Statistics

Let the missing observations be $x$ and $y$.

$ \text { Mean }(\bar{x})=9: \frac{2+3+5+10+11+13+15+21+x+y}{10}=9 $

$ 80+x+y=90 \Rightarrow \mathbf{x}+\mathbf{y}=\mathbf{1 0} $

$ \text { Variance }\left(\sigma^2\right)=34.2 : \sigma^2=\frac{\sum x_i^2}{n}-(\bar{x})^2 $

$ \frac{2^2+3^2+5^2+10^2+11^2+13^2+15^2+21^2+x^2+y^2}{10}-9^2=34.2 $

$ \frac{4+9+25+100+121+169+225+441+x^2+y^2}{10}-81=34.2 $

$ \frac{1094+x^2+y^2}{10}=115.2 $

$ 1094+x^2+y^2=1152 \Rightarrow \mathbf{x}^2+\mathbf{y}^2=\mathbf{5 8} $

Solving $x+y=10$ and $x^2+y^2=58$

To solve the system $x+y=10$ and $x^2+y^2=58$ by finding $x-y$ use the formula $x-y=\sqrt{2\left(x^2+y^2\right)-(x+y)^2}$ :

$ x-y=\sqrt{2\left(x^2+y^2\right)-(x+y)^2} $

$ x-y=\sqrt{2(58)-(10)^2}=\sqrt{116-100}=\sqrt{16} $

$ x-y=4 $

Now add these equations $x+y=10 \& x-y=4$

$ 2 x=14 \Longrightarrow x=7 $

Put $x=7$ in $x+y=10$

$ 7+y=10 \Longrightarrow y=3 $

missing numbers are 3 and 7 .

Find the Median

The 10 observations in ascending order are: $2,3,3,5,7,10,11,13,15,21$.

• Since $n=10$, the median is the average of the 5th and 6th terms :
  
  Median $=\frac{7+10}{2}=$ 8.5

Calculate Mean Deviation about Median

$ \begin{aligned} & \text { M.D. }(\text { median })=\frac{\sum \mid x_i-\text { Median } \mid}{n} \\ & = \frac{|2-8.5|+|3-8.5|+|3-8.5|+|5-8.5|+|7-8.5|+|10-8.5|+|11-8.5|+|13-8.5|+|15-8.5|+|21-8.5|}{10} \\ & =\frac{6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5}{10}=\frac{50}{10}=\mathbf{5} \end{aligned} $

$ Y=\{a x+b: x \in X\} . \text { and } X=\{x \in \mathbb{N}: 1 \leq x \leq 19\} $

mean for $Y$ is 30.

$\therefore $ $(\bar{Y}=30)$

$\bar{Y}=\frac{\sum\limits_{i=1}^{19} Y_i}{19}$

$\Rightarrow \frac{\sum\limits_{i=1}^{19} a x_i+b}{19}=30$

$\Rightarrow $ $\sum\limits_{i=1}^{19} a x_i+\sum\limits_{i=1}^{19} b=19 \times 30$

$\Rightarrow $ $a(1+2+\cdots+19)+19 b=19 \times 30$

$\Rightarrow $ $a \times \frac{19 \times 20}{2}+19 b=570$

$\Rightarrow $ $ 10 a+b=30 \Rightarrow a=\frac{30-b}{10} $ .......(1)

Variance for $Y$ is 750

$ \sigma^2=\frac{\sum\limits_{i=1}^{19} Y_i^2}{19}-(\bar{Y})^2=750 $

$\Rightarrow $ $\frac{\sum\limits_{i=1}^{19}\left(a x_i+b\right)^2}{19}-(30)^2=750$

$\Rightarrow $ $\sum\limits_{i=1}^{19} a^2 x_i^2+b^2+2 a x_i=(750+900) 19$

$\Rightarrow $ $a^2 \sum\limits_{i=1}^{19}\left(x_i^2+b^2+2 a \sum\limits_{i=1}^{19} x_i\right)=1650 \times 19$

$\Rightarrow $ $a^2\left(1^2+2^2+\cdots+19^2\right)+\sum\limits_{i=1}^{19} b^2+2 a b(1+2+\cdots+19)=1650 \times 19$

