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 $44$.

Given $12+f_1+f_2=19$

Median = 6

$\Rightarrow \quad f_1+f_2=7$

Median $=\left(\frac{N+1}{2}\right)^{\text {th }}$ term $=\left(\frac{19+1}{2}\right)^{\text {th }}$ term $=10^{\text {th }}$ term

$ \begin{aligned} & \Rightarrow \quad 6+f_1=10 \Rightarrow f_1=4 \text { and } f_2=3 \\ & \therefore \quad 7 f_1+9 f_2=7 \times 4+9 \times 3 \\ & =55 \Rightarrow P \rightarrow 5 \end{aligned} $

Mean $(\bar{X})=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{133}{19}=7$

$ \begin{aligned} & \text { Mean deviation about mean }(\alpha)=\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i} \\ & =\frac{5(3)+4(2)+1(1)+3(1)+2(2)+3(4)+1(5)}{19}=\frac{48}{19} \\ & \Rightarrow 19 \alpha=48 \quad Q \rightarrow 3 \end{aligned} $

Mean deviation about median $(\beta)$

$ \begin{aligned} & =\frac{\Sigma f_i\left|x_i-M\right|}{\Sigma f_i} \\ & =\frac{5(2)+4(1)+1(0)+3(2)+2(3)+3(5)+1(6)}{19} \\ & =\frac{47}{19} \\ & 19 \beta=47 \quad R \rightarrow 2 \end{aligned} $

There are 27 numbers in the list: $3x, 6x, 9x, \ldots, 81x$.

Because there are an odd number of numbers, the median is the $14^\text{th}$ number in the list.

Each number increases by $3x$, so the $14^\text{th}$ number is: $3x + (14-1) \times 3x = 3x + 39x = 42x$ So, the median is $42x$.

The mean deviation about the median means we find the average of the distances from each value to $42x$: $\frac{\,|3x-42x| + |6x-42x| + \ldots + |81x-42x|\,}{27} = 91$

So, $|3x-42x| + |6x-42x| + \ldots + |81x-42x| = 27 \times 91$

Every term inside the absolute value is a multiple of $|x|$.
The differences $|3x-42x|, |6x-42x|, \ldots, |81x-42x|$ will be $|39x|, |36x|, \ldots, |0x|, \ldots, |39x|$.

So the total sum is:
$|x| \times (39 + 36 + \ldots + 0 + 3 + 6 + \ldots + 39) = 27 \times 91$

The numbers in the bracket add up to $546$, so: $|x| \times 546 = 27 \times 91$

$|x| = \frac{27 \times 91}{546} = \frac{9}{2}$

Let us take assume mean $(A)=5$ and $h=2$

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$ \begin{aligned} \because \quad \bar{x} & =A+\frac{\Sigma f_i \mu_i}{\Sigma f_i} \times h \\ \bar{x} & =5+0=5 \\ \mathrm{SD} & =\sigma=h \times \sqrt{\frac{\Sigma f_i \mu_i^2}{N}-\left(\frac{\Sigma f_i \mu_i}{N}\right)^2} \\ \sigma & =2 \times \sqrt{\frac{22}{15}} \\ \mathrm{CV} & =\frac{\sigma}{\bar{x}} \times 100=\frac{2 \times \sqrt{\frac{22}{15}} \times 100}{5} \\ & =\frac{2 \times \sqrt{2} \times \sqrt{11}}{\sqrt{3} \times \sqrt{5} \times 5 \times \sqrt{5}} \times 100 \times \sqrt{5} \\ & =\sqrt{\frac{110}{3}} \times 8 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \bar{X}=\frac{2+3+5+7+11+13+17+19+22}{9} \\ & \quad=11 \\ & \left|x_i-\bar{x}\right|=9,8,6,4,0,2,6,8,11 \end{aligned}\\ &\text { ∴ Mean deviation from mean }\\ &\begin{aligned} & =\frac{\Sigma\left|x_i-\bar{x}\right|}{9} \\ & =\frac{9+8+6+4+0+2+6+8+11}{9} \end{aligned} \end{aligned} $

$ =\frac{54}{9}=6 $

Step 1: List all values and probabilities

We have $X$ taking the values 2, 3, 5, 7, and 12. Their probabilities are $3k$, $k$, $k$, $2k$, and $k$.

