Solve the Following Question.(3 Marks)

Question 13 Marks

The probability distribution of discrete random variable $X$ is as follows:

$X =x$	1	2	3	4	5	6
$P ( X =x)$	$k$	$2k$	$3k$	$4k$	$5k$	$6k$

Find (1) the value of $k$
(ii) $P (2< X <4)$
(iii) $P ( X \geq 3$ )

Answer

(i) Since the table gives probability distribution, of a discrete random variable $X$
$
\begin{aligned}
\therefore P ( X =1)+ P ( X =2) & + P ( X =3)+ P ( X =4) \\
& + P ( X =5)+ P ( X =6)=1
\end{aligned}
$
$
\therefore k+2 k+3 k+4 k+5 k+6 k=1
$
$
\begin{aligned}
\therefore21 k & =1 \\
\therefore k & =\frac{1}{21}
\end{aligned}
$
$
\begin{aligned}
(ii) P (2< X <4)= P ( X =3) & =3 k=3 \times \frac{1}{21} \\
& =\frac{1}{7}
\end{aligned}
$
$
\begin{aligned}
(iii) P ( X \geq 3) F & = P ( X =3)+ P ( X =4)+ P ( X =5) \\
& + P ( X =6) \\
= & 3 k+4 k+5 k+6 k \\
= & 18 k=18 \times \frac{1}{21}=\frac{6}{7}
\end{aligned}
$

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Question 23 Marks

The probability distribution of $X$, the number of defects per $10$ metres of a fabric is given by

$x$	$0$	$1$	$2$	$3$	$4$
$P ( X =x)$	$0.45$	$0.35$	$0.15$	$0.03$	$0.02$

Find the varlance of $X$.

Answer

$x_i$	$p_i$	$p_i x_i$	$p_i x_i^2$
$0$	$0.45$	$0$	$0$
$1$	$0.35$	$0.35$	$0.35$
$2$	$0.15$	$0.30$	$0.60$
$3$	$0.03$	$0.09$	$0.27$
$4$	$0.02$	$0.08$	$0.32$
Total		$0.82$	$1.54$

From the table$, \sum p_i x_i=0.82$ and $\sum p_i x_i^2=1.54$
$\therefore \operatorname{Var}( X )=\sum p_i x_i^2-\left(\sum p_i x_i\right)^2$
$= 1.54 − (0.82)^2$
$= 1.54 − 0.6724$
$\therefore $ Var $(X) = 0.8676$

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Question 33 Marks

Two dice are thrown simultaneously. If $X$ denotes the number of sixes, find the expectation of $X$.

Answer

Here the r.v.X has values $0,1,2$
The sample space contains 36 points. Favourable cases for $X =1$ are
$
\begin{aligned}
& (1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2) \\
& (6,3),(6,4),(6,5) \\
\therefore & P(X=1)=\frac{10}{36} .
\end{aligned}
$
Favourable case for $X =2$ is $(6,6)$
$
\therefore P ( X =2)=\frac{1}{36}
$
Favourable case for $X =0$ are 25 .
$
\therefore P ( X =0)=\frac{25}{36}
$

$X =x$	0	1	2
$P ( X =x)$	$\frac{25}{36}$	$\frac{10}{36}$	$\frac{1}{36}$
$x P (x)$	0	$\frac{10}{36}$	$\frac{2}{36}$

$
E ( X )=\Sigma x \cdot P (x) .
$
$
\begin{aligned}
\therefore E ( X ) & =0 \times \frac{25}{36}+(1)\left(\frac{10}{36}\right)+2\left(\frac{1}{36}\right) \\
& =\frac{12}{36}=\frac{1}{3}
\end{aligned}
$

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Question 43 Marks

If $f(x)=k x \quad$, for $0<x<2$
$=0$, otherwise,
is the probability density function of a random variable $X$. then find :
(1) Value of $k$,
(ii) $P (1< X <2)$

Answer

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Question 53 Marks

For the following probability density function (p. d. f) of X, find P(X < 1) and P(|x| < 1)
$f(x)=\frac{x^2}{18},-3<x<3$
= 0, otherwise

Answer

We have
$P ( X <1)=\int_{-3}^1 f(x) d x$
$\begin{aligned} & =\int_{-3}^1 \frac{x^2}{18} d x \\ & =\left[\frac{x^3}{54}\right]_{-3}^1\end{aligned}$
$\begin{aligned} & =\frac{1}{54}-\left(\frac{-27}{54}\right) \\ & =\frac{28}{54} \\ & =\frac{14}{27}\end{aligned}$
= 0.5185
Now |X| < 1
± X < 1
∴ X < 1, – X < 1,
i.e. X > – 1
i.e. – 1 < X < 1
$\therefore$ Required Probability $=\int_{-1}^1 f(x) d x$
$\begin{aligned} & =\int_{-1}^1 \frac{x^2}{18} d x \\ & =\left[\frac{x^3}{54}\right]_{-1}^1 \\ & =\frac{1}{54}+\frac{1}{54} \\ & =\frac{2}{54} \\ & =\frac{1}{27} \\ & =0.03704\end{aligned}$

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Question 63 Marks

The probability distribution of a random variable $X$, the number of defects per 10 meters of a fabric is given by

$x$	0	1	2	3	4
$P ( X =x)$	0.45	0.35	0.15	0.03	0.02

Find the variance of $X$.