$\Rightarrow $ $a^2 \cdot \frac{19 \times 20 \times 39}{6}+19 b^2+\frac{2 a b \times 19 \times 20}{2}=1650 \times 19$

$\Rightarrow $ $ 130 a^2+b^2+20 a b=1650 $ ......(2)

Substitute value of $a$ from (1) in equation (2)

$ 130\left(\frac{30-b}{10}\right)^2+b^2+20 b\left(\frac{30-b}{10}\right)=1650 $

$\Rightarrow $ $\frac{13}{10}\left(900+b^2-60 b\right)+b^2+60 b-2 b^2=1650$

$\Rightarrow $ $1170+1.3 b^2-78 b+10 b^2+600 b-20 b^2=16500$

$\Rightarrow $ $3 b^2-180 b-4800=0$

$\Rightarrow $ $b^2-60 b-1600=0$

Let roots of quadratic equation is $b_1, b_2$

So sum of possible values of $b=$ sum of roots

$ b_1+b_2=\frac{- \text { coefficient of } b}{\text { coefficient of } b^2}=60 $

Given, total observation $n=10$

Initial mean $\bar{x}_1=10 $

$\Rightarrow \frac{\sum\limits_{i=1}^{10} x_i}{10}=10$

Sum of 10 observation $\sum_{i=1}^{10} x_i=10 \times 10=100$

Initial variance $\sigma _1^2$$=2$

$\begin{aligned} & \Rightarrow \frac{\sum\limits_{i=1}^{10} x_i^2}{n}-\left(\bar{x}_1\right)^2=2 \\ & \Rightarrow \sum\limits_{i=1}^{10} x_i^2=n\left[2+\left(\bar{x}_1\right)^2\right] =10\left[2+\left(10^2\right)\right]=1020\end{aligned}$

It is given that if $\alpha$ is replace by $\beta$

then, new mean $\left(\bar{x}_2\right)=10.1$

$ \begin{aligned} & \bar{x}_2=\frac{100-\alpha+\beta}{10}=10.1 \\ & \Rightarrow 100-\alpha+\beta=101 \\ & \Rightarrow \beta-\alpha=1......(1) \end{aligned} $

Also new variance $\sigma_2^2=1.99$ (given)

$\Rightarrow $ $\sigma_2^2=\frac{1020-\alpha^2+\beta^2}{10}-(10.1)^2=1.99$

$\Rightarrow $ $1020-\alpha^2+\beta^2=10[1.99+102.01]$=10 \times 104=1040$

$\begin{aligned} & \Rightarrow \beta^2-\alpha^2=20 \\\\ & \Rightarrow (\beta+\alpha)(\beta-\alpha)=20\end{aligned}$

$\Rightarrow $$(\beta+\alpha)(1)=20$ (substitute $\beta-\alpha=1)$

$\Rightarrow $ $\alpha+\beta=20$

$\begin{aligned} & \sum f_i=14+\lambda \\ & \sum f_i x_i=18+10 \lambda+56+126=200+10 \lambda \\ & \text { Mean } \mu=\frac{\sum f_i x_i}{\sum f_i}=\frac{200+10 \lambda}{14+\lambda} \\ & \sum f_i x_i^2=3\left(6^2\right)+\lambda\left(10^2\right)+4\left(14^2\right)+7\left(18^2\right)=108+100 \lambda+784+2268=3160+100 \lambda\end{aligned}$

$\begin{aligned} & \text { Variance }=\frac{\sum f_i x_i^2}{\sum f_i}-\mu^2=19 \\ & 19=\frac{3160+100 \lambda}{14+\lambda}-\left(\frac{200+10 \lambda}{14+\lambda}\right)^2 \\ & 19(14+\lambda)^2=(3160+100 \lambda)(14+\lambda)-(200+10 \lambda)^2 \\ & 19\left(196+28 \lambda+\lambda^2\right)=44240+4560 \lambda+100 \lambda^2-40000-4000 \lambda-100 \lambda^2 \\ & 3724+532 \lambda+19 \lambda^2=4240+560 \lambda \\ & 19 \lambda^2-28 \lambda-516=0\end{aligned}$