Step 2: Make a table with useful information

$ \begin{array}{lllll} \hline x_i & p_i & p_ix_i & x_i^2 & p_ix_i^2 \\ \hline 2 & 3k & 6k & 4 & 12k \\ 3 & k & 3k & 9 & 9k \\ 5 & k & 5k & 25 & 25k \\ 7 & 2k & 14k & 49 & 98k \\ 12 & k & 12k & 144 & 144k \\ \hline & & 40k & & 288k \\ \hline \end{array} $

Step 3: Find value of $k$

Add up all probabilities:

$ 3k + k + k + 2k + k = 8k $

Set equal to 1:

$ 8k = 1 $

So, $ k = \frac{1}{8} $

Step 4: Find the mean and expected value of $x^2$

$ \Sigma p_i x_i = 40k $

$ \Sigma p_i x_i^2 = 288k $

Step 5: Write the formula for standard deviation

$ \mathrm{SD} = \sqrt{ \Sigma p_i x_i^2 - \left( \Sigma p_i x_i \right)^2 } $

Step 6: Substitute the values and solve

$ \mathrm{SD} = \sqrt{ 288k - (40k)^2 } $ $ = \sqrt{ \frac{288}{8} - (5)^2 } $ $ = \sqrt{36 - 25} $ $ = \sqrt{11} $

Data $x_i: 3,4,5,6,7,8,10,13$

$ \begin{aligned} & \begin{array}{l} \Sigma x_i=3+4+5+6+7+8+10+13=56 \\ \Sigma x_i^2=9+16+25+36+49+64 +100+169 \end{array} \\ & \quad=468 \\ & \text { Variance }=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2 \\ & \qquad=\frac{468}{8}-\left(\frac{56}{8}\right)^2=58.5-49=9.5 \end{aligned} $

$ \begin{aligned} & \because P(X=K)=\frac{e^{-\lambda} \lambda^K}{K!} \\ & \because \quad P(X=3)=P(X=5) \\ & \Rightarrow \quad \frac{e^{-\lambda} \lambda^3}{3!}=\frac{e^{-\lambda} \lambda^5}{5!} \\ & \Rightarrow \quad \lambda^2=\frac{120}{6}=20 \\ & \Rightarrow \lambda=\sqrt{20}=2 \sqrt{5} \\ & \therefore P(X=4)=\frac{e^{-\lambda} \lambda^4}{4!}=\frac{e^{-2 \sqrt{5}},(\sqrt{20})^4}{24} \\ & \quad=\frac{400}{24 e^{2 \sqrt{5}}}=\frac{50}{3 e^{2 \sqrt{5}}} \end{aligned} $

Given, $a=9, d=6$

$\therefore$ Variance,

$ \begin{aligned} P & =\frac{6^2\left(n^2-1\right)}{12} \\ & =3\left(n^2-1\right) \\ \Rightarrow \quad n^2-1 & =\frac{P}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, variance of $2,4,6, \ldots, 2 n$

$ \begin{gathered} a=2, d=2 \\ \text { Variance, } \frac{\left(4^2\left(n^2-1\right)\right.}{12} \\ =\frac{4\left(n^2-1\right)}{12}=\frac{n^2-1}{3} \\ =\frac{\frac{P}{3}}{3}=\frac{P}{9} \end{gathered} $

Given data,

$ \begin{array}{lllllll} \hline \boldsymbol{x}_{\boldsymbol{i}} & 2 & 9 & 8 & 3 & 5 & 7 \\ \hline \boldsymbol{f}_{\boldsymbol{i}} & 5 & 3 & 1 & 6 & 6 & 1 \\ \hline \end{array} $

