Solve the Following Question.(4 Marks)

Question 14 Marks

Find the expected value, variance and standard derivation of random variable $X$ whose probability mass function $(p.m.f.)$ is given below

$X =x$	$1$	$2$	$3$
$P ( X =x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

Answer

$E(X)=\sum x_i P\left(x_i\right)$
$=1(1/5)+2(2/5)+3(2/5)$
$=(1+4+6)/5=11/5$
$=2.2$
$E\left(X^2\right)=\sum x_i^2 P\left(x_i\right)$
$=1^2(1/5)+2^2(2/5)+3^2(2/5)$
$=(1+8+18)/5=27/5$
$=5.4$
Var $(X) = E(X^2) - [E(X)]^2$
$= 5.4 - (2.2)^2$
$= 5.4 - 4.84$
$= 0.56$
$S.D. =\sqrt{\operatorname{Var}(X)}=\sqrt{0.56}=0.7483$ bnhyt $5r$

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Question 24 Marks

A bakerman sells $5$ types of cakes. Profits due to the sale of each type of cake is respectively $₹ 3,₹ 2.5,₹ 2, ₹ 1.5, ₹ 1 . $ The demands for these cakes are $10 \%, 5 \%, 25 \%, 45 \%$ and $15 \%$ respectively. What is the expected profit per cake?

Answer

Let $X =$ demand for each type of cake $($according to the profit$)$
$P(X = 3) = 10\% = 10/100 = 0.1$
$P(X = 2.5) = 5\% = 5/100= 0.05$
$P(X = 2) = 20\% = 20/100= 0.2$
$P(X = 1.5) = 50\% = 50/100= 0.5$
$P(X = 1) = 15\% = 15/100 = 0.15$
$\therefore$ The probability distribution table is as follows:

$X$	$3$	$2.5$	$2$	$1.5$	$1$
$P(X)$	$0.1$	$0.05$	$0.2$	$0.5$	$0.15$

$E(X) = ∑x_i ⋅P(x_i)$
$= 3(0.1) + 2.5(0.05) + 2(0.2) + 1.5(0.5) + 1(0.15)$
$= 0.3 + 0.125 + 0.4 + 0.75 + 0.15 = 1.725$
$= 1.73$
$\therefore$ Expected profit per cake $= Rs. 1.73$

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Question 34 Marks

Find the variance and standard deviation of the random variable $X$ whose probability distribution is given below:

$x$	0	1	2	3
$P ( X =x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Answer

$x_i$	$p_i$	$p_i x_i$	$p_i x_i^2$
0	$1 / 8$	0	0
1	$3 / 8$	$3 / 8$	$3 / 8$
2	$3 / 8$	$6 / 8$	$12 / 8$
3	$1 / 8$	$3 / 8$	$9 / 8$
	Total	$12 / 8$	$24 / 8$= 3

$E(X)=\mu=\sum p_i x_i=\frac{12}{8}=\frac{3}{2}$
$\operatorname{Var}(X)=\sum_{i=1}^n p_i x_i^2-\mu^2$
$\begin{aligned} & =3-\left(\frac{3}{2}\right)^2 \\ & =3-\frac{9}{4} \\ & =\frac{3}{4} \\ & \therefore \operatorname{Var}(X)=\sigma^2=\frac{3}{4}\end{aligned}$
Standard deivation of $(X)=\sigma_x=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$

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Question 44 Marks

Given the probability density function (p.d.f.) of a continuous random variable $x$ as:
$\begin{aligned}
f(x) & =\frac{x^2}{3}, & -1<x<2 \\
& =0, & \text { otherwise }
\end{aligned}$
Determine the cumulative distribution function (c.d.f.) of $X$ and hence find
$P ( X <1), P ( X >0), P (1< X <2) \text {. }$

Answer

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Question 54 Marks

The following is the p.d.f. (Probability Density Function) of a continuous random variable $X$ :
$\begin{aligned}
f(x) & =\frac{x}{32}, & 0& =0, & \text { otherwise }
\end{aligned}$
(a) Find the expression for c.d.f. (Cumulative Distribution Function) of $X$.
(b) Also find its value at $x=0.5$ and 9 .

Answer

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Question 64 Marks

Given the p.d.f. (probability density function) of a continuous random variable $x$ as:
$\begin{aligned}
f(x) & =\frac{x^2}{3}, & -1& =0, & \text { otherwise }
\end{aligned}$
Determine the c.d.f. (cumulative distribution function) of $x$ and hence find
$P (x<1), P (x \leq-2), P (x>0), P (1

Answer

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Question 74 Marks

A random variable $X$ has the following distribution :

$x$	0	1	2	3	4	5	6
$P ( X =x)$	$k$	$3k$	$5k$	$7k$	$9k$	$11k$	$13k$

(a) Find $k$,
(b) Find $P (0< X <4)$
(c) Obtain the cumulative distribution function (c.d.f.) of X.

Answer

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Question 84 Marks

Find $k$ if the function $f(x)$ is defined by :
$f(x)=k x(1-x), \quad$ for $0<x<1$
$=0$, otherwise.
is the probability density function (p.d.f.) of a random variable (r.v.) X. Also find $P\left(X<\frac{1}{2}\right)$

Answer

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Maths Solve the Following Question.(4 Marks)

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