Solving for $\lambda$ using the quadratic formula:

$ \lambda=\frac{28 \pm \sqrt{28^2-4(19)(-516)}}{38}=\frac{28 \pm 200}{38} $

Taking the positive value, $\lambda=\frac{228}{38}=6$

Finding the mean $\mu$ :

$ \mu=\frac{200+10(6)}{14+6}=\frac{260}{20}=13 $

The value of $\lambda+\mu$ is:

$ 6+13=19 $

$ \begin{aligned} & \text { Mean }=\frac{7}{2} \\ & \frac{-10-7-1+x+y+16+2+9}{8}=\frac{7}{2} \\ & \Rightarrow \mathrm{x}+\mathrm{y}+9=28 \\ & \Rightarrow x+y=19 \\ & \text { Variance }=\frac{293}{4} \\ & \frac{100+49+1+x^2+y^2+256+4+81}{8}-\frac{49}{4}=\frac{293}{4} \\ & \frac{491+x^2+y^2}{8}=\frac{342}{4} \\ & x^2+y^2=193 \end{aligned} $

$ \begin{aligned} & \Rightarrow(x, y)=(7,12) \text { or }(12,7) \\ & \text { Mean }=\frac{x+y+x+y+1+(x-y)}{4} \\ & =\frac{19+20+5}{4} \\ & =11 \end{aligned} $

$ \begin{aligned} & \because \text { median }=\frac{1001 k}{2}=X_M \\ & \therefore \text { mean deviation about median }=\frac{\sum\left|X_i-X_M\right|}{n} \\ & =\frac{2\left(\frac{k}{2}+\frac{3 k}{2}+\frac{5 k}{2}+\ldots . . .500 \text { terms }\right)}{1000}=\frac{2 . \frac{k}{2}(500)^2}{1000}=\frac{500 k}{2}=500(\text { given }) \\ & \therefore k=2 \\ & \therefore k^2=4 \end{aligned} $

$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline p(n) & \frac{2 a+1}{30} & \frac{8 a-1}{30} & \frac{4 a+1}{30} & \mathrm{~b} \\ \hline \end{array} $

$ \begin{aligned} &\begin{aligned} & \mu=\Sigma \times p(n)=0+\frac{8 a-1}{30}+2 \times \frac{4 a+1}{30}+3 b=\frac{16 a+90 b+1}{30} \\ & \sigma^2=\Sigma x^2 p(n)-\mu^2 \Rightarrow \sigma^2+\mu^2=\Sigma x^2 p(n)=2 \\ & \frac{8 a-1}{30}+\frac{4(4 a+1)}{30}+9 b=2 \\ & 8 a+90 b-19=0 \ldots \ldots \ldots(i) \\ & \Sigma p(n)=1 \Rightarrow 14 a+1+30 b=30 \\ & 14 a+30 b-29=0 \ldots \ldots . .(i i) \end{aligned}\\ &\text { From (i) and (ii) }\\ &a=2 \quad b=\frac{1}{30}=\frac{a}{b}=60 \end{aligned} $

$\begin{aligned} &\text { Let the observations be } x_1, x_2, \ldots, x_{99}, 50\\ &\begin{aligned} & \text { Mean }=\frac{x_1+x_2+\ldots+x_9+50}{100}=40 \\ & \Rightarrow x_1+x_2+\ldots+x_{99}=4000-50 \\ & \Rightarrow x_1+x_2+\ldots+x_{99}=3950 \\ & \text { Current Mean }=\frac{3950+40}{100} \\ & \mu=\frac{399}{10}=39.9 \\ & (\text { S.D })^2=\sum_{i=1}^{99} \frac{\left(x_i\right)^2+2500}{100}-(40)^2 \\ & \sum_{i=1}^{99} x_i^2=160101 \\ & (\text { Correct S.D })^2=\frac{160101+1600}{100}-\left(\frac{399}{10}\right)^2 \\ & \sigma=5 \\ & 10(\mu+\sigma)=10(39.9+5)=449 \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \frac{2+3+3+4+5+7+a+b}{8}=4 \\ & \Rightarrow a+b=8 \\ & (\sqrt{2})^2=\frac{2^2+3^2+3^2+4^2+5^2+7^2+a^2+b^2}{8}-16 \\ & 112+a^2+b^2=18 \times 8 \\ & \Rightarrow a^2+b^2=32 \\ & \Rightarrow a=b=4 \end{aligned}\\ &\text { Now numbers be }\\ &2,3,3,4,4,4,5,7 \end{aligned}$