$ \begin{array}{ccc} \hline \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} & \begin{array}{c} \text { Cumulative } \\ \text { frequency } \end{array} \\ \hline 2 & 5 & 5 \\ \hline 9 & 3 & 8 \\ \hline 8 & 1 & 9 \\ \hline 3 & 6 & 15 \\ \hline 5 & 6 & 21 \\ \hline 7 & 1 & 22 \\ \hline \end{array} $

Total frequency $N=22$ (even)

$ \Rightarrow \quad \frac{N}{2}=\frac{22}{2}=11 $

Since, the cumulative frequency just greater than $\frac{N}{2}$ i.e., 11 is 15 and the value of $x$ corresponding to 15 is 3 .

$\therefore$ Median $=3$

Now, mean deviation

$ \left.=\frac{1}{N} \Sigma f_i \right\rvert\, x_i-\text { Median } \mid $

So, $\Sigma f_i\left|x_i-3\right|=44$

So, mean deviation $=\frac{44}{22}=2$

The numbers on a single die are 1 , $2,3,4,5$ and 6 .

$ \begin{aligned} \text { So, the mean } & =\frac{1+2+3+4+5+6}{6} \\ & =\frac{21}{6}=3.5 \end{aligned} $

When multiple dice are thrown, the mean of the sum of the numbers appearing on them = the sum of means of individual dice.

Since, each die has mean 3.5.

So, the mean of the sum of the numbers appearing on three dice

$ =3.5+3.5+3.5=10.5 $

$ \begin{array}{lcccc} \hline \begin{array}{l} \text { Class } \\ \text { interval } \end{array} & \begin{array}{l} \text { Frequency } \\ \left(\boldsymbol{f}_{\boldsymbol{i}}\right) \end{array} & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} x_{\boldsymbol{i}}^2 \\ \hline 0-4 & 1 & 2 & 2 & 4 \\ \hline 4-8 & 2 & 6 & 12 & 72 \\ \hline 8-12 & 2 & 10 & 20 & 200 \\ \hline 12-16 & 1 & 14 & 14 & 196 \\ \hline & \begin{array}{c} \sum f_i=N \\ =6 \end{array} & & \begin{array}{c} \sum f_i x_i \\ =48 \end{array} & \begin{array}{c} \sum f_i x_i^2 \\ =472 \end{array} \\ \hline \end{array} $

$ \begin{aligned} & \text { So, mean }(\bar{x})=\frac{\sum f_i x_i}{N}=\frac{48}{6}=8 \\ & \text { and variance }\left(\sigma^2\right)=\frac{\sum f_i x_i^2}{N}-(\bar{x})^2 \\ & \qquad=\frac{472}{6}-8^2=\frac{236}{3}-64 \\ & =\frac{236-192}{3}=44 / 3 \end{aligned} $

We have, $\sum_{i=1}^9 x_i-5=9$

$ \begin{aligned} & \text { and } \sum_{i=1}^9\left(x_i-5\right)^2=45 \\ & \qquad \begin{array}{l} \Sigma x_i=9+45=54 \\ \Sigma x_i^2-10 \Sigma x_i+25 \times 9=45 \\ \Sigma x_i^2-540+225=45 \\ \Sigma x_i^2=585-225=360 \end{array} \end{aligned} $

∴ Standard deviation $=\sqrt{\text { Variance }}$

$ \begin{aligned} & =\sqrt{\frac{\Sigma x_i^2}{9}-\left(\frac{\Sigma x_i}{9}\right)^2}=\sqrt{\frac{360}{9}-\left(\frac{54}{9}\right)^2} \\ & =\sqrt{40-36}=2 \end{aligned} $

