x_i - \overline{x}|$ = 1128.8

$\therefore$ $M.D.(\overline{x}) = \frac{1}{100}\sum_{i=1}^{6}f_i|x_i - \overline{x}|$

$= \frac{1128.8}{100} = 11.29$

Hence, the mean deviation from the mean is 11.29

Question 11: Find the mean deviation about the median for the following data :

Answer:

Now, N = 50, which is even.

The class interval containing $\left (\frac{N}{2} \right)^{th}$ or $25^{th}$ item is 20-30. Therefore, 20-30 is the median class.

We know,

Median $= l + \frac{\frac{N}{2}- C}{f}\times h$

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median $= 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85$

Now, we calculate the absolute values of the deviations from the median, $|x_i - M|$, and

$\sum f_i|x_i - M|$ = 517.1

$\therefore$ $M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|$

$= \frac{517.1}{50} = 10.34$

Hence, the mean deviation from the median is 10.34

Question 12: Calculate the mean deviation about the median age for the age distribution of $\small 100$ persons given below:

[Hint: Convert the given data into continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval]

Answer:

Now, N = 100, which is even.

The class interval containing $\left (\frac{N}{2} \right)^{th}$ or $50^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median $= l + \frac{\frac{N}{2}- C}{f}\times h$

Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

Therefore, Median $= 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$

Now, we calculate the absolute values of the deviations from the median, $|x_i - M|$, and

$\sum f_i|x_i - M|$ = 735

$\therefore$ $M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|$

$= \frac{735}{100} = 7.35$

Hence, the mean deviation from the median is 7.35.

Question 1: Find the mean and variance for each of the data.

$\small 6, 7, 10, 12, 13, 4, 8, 12$

Answer:

Mean ( $\overline{x}$ ) of the given data:

$\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{6+ 7+ 10+ 12+ 13+ 4+ 8+ 12}{8} = \frac{72}{8} = 9$

The respective values of the deviations from mean, $(x_i - \overline{x})$ are

-3, -2, 1, 3, 4, -5, -1, 3

$\therefore$ $\sum_{i=1}^{8}(x_i - \overline{x})^2 = 74$

$\therefore$ Variance, $\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

So, $\frac{1}{8}\sum_{i=1}^{8}(x_i - \overline{x})^2= \frac{74}{8} = 9.25$

Hence, Mean = 9 and Variance = 9.25

Question 2: Find the mean and variance for each of the data:

First natural numbers.

Answer:

Mean ( $\overline{x}$ ) of first n natural numbers:

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

We know, Variance $\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}\left (x_i - \frac{n+1}{2} \right )^2$

We know that $(a-b)^2 = a^2 - 2ab + b^2$

$\\ \therefore n\sigma^2 = \sum_{i=1}^{n}x_i^2 + \sum_{i=1}^{n}(\frac{n+1}{2})^2 - 2\sum_{i=1}^{n}x_i\frac{n+1}{2} \\ = \frac{n(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4}\times n - 2.\frac{(n+1)}{2}.\frac{n(n+1)}{2}$

$⇒ \sigma^2 = \frac{(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4} - \frac{(n+1)^2}{2}$

$⇒ \sigma^2= \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \\ = (n+1)\left [\frac{4n+2 - 3n -3}{12} \right ] \\ = (n+1).\frac{(n-1)}{12} \\ = \frac{n^2-1}{12}$

Hence, Mean = $\frac{n+1}{2}$ and Variance = $\frac{n^2-1}{12}$

Question 3: Find the mean and variance for each of the data:

First 10 multiples of 3.

Answer:

The first 10 multiples of 3 are:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Mean ( $\overline{x}$ ) of the above values:

$\\ \overline{x} = \frac{1}{10}\sum_{i=1}^{10}x_i = \frac{3+ 6+ 9+ 12+ 15+ 18+ 21+ 24+ 27+ 30}{10} \\ = 3.\frac{\frac{10(10+1)}{2}}{10} = 16.5$

The respective values of the deviations from mean, $(x_i - \overline{x})$ are:

-13.5, -10.5, -7.5, -4.5, -1.5, 1.5, 4.5, 7.5, 10.5, 13.5

$\therefore$ $\sum_{i=1}^{10}(x_i - \overline{x})^2 = 742.5$

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

So, $\frac{1}{10}\sum_{i=1}^{10}(x_i - \overline{x})^2= \frac{742.5}{10} = 74.25$

Hence, Mean = 16.5 and Variance = 74.25

Question 4: Find the mean and variance for each of the data.

Answer:

$N = \sum_{i=1}^{7}{f_i} = 40 ; \sum_{i=1}^{7}{f_ix_i} = 760$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{760}{40} = 19$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$⇒ \sigma^2 = \frac{1736}{40} = 43.4$

Hence, Mean = 19 and Variance = 43.4

Question 5: Find the mean and variance for each of the data.