Mode $=4$

Mean deviation about mode :

$\begin{aligned} & \frac{|2-4|+|3-4|+|3-4|+0+0+0+|5-4|+|4-7|}{8} \\ = & \frac{2+1+1+1+3}{8}=\frac{8}{8}=1 \end{aligned}$

First, find the values of $a$ and $b$ using the given conditions:

Mean Condition:

$ 5 = \frac{1 + 3 + a + 7 + b}{5} $

This implies:

$ a + b = 14 $

Variance Condition:

$ \frac{1 + 9 + a^2 + 49 + b^2}{5} - (5)^2 = 10 $

Simplifying, we get:

$ a^2 + b^2 = 116 $

Using $a + b = 14$, we have:

$ b = 14 - a $

Substitute $b$ into the equation for $a^2 + b^2$:

$ a^2 + (14 - a)^2 = 116 $

Expand and simplify:

$ a^2 + 196 - 28a + a^2 = 116 $

$ 2a^2 - 28a + 80 = 0 $

Simplify further:

$ a^2 - 14a + 40 = 0 $

Solve the quadratic equation:

$ (a - 10)(a - 4) = 0 $

Thus, $a = 10$ and $b = 4$ (since $a > b$).

New Observations:

The transformed observations are:

$ 1+1=2, \quad 3+2=5, \quad 10+3=13, \quad 7+4=11, \quad 4+5=9 $

Calculate the Variance of the New Observations:

First, find the mean of the new data:

$ \text{Mean} = \frac{2 + 5 + 13 + 11 + 9}{5} = 8 $

Calculate the variance:

$ \text{Variance} = \frac{(2-8)^2 + (5-8)^2 + (13-8)^2 + (11-8)^2 + (9-8)^2}{5} $

$ = \frac{36 + 9 + 25 + 9 + 1}{5} $

$ = \frac{80}{5} = 16 $

Therefore, the variance of the transformed observations is 16.

Let’s set up two equations from the given mean and variance:

Mean = 9 ⇒ total sum = 8·9 = 72

Known values sum to 6+4+8+12+10+13 = 53, so

$a + b = 72 - 53 = 19.$

Population variance = 9.25 ⇒

$\frac{\sum x_i^2}{8} - 9^2 = 9.25\quad\Longrightarrow\quad \sum x_i^2 = 8\,(81 + 9.25) = 722.$

Known squares sum to 6²+4²+8²+12²+10²+13² = 529, so

$a^2 + b^2 = 722 - 529 = 193.$

Now use

$(a + b)^2 = a^2 + 2ab + b^2 \quad\Longrightarrow\quad 19^2 = 193 + 2ab$

so

$361 = 193 + 2ab\quad\Longrightarrow\quad ab = 84.$

Finally,

$a + b + ab = 19 + 84 = 103.$

Answer: 103 (Option A).

$\begin{aligned} & \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=50, \quad \therefore \text { mean }=5 \\ & \text { Variance }=\frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & \frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-25 \\ & \Rightarrow \sum \mathrm{x}_{\mathrm{i}}^2=258 \quad\text{.... (1)}\\ & \text { Now } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\beta\right)^2=98 \\ & \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}^2-2 \beta \cdot \mathrm{x}_{\mathrm{i}}+\beta^2\right)=98 \\ & 258-2 \beta(50)+10 \beta^2=98 \\ & (\beta-8)(\beta-2)=0 \\ & \beta=\text { or } \beta=2 \quad(\text { as } \beta>2) \\ & \therefore \beta=8\quad\text{..... (2)} \end{aligned}$