$ \begin{array}{|c|c|c|c|c|} \hline \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} & \boldsymbol{c f} & \left|\boldsymbol{x}_{\boldsymbol{i}}-\mathbf{3}\right| & \boldsymbol{f}_{\boldsymbol{i}}\left|\boldsymbol{x}_{\boldsymbol{i}}-\mathbf{3}\right| \\ \hline 2 & 8 & 8 & 1 & 8 \\ \hline 3 & 6 & 14 & 0 & 0 \\ \hline 5 & 4 & 18 & 2 & 8 \\ \hline 7 & 2 & 20 & 4 & 8 \\ \hline 9 & 1 & 21 & 6 & 6 \\ \hline \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad N=21 \Rightarrow \frac{N}{2}=10.5 \\ & \therefore \text { Median }=M=3 \end{aligned}\\ &\text { and Mean deviation about median }\\ &=\frac{\sum f_i\left|x_i-M\right|}{\sum f_i}=\frac{30}{21}=\frac{10}{7} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \sum f_i & =11, \sum f_i x_i=55 \\ \bar{X} & =\frac{\sum f_i x_i}{\sum f_i}=\frac{55}{11}=5 \\ \sum f_i\left|x_i-\bar{x}\right| & =20 \end{aligned}\\ &\therefore \text { Mean deviation about mean is }\\ &\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i}=\frac{20}{11} \end{aligned} $

$ \begin{aligned} & \text { } \sum_{i=1}^{11}\left(x_i-4\right)=22 \\ & \Rightarrow \quad \sum_{i=1}^{11} x_i-4 \times 11=22 \\ & \Rightarrow \quad \sum_{i=1}^{11} x_i=66 \end{aligned} $

So, mean $(\bar{x})=\frac{66}{11}=6$ and                                      ...(i)

$ \begin{aligned} \because \quad x_i-\bar{x} & =x_i-6 \\ & =\left(x_i-4\right)-2 \\ \therefore \quad\left[x_i-\bar{x}\right]^2 & =\left[\left(x_i-4\right)-2\right]^2 \\ & =\left(x_i-4\right)^2-4\left(x_i-4\right)+4 \end{aligned} $

$ \begin{aligned}So,\,\, \sum_{i=1}^{11}\left(x_i-\bar{x}\right)^2 & =\sum_{i=1}^{11}(x-4)^2 \\ - & 4 \sum_{i=1}^{11}\left(x_i-4\right)+11 \times 4 \\ =154- & 4 \times 22+44=110 \end{aligned} $

$ \text { } \begin{aligned}Variance\,\, (\sigma) & =\frac{1}{n} \sum\left(x_i-\bar{x}\right)^2 \\ & =\frac{1}{11} \times 110=10 \end{aligned} $

Thus, $\alpha=6, \beta=10$

Therefore, $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{6}{10}+\frac{10}{6}=\frac{3}{5}+\frac{5}{3}$

$ =\frac{9+25}{15}=34 / 15=\text { Sum of roots } $

and $\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}=1=$ product of roots

Hence, the quadratic is

$ \begin{aligned} & x^2-\left(\frac{34}{15}\right) x+1=0 \\ \Rightarrow & 15 x^2-34 x+15=0 \end{aligned} $

Use formula for combined variance

Let $n_1=15, \mu_1=2, \sigma_1^2=4$

$ n_2=10, \mu_2=2, \sigma_2^2=5 $

$ \begin{aligned} \text { Combined mean } & =\mu=\frac{n_1 \mu_1+n_2 \mu_2}{n_1+n_2} \\ & =\frac{15 \times 2+10 \times 2}{25}=2 \end{aligned} $

Combined variance $=\sigma^2$

$ =\frac{n_1 \sigma_1^2+n_2 \sigma_2^2+n_1\left(\mu_1-\mu\right)^2+n_2\left(\mu_2-\mu\right)^2}{n_1+n_2} $

Since, $\mu_1=\mu_2=$ the last two terms becomes zero.