Answer:

$N = \sum_{i=1}^{7}{f_i} = 22 ; \sum_{i=1}^{7}{f_ix_i} = 2200$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{2200}{22} = 100$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$⇒ \sigma^2 = \frac{640}{22} = 29.09$

Hence, Mean = 100 and Variance = 29.09

Question 6: Find the mean and standard deviation using the shortcut method.

Answer:

Let the assumed mean, A = 64 and h = 1

$N = \sum_{i=1}^{9}{f_i} = 100 ; \sum_{i=1}^{9}{f_iy_i} = 0$

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =64 + \frac{0}{100} = 64$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$⇒ \sigma^2 = \frac{1}{(100)^2}\left [100(286) - (0)^2 \right ] \\ = \frac{28600}{10000} = 2.86$

We know, Standard Deviation = $\sigma = \sqrt{\text{Variance}}$

$\therefore \sigma = \sqrt{2.86} = 1.691$

Hence, Mean = 64 and Standard Deviation = 1.691

Question 7: Find the mean and variance for the following frequency distributions.

Answer:

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{3210}{30} = 107$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$⇒ \sigma^2 = \frac{68280}{30} = 2276$

Hence, Mean = 107 and Variance = 2276

Question 8: Find the mean and variance for the following frequency distributions.

Answer:

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{1350}{50} = 27$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$⇒ \sigma^2 = \frac{6600}{50} = 132$

Hence, Mean = 27 and Variance = 132

Question 9: Find the mean, variance, and standard deviation using the short-cut method.

Answer:

Let the assumed mean, A = 92.5 and h = 5

Mean,

$\overline{y} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =92.5 + \frac{6}{60}\times5 = 93$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$⇒ \sigma^2 = \frac{1}{(60)^2}\left [60(254) - (6)^2 \right ] \\ = \frac{1}{(60)^2}\left [15240 - 36 \right ] \\ = \frac{15204}{144} = 105.583$

We know, Standard Deviation = $\sigma = \sqrt{\text{Variance}}$

$\therefore \sigma = \sqrt{105.583} = 10.275$

Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

Question 10: The diameters of circles (in mm) drawn in a design are given below:

Calculate the standard deviation and mean diameter of the circles.
[Hint: First, make the data continuous by making the classes as $32.5-36.5,36.5-40.4,40.5-44.5,44.5-48.5,48.5-52.5$ and then proceed.]

Answer:

Let the assumed mean, A = 92.5 and h = 5

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =42.5 + \frac{25}{100}\times4 = 43.5$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\begin{aligned} & ⇒ \sigma^2=\frac{1}{(100)^2}\left[100(199)-(25)^2\right] \times 4^2 \\ & ⇒ \sigma^2=\frac{1}{625}[19900-625] \\ &⇒ \sigma^2 =\frac{19275}{625}=30.84\end{aligned}$

We know, Standard Deviation = $\sigma = \sqrt{\text{Variance}}$

$\therefore \sigma = \sqrt{30.84} = 5.553$

Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553

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

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

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

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

$⇒ x^2+ y^2 = 80$ -------(ii)

Squaring (i), we get,

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

(iii) - (ii):

2xy = 64 ---------------(iv)

Now, (ii) - (iv):

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

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

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

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

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

$⇒ x^2+ y^2 = 112- 12 =100$ ----------(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$
$⇒(x-y)^2 = 4$
$⇒ x-y = \pm 2$ ----------(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 $\small 8$ 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 rating observations if each observation is multiplied by 3:

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

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 = $\sigma = \sqrt{\text{Variance}}$

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

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

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

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

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

Therefore, Standard Deviation = $\sigma = \sqrt{\text{Variance}}$ = $\sqrt{144}$ = 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 resulting observations if each observation is multiplied by a:

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

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

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

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

We know,

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

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

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

Hence, it is proven.

Question 5: 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:

(i) If the wrong item is omitted.

(ii) If it is replaced by $\small 12.$

Answer(i):

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

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

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

Thus, the incorrect sum = 200

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

Therefore, Correct Mean = $\frac{\text{(Correct Sum)}}{19}$

$=\frac{192}{19}$

$= 10.1$

Now, Standard Deviation,

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

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

Thus, New sum = Old sum - (8 × 8)

= 2080 – 64

= 2016

Hence, Correct Standard Deviation =

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

$⇒\sigma'= \sqrt{106.1 - 102.01} = \sqrt{4.09}=2.02$

Answer(ii):

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

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

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

Thus, the incorrect sum = 200

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

Therefore, Correct Mean = $\frac{\text{(Correct Sum)}}{20}=\frac{204}{20}=10.2$

Now, Standard Deviation,

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

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

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

= 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

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

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

Thus, incorrect sum = 2000

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

New number of observations, n' $= 100-3 =97$

Therefore, New Mean = $\frac{\text{(New Sum)}}{100}$

$=\frac{1940}{97}$

= 20

Now, Standard Deviation,

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

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

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

= 40900 - 441 - 441 - 324

= 39694

Hence, Correct Standard Deviation =

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

$⇒\sigma'= \sqrt{108 - 104.04} = \sqrt{3.96}=3.036$