Now as per the question

$ \begin{aligned} &2\left(\mathrm{x}_1-1\right)+4 \beta, 2\left(\mathrm{x}_2-1\right)+4 \beta, \ldots .2\left(\mathrm{x}_{10}-1\right)+4 \beta\\ &\text { can be simplified to }\\ &2 \mathrm{x}_1+30,2 \mathrm{x}_2+30, \ldots . .2 \mathrm{x}_{10}+30 \text { using eq. (2) }\\ &\mu=2(5)+30=40\quad\text{..... (3)}\\ &\sigma^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5}\\ &\because \frac{\beta \mu}{\sigma^2}=\frac{8 \times 40}{16 / 5}=100 \end{aligned}$

$\begin{aligned} & \text { Mean } \overline{\mathrm{x}}=5.5 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=5.5 \times 10=55 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2=371 \\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=55-(4+5)+(6+8)=60 \\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430 \\ & \\ & \text { Variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & \\ & \sigma^2=\frac{430}{10}-\left(\frac{60}{10}\right)^2 \\ & \sigma^2=43-36 \\ & \sigma^2=7 \end{aligned}$

The median for grouped data is given by:

$ \text{Median} = L + \left(\frac{\frac{n}{2} - CF}{f}\right) \times h $

where

$L$ is the lower limit (or boundary) of the median class.

$CF$ is the cumulative frequency of all classes preceding the median class.

$f$ is the frequency of the median class.

$h$ is the class width.

$n$ is the total number of students.

Given:

Median $= 14$

Median class interval is $12-18$, so $L = 12$ and the class width $h = 18 - 12 = 6$.

Frequency of median class $f = 12$.

Cumulative frequency below the median class $= 18$ (i.e., $CF = 18$).

Plugging these into the formula:

$ 14 = 12 + \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6 $

Step 1: Subtract 12 from both sides:

$ 2 = \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6 $

Step 2: Simplify the multiplication factor:

$ \left(\frac{6}{12}\right) = \frac{1}{2} $

So the equation becomes:

$ 2 = \frac{1}{2}\left(\frac{n}{2} - 18\right) $

Step 3: Multiply both sides by 2 to remove the fraction:

$ 4 = \frac{n}{2} - 18 $

Step 4: Solve for $\frac{n}{2}$:

$ \frac{n}{2} = 4 + 18 = 22 $

Step 5: Multiply both sides by 2 to find $n$:

$ n = 44 $

Thus, the total number of students is $44$.

$\begin{aligned} & \sigma^2=E\left(x^2\right)-[E(x)], \sum f\left(x_i\right)=7 \\ & E(x)=\sum x f(x)=22 C \\ & E\left(x^2\right)=\sum x^2 f(x)=92 C^2 \end{aligned}$

$\begin{aligned} & \sigma^2=160=\frac{92 C^2}{7}-\left(\frac{22 C}{7}\right)^2 \\ & \Rightarrow C= \pm 7 \text { but } C \in N \\ & \Rightarrow C=7 \end{aligned}$

$\begin{aligned} & 30+x+y=40 \\ & x+y=10 \end{aligned}$

Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ observation

$=\frac{40+1}{2}=\frac{41}{2}$

Median $=18$

Mean deviation about median

$\begin{aligned} & 5.3+8.2+5.1+12.0+x \cdot 1+y \cdot 2=1.25 \times 40 \\ & 15+16+5+x+2 y=50 \\ & x+2 y=14 \\ & \quad x+y=10 \\ & \Rightarrow \quad x=4 \\ & \Rightarrow \quad y=6 \\ & \begin{aligned} 4 x+5 y & =24+20 \\ & =44 \end{aligned} \end{aligned}$

To find the correct standard deviation, we first need to adjust the mean and sum of squares of the observations based on the correction of the erroneous entry. The original erroneous observation was 8, and the correct observation is 12. The mean and standard deviation of the 20 observations before correction were 10 and 2, respectively.