So, $\sigma^2=\frac{15 \times 4+10 \times 5}{25}=\frac{60+50}{25}=\frac{110}{25}$

$ =4.4 $

$ \begin{array}{cccc} \hline \text { Class interval } & \text { Frequency } & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}} \\ \hline 0-2 & 2 & 1 & 2 \\ \hline 2-4 & 3 & 3 & 9 \\ \hline 4-6 & 5 & 5 & 25 \\ \hline 6-8 & 3 & 7 & 21 \\ \hline 8-10 & 2 & 9 & 18 \\ \hline & \Sigma \boldsymbol{f}_{\boldsymbol{i}}=15 & & 75 \\ \hline \end{array} $

$ \begin{aligned} & \text { Mean }(\mu)=\frac{75}{15}=5 \\ & \qquad \begin{aligned} \sigma^2 & =\frac{\Sigma\left(x_i-\mu\right)^2 \cdot f_i}{N} \\ & =\frac{32+12+0+12+32}{15}=\frac{88}{15} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, } \Sigma f=20\\ &\begin{aligned} \therefore \quad(\alpha+2)+(\alpha-1)^2+4 & +(\alpha-1)+2+\alpha=20 \end{aligned} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \alpha^2+\alpha-12=0 \\ & (\alpha+4)(\alpha-3)=0 \Rightarrow \alpha=+3,-4 \\ \therefore & \alpha=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\because \alpha \neq-4]\end{array} $

$ \begin{array}{ccccc} \hline x_i & f_i & f_i x_i & \left|x_i-\bar{x}\right| & f \cdot\left|x_i-\bar{x}\right| \\ \hline 3 & 5 & 15 & 3 & 15 \\ \hline 4 & 4 & 16 & 2 & 8 \\ \hline 5 & 4 & 20 & 1 & 4 \\ \hline 8 & 2 & 16 & 2 & 4 \\ \hline 10 & 2 & 20 & 4 & 8 \\ \hline 11 & 3 & 33 & 5 & 15 \\ \hline & 20 & 120 & & \sum\left|x_i-\bar{x}\right|=54 \\ \hline \end{array} $

$ \begin{aligned} \bar{x} & =\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{120}{20}=6 \\ \mathrm{MD} & =\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f}=\frac{54}{20}=27 \end{aligned} $

Variance of Ist $n$ natural number is

$ =\frac{n^2-1}{12}=10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and variance of Ist $m$ even natural number is

$ \frac{m^2-1}{3}=16 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On solving Eqs. (i) and (ii), we get

$ \begin{aligned} & n^2=121 \text { and } m^2=49 \\ & n: m=11: 7 \end{aligned} $

$ \begin{aligned} &\text { Given, } 2,12,3,11,5,10,6,7\\ &\begin{aligned} \text { Mean } \bar{x} & =\frac{56}{8}=7 \\ \Sigma\left(x_i-\bar{x}\right)^2 & =25+25+16+16+4 +9+1+0 \\ & =96 \\ \operatorname{var}(x) & =\frac{\left(x_i-\bar{x}\right)^2}{n}=\frac{96}{8}=12 \end{aligned} \end{aligned} $

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

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

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

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

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

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

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

$\bar{x}=\frac{1+2+3+5+8+13+17}{7} $

$ =\frac{49}{7}=7 $

$ \begin{aligned} \text { Variance } & =\frac{\Sigma\left(x_{1}-x\right)^{2}}{7} \\\\ & =\frac{218}{7}=3114 \end{aligned} $

To calculate the variance of the first 10 natural numbers that are multiples of 3, we use the formula for variance:

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

Here, let $ x_i = 3k $ where $ k \in [1, 10] $. So the numbers are $ 3 \times 1, 3 \times 2, \ldots, 3 \times 10 $.