Answer

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Question 73 Marks

Let the p.m.f. (probablity mass function) of random variable $x$ be.
$\begin{array}{rlr}
P (x) & =\left(\frac{4}{x}\right)\left(\frac{5}{9}\right)^x\left(\frac{4}{9}\right)^{4-x}, & x=0,1,2,3,4 \\
& =0, & \text { otherwise }
\end{array}$
Find $E (x)$ and $\operatorname{Var}(x)$.

Answer

$p(x)=\left(\frac{4}{x}\right)\left(\frac{5}{9}\right)^x\left(\frac{4}{9}\right)^{4-x}, x=0,1,2, \ldots, 4$
Comparing with $p ( x )=\left(\frac{ n }{ x }\right)( p )^{ x }( q )^{ n - x }$
$\therefore n =4, p =\frac{5}{9}, q =\frac{4}{9}$
$\begin{aligned} & E(x)=n p=4 \times \frac{5}{9}=\frac{20}{9} \\ & V(x)=n p q=4 \times \frac{5}{9} \times \frac{4}{9}=\frac{80}{81}\end{aligned}$

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Question 83 Marks

The time (in minutes) for a lab assistant to prepare the equipment for a certain experiment is a random variable X taking values between 25 and 35 minutes with p.d.f.
$\begin{aligned}
f(x) & =\frac{1}{10}, & & 25 \leq x \leq 35 \\
& =0, & & \text { otherwise }
\end{aligned}$
What is the probability that preparation time exceeds 33 minutes ? Also find the c.d.f. of X.

Answer

Required probability = P(X > 33)
$\begin{aligned} & =\int_{33}^{\infty} f(x) d x \\ & =\int_{33}^{35} f(x) d x+\int_{35}^{\infty} f(x) d x \\ & =\int_{33}^{35} f(x) d x+0 \quad \ldots[ f ( x )=0, \text { when } x >35] \\ & =\int_{33}^{35} \frac{1}{10} d x\end{aligned}$
$\begin{aligned} & =\frac{1}{10} \int_{33}^{35} 1 d x \\ & =\frac{1}{10}[x]_{33}^{35} \\ & =\frac{1}{10}[35-33] \\ & =\frac{2}{10} \\ & =\frac{1}{5}\end{aligned}$
Let F (x) be the c.d.f. of X
∴ F(x) = P[X ≤ x]
$\begin{aligned} & =\int_{-\infty}^x f(x) d x \\ & =\int_{-\infty}^{25} f(x) d x+\int_{25}^x f(x) d x\end{aligned}$
$\begin{aligned} & =0+\int_{25}^x f(x) d x \quad \ldots[\because f( x )=0 \text {, when } f ( x )<25] \\ & =\int_{25}^x \frac{1}{10} d x \\ & =\frac{1}{10} \int_{25}^x 1 d x \\ & =\frac{1}{10}[x]_{25}^x\end{aligned}$
$\begin{aligned} & =\frac{1}{10}[x-25] \\ & \therefore F ( x )=\frac{x-25}{10}\end{aligned}$

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Question 93 Marks

Probability distribution of $X$ is given by

$X =x$	1	2	3	4
$P ( X =x)$	0.1	0.3	0.4	0.2

Find $P(x \geq 2)$ and obtain cumulative distribution function of $X$.

Answer

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Question 103 Marks

The probability mass function for $X=$ number of major defects in a randomly selected appliance of a certain type is

$X =x$	0	1	2	3	4
$P ( X =x)$	0.08	0.15	0.45	0.27	0..5

Find the expected value and variance of $X$.

Answer

$E(X)=\sum x_i . P\left(x_i\right)$
$=0(0.08)+1(0.15)+2(0.45)+3(0.27)+4(0.05)$
$=0+0.15+0.9+0.81+0.2=2.06$
$E\left(X^2\right)=\sum x_i^2 \cdot P\left(x_i\right)$
$=0(0.08)+1^2(0.15)+2^2(0.45)+3^2(0.27)+4^2(0.05)$
$=0+0.15+1.8+2.43+0.8=5.18$
$\operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2$
$=5.18-(2.06)^2$
$=5.18-4.2436$
$=0.9364$

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Maths Solve the Following Question.(3 Marks)

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