The sum of all 20 original observations ($S_{\text{original}}$) can be calculated from the mean formula:

$ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} $

$ 10 = \frac{S_{\text{original}}}{20} $

$ S_{\text{original}} = 10 \times 20 = 200 $

The corrected sum of observations ($S_{\text{correct}}$) will replace the incorrect observation (8) with the correct one (12):

$ S_{\text{correct}} = S_{\text{original}} - 8 + 12 = 200 - 8 + 12 = 204 $

The corrected mean ($\mu_{\text{correct}}$) is:

$ \mu_{\text{correct}} = \frac{S_{\text{correct}}}{20} = \frac{204}{20} = 10.2 $

To find the corrected standard deviation, we need the sum of squares of the deviations from the mean for both the original and corrected data. The original sum of squares ($SS_{\text{original}}$) is calculated from the original standard deviation formula, where $\sigma = 2$:

$ \sigma^2 = \frac{SS}{n} $

$ 4 = \frac{SS_{\text{original}}}{20} $

$ SS_{\text{original}} = 4 \times 20 = 80 $

To calculate the corrected sum of squares ($SS_{\text{correct}}$), we need to adjust $SS_{\text{original}}$ by removing the square of the deviation of the incorrect observation and adding the square of the deviation of the correct observation:

$ SS_{\text{correct}} = SS_{\text{original}} - (8 - 10)^2 + (12 - 10.2)^2 $

$ SS_{\text{correct}} = 80 - (-2)^2 + (1.8)^2 $

$ SS_{\text{correct}} = 80 - 4 + 3.24 = 79.24 $

Finally, the corrected standard deviation ($\sigma_{\text{correct}}$) is:

$ \sigma_{\text{correct}} = \sqrt{\frac{SS_{\text{correct}}}{20}} $

$ \sigma_{\text{correct}} = \sqrt{\frac{79.24}{20}} $

$ \sigma_{\text{correct}} = \sqrt{3.962} $

The corrected standard deviation is closest to the value given in Option B, which is

$ \sqrt{3.96} $

$\begin{aligned} & \text { Mean }=\frac{-3+4+7+(-6)+\alpha+\beta}{6}=2 \\ & \Rightarrow \alpha+\beta=10 \\ & \text { Variance }=\frac{\sum x_i^2}{n}-\left(\frac{\bar{x}}{n}\right)^2=23 \\ & \Rightarrow \sum x_i^2=27 \times 6 \\ & \Rightarrow 9+16+49+36+\alpha^2+\beta^2=162 \\ & \Rightarrow \alpha^2+\beta^2=52 \end{aligned}$

We get $\alpha$ and $\beta$ as 4 and 6

So, mean deviation about mean

$\begin{aligned} & =\frac{|-3-2|+|4-2|+|7-2|+|-6-2|+|4-2|+|6-2|}{6} \\ & =\frac{5+2+5+8+2+4}{6} \\ & =\frac{13}{3} \end{aligned}$

$\begin{aligned} & a, b, 68,44,48,60 \\ & \text { Mean }=55 \quad a>b \\ & \text { Variance }=194 \quad a+3 b \\ & \frac{a+b+68+44+48+60}{6}=55 \\ & \Rightarrow 220+a+b=330 \\ & \therefore a+b=110 \ldots . .(1) \end{aligned}$

Also,

$\begin{aligned} & \sum \frac{\left(x_i-\bar{x}\right)^2}{n}=194 \\ & \Rightarrow(a-55)^2+(b-55)^2+(68-55)^2+(44-55)^2 \\ & +(48-55)^2+(60-55)^2=194 \times 6 \\ & \Rightarrow(a-55)^2+(b-55)^2+169+121+49+25=1164 \\ & \Rightarrow(a-55)^2+(b-55)^2=1164-364=800 \\ & a^2+3025-110 a+b^2+3025-110 b=800 \\ & \Rightarrow a^2+b^2=800-6050+12100 \\ & a^2+b^2=6850 \ldots \ldots . .(2) \end{aligned}$

Solve (1) & (2);

$\begin{aligned} & a=75, b=35 \\ & \therefore a+3 b=75+3(35)=75+105=180 \end{aligned}$