First, we find the sum of squares of these numbers:

$ \sigma^2 = \frac{1}{n} \times 9 \sum_{k=1}^{10} k^2 - 9(\bar{k})^2 $

Calculating step-by-step:

The formula for the sum of squares of the first $ n $ natural numbers is:

$ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} $

Substituting $ n = 10 $:

$ \sum_{k=1}^{10} k^2 = \frac{10 \times 11 \times 21}{6} $

The mean $ \bar{k} $ of the first 10 natural numbers is:

$ \bar{k} = \frac{1}{10} \sum_{k=1}^{10} k = \frac{10 \times 11}{2 \times 10} $

Plugging these into the variance formula:

$ \sigma^2 = \frac{9}{10} \times \frac{10 \times 11 \times 21}{6} - 9 \left(\frac{10 \times 11}{20}\right)^2 $

This simplifies to:

$ \sigma^2 = \frac{9}{10} \times 385 - 9 \times 30.25 $

Finally, calculating the result:

$ \sigma^2 = \frac{9}{10} \times 385 - 272.25 = 74.25 $

Thus, the variance of the first 10 multiples of 3 is $ 74.25 $.

We have, data $ 44,5,27,20,8,54,9 $, 14, 35

Data in ascending order is

$ 5,8,9,14,20,27,35,44,54 $

Mean

$ \begin{aligned} (\bar{x}) & =\frac{5+8+9+14+20+27+35+44+54}{9} \\\\ & =\frac{216}{9}=24 \end{aligned} $

Median $ (M)=20 $

$ M_{1}=\frac{1}{\eta} \Sigma\left|x_{i}-\bar{x}\right| $

$ =\frac{1}{9}[19+16+15+10+4+3+11+20+30] $ $ =\frac{128}{9} $

$ M_{2}=\frac{1}{\eta} \Sigma\left|x_{i}-M\right| $

$ =\frac{1}{9}[15+12+11+6+0+7+15+24+34] $

$ =\frac{124}{9} $

$ M_{1}-M_{2}=\frac{128}{9}-\frac{124}{9}=\frac{4}{9} $

Mean $ =1 $

$ \begin{array}{l} \Rightarrow n p=1 \Rightarrow P(X=2)=\frac{27}{128} \\\\ \Rightarrow{ }^{n} C_{2} p^{2} q^{n-2}=\frac{27}{128} \\\\ \Rightarrow \frac{n(n-1)}{2} p^{2} \cdot q^{n-2}=\frac{27}{128} \\\\ \Rightarrow \frac{n(n-1)}{2}\left(\frac{1}{n}\right)^{2} \times\left(1-\frac{1}{n}\right)^{n-2}=\frac{27}{128} \\\\ \Rightarrow\left(\frac{n-1}{2 n}\right)\left(\frac{n-1}{n}\right)^{n-2}=\frac{27}{128} \end{array} $

$ \begin{array}{l} \Rightarrow \quad \frac{(n-1)^{n-1}}{(n)^{n-1} \cdot 2}=\frac{27}{128}=\left(\frac{3}{4}\right)^{3} \times \frac{1}{4} \\\\ \therefore \quad n=4 \\\\ \text { Variance }=n p q=\frac{3}{4} \end{array} $

Mean, $\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{56}{8}=7$

Variance, $\sigma^2=\frac{\Sigma f_i\left(x_i-\bar{x}\right)^2}{\Sigma f_i}$

$ \begin{aligned} & =\frac{2(2-7)^2+\frac{3(6-7)^2+2(10-7)^2+(1)(14-7)^2}{8}}{=\frac{50+3+18+49}{8}=\frac{120}{8}=15} \end{aligned} $

Let assumed mean $A=5$ and $h=2$

Here, $N=13, \Sigma f_i d_i=0$

Mean $\bar{X}=A+h\left(\frac{1}{N} \Sigma f_i d_i\right)=5$

$\therefore \quad$ Mean deviation about mean $=\frac{1}{N} \Sigma f_i\left|x_i-\bar{X}\right|=\frac{20}{13}$