$\begin{aligned} & \mathrm{M}=1+\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\right) \mathrm{h} \\ & \mathrm{M}=8+\frac{18-12}{10} \times 4 \\ & \mathrm{M}=10.4 \\ & 20 \mathrm{M}=208 \end{aligned}$

$\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}$

Let first four observation be $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$

Here, $\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$. ..... (1)

Also, $\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}$

$\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14$

Now from eqn -1

$\mathrm{x}_5=10$

Now, $\sigma^2=\frac{194}{25}$

$\begin{aligned} & \frac{\mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2+\mathrm{x}_5^2}{5}-\frac{576}{25}=\frac{194}{25} \\ & \Rightarrow \mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2=54 \end{aligned}$

Now, variance of first 4 observations

$\begin{aligned} \operatorname{Var} & =\frac{\sum_\limits{i=1}^4 x_i^2}{4}-\left(\frac{\sum_\limits{i=1}^4 x_i}{4}\right)^2 \\ & =\frac{54}{4}-\frac{49}{4}=\frac{5}{4} \end{aligned}$

$\begin{aligned} & \sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50 \\ & \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50 \quad \text{.... (i)}\\ & \sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=1100 \quad \text{.... (ii)}\\ & \text { If } \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50 . \\ & \left(\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}\right)^2=2500 \\ & \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2+2 \sum_{\mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=2500 \\ & \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=2500-2(1100) \\ & \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=300 \text {, Standard deviation ‘ } \sigma \text { ’ } \end{aligned}$

$\begin{aligned} & =\sqrt{\frac{\sum \mathrm{a}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{a}_{\mathrm{i}}}{10}\right)^2}=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2} \\ & =\sqrt{30-25}=\sqrt{5} \end{aligned}$

Given that the mean of the observations ${x, y, 1, 2, 4, 5}$ is 5, we get the equation:

$x + y + 1 + 2 + 4 + 5 = 6 \cdot 5 \Rightarrow x + y = 18$ ..........$(1)$

We are also given that the variance of the observations is 10. Using the formula for variance, we have:

$V = \frac{\Sigma x_i^2}{n} - \bar{x}^2$, where $n$ is the number of observations, and $\bar{x}$ is the mean.

Substituting the given values in the variance formula, we get:

$10 = \frac{\Sigma x_i^2}{6} - 5^2 = \frac{\Sigma x_i^2}{6} - 25 \Rightarrow \Sigma x_i^2 = 210$.

Here, $\Sigma x_i^2$ is the sum of the squares of all the observations, thus:

$x^2 + y^2 + 1 + 4 + 16 + 25 = 210 \Rightarrow x^2 + y^2 = 164$ .........$(2)$

By solving the system of equations $(1)$ and $(2)$, we get $x = 8$ and $y = 10$.

Now, the mean deviation about the mean ($\bar{x} = 5$) is calculated by taking the average of the absolute differences of each observation from the mean:

$MD(5) = \frac{1}{6} [|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|] = \frac{16}{6} = \frac{8}{3}$.

So, the mean deviation about the mean is $\frac{8}{3}$, and the correct answer is Option B, $\frac{8}{3}$.

To solve this problem, let's break it down step by step :

Step 1 : Determine the mean of set C

The mean of set A = $5$

The mean of set B = $8$

After subtracting 3 from each element in set A, the new mean becomes $5 - 3 = 2$. After adding 2 to each element in set B, the new mean becomes $8 + 2 = 10$.

When we combine both modified sets to form set C, the mean of set C is the weighted average of the means of these modified sets:

Mean of C = $\frac{5 \times 2 + 5 \times 10}{10} = \frac{10 + 50}{10} = 6$

Step 2 : Determine the variance of set C

Variance is defined as the expectation of the squared deviation of a random variable from its mean. One property of variance is that, if you add (or subtract) a constant from each data point in a set, the variance of the set does not change.

Thus, the variance of the elements in set A remains $12$ even after subtracting 3 from each element, and the variance of the elements in set B remains $20$ even after adding 2 to each element.

Now, when combining variances from two datasets into one :

Variance of C

= $\frac{n_1 \times \text{Variance of A} + n_1 \times (\text{Mean of modified A} - \text{Mean of C})^2 + n_2 \times \text{Variance of B} + n_2 \times (\text{Mean of modified B} - \text{Mean of C})^2}{n_1 + n_2}$

Given :

$n_1 = n_2 = 5$

Variance of A = $12$, Variance of B = $20$

Mean of modified A = $2$, Mean of modified B = $10$, Mean of C = $6$

Plugging in the values, we get :

Variance of C = $\frac{5 \times 12 + 5 \times (2 - 6)^2 + 5 \times 20 + 5 \times (10 - 6)^2}{10}$

Variance of C = $\frac{60 + 80 + 100 + 80}{10} = \frac{320}{10} = 32$

Step 3 : Sum of the mean and variance of set C

Sum = Mean of C + Variance of C = $6 + 32 = 38$

So, the correct answer is :

Option C : 38.

Let ${a_1} = a \Rightarrow {a_2} = a + 1,\,.....\,{a_{100}} = a + 90$

$\mu = {{{{100a + (99 \times 100)} \over 2}} \over {100}} = a + {{99} \over 2}$

M.D $ = {{\sum {|{a_1} - \mu |} } \over {100}} = {{{{\left( {{{99} \over 2} + {{97} \over 2}\, + \,...\, + \,{1 \over 2}} \right)}^2}} \over {100}} = {{2500} \over {100}} = 25$

$\therefore$ $a \to z$

3 rotten apples are mixed with 7 good apples.

$\therefore$ Total apples = 10

Among those 10 apples 4 are chosen randomly.

${x_i}$ = Number of rotten apples drawn.

${p_i}$ = Probability of rotten apple.

We know,

Mean $(\mu ) = \sum {{p_i}{x_i}} $

$ = 0 + {{105} \over {210}} + {{126} \over {210}} + {{21} \over {210}}$

$ = {{252} \over {210}} = {6 \over 5}$

Also,

Variance $({\sigma ^2}) = \left( {\sum {{p_i}{{({x_i})}^2}} } \right) - {\mu ^2}$

$ = {{105} \over {210}} + {{252} \over {210}} + {{63} \over {210}} - {{36} \over {25}}$

$ = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {{36} \over {25}} = {{14} \over {25}}$

$\therefore$ $10(\mu^2 + {\sigma ^2})$

$ = 10\left( {({6 \over 5})^2 + {{14} \over {25}}} \right)$

$ = 10\left( {{{36 + 14} \over {25}}} \right)$

$ = 10 \times {{50} \over {25}} = 20$

${a_1},{a_2},{a_3},{a_4},{a_5},{a_6}$ are in AP.

Let

${a_1} = a$

${a_2} = a + d$

${a_3} = a + 2d$

${a_4} = a + 3d$

${a_5} = a + 4d$

${a_6} = a + 5d$

Now Mean of ${a_1},{a_2},{a_3},{a_4},{a_5}$ and ${a_6}$ is

$ = {{{a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6}} \over 6} = {{19} \over 2}$

$ \Rightarrow {{6a + 15d} \over 6} = {{19} \over 2}$

$ \Rightarrow 2a + 5d = 19$ ...... (1)

Also, given,

${a_1} + {a_3} = 10$

$ \Rightarrow a + a + 2d = 10$

$ \Rightarrow 2a + 2d = 10$

$ \Rightarrow a + d = 5$ ..... (2)

From equation (1) and (2), we get $a = 2$ and $d = 3$

$\therefore$ AP is 2, 5, 8, 11, 14, 17

Now, Variance $({\sigma ^2}) = {{\sum {x_i^2} } \over 6} - {\left( {\overline x } \right)^2}$

$ = {{{2^2} + {5^2} + {8^2} + {{11}^2} + {{14}^2} + {{17}^2}} \over 6} - {\left( {{{19} \over 2}} \right)^2}$

$ = {{105} \over 4}$

$\therefore$ $8{\sigma ^2} = 8 \times {{105} \over 4} = 210$