Answer the following questions in short.

Question 11 Mark

Solve the following:
Let $X ~ B(n, p).$
If $E(X) = 5$ and $Var(X) = 2.5,$ find $n$ and $p.$

Answer

Given: $E(X)=5$ and $\operatorname{Var}(X)=2.5$
$ \therefore n p=5 \text { and } n p q=2.5$
$\therefore \frac{n p q}{n p}=\frac{2.5}{5}$
$\therefore q=0.5=\frac{5}{10}=\frac{1}{2}$
$\therefore \mathrm{p}=1-\mathrm{q}=1-\frac{1}{2}=\frac{1}{2} $
Substituting $p=\frac{1}{2}$ in $n p=5$, we get
$ \mathrm{n}\left(\frac{1}{2}\right)=5$
$\therefore \mathrm{n}=10 $
Hence, $n=10$ and $p=\frac{1}{2}$

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Question 21 Mark

Solve the following:
Let $X \sim B(n, p).$
If $n = 10, E(X) = 5,$ find p and Var(X).

Answer

Given: $n=10$ and $E(X)=5$
But $E(X)=n p$
$ \therefore \mathrm{np}=5 .$
$\therefore 10 \mathrm{p}=5$
$\therefore \mathrm{p}=\frac{1}{2}$
$\therefore \mathrm{q}=1-\mathrm{p}=1-\frac{1}{2}=\frac{1}{2}$
$\operatorname{Var}(X)=\mathrm{npq}=10\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=2.5 .$
$\text { Hence, } \mathrm{p}=\frac{1}{2} \text { and } \operatorname{Var}(X)=2.5 $

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Question 31 Mark

If $E(X)=6$ and $\operatorname{Var}(X)=4 \cdot 2$, find $n$ and $p$.

Answer

$E(X)=6$ therefore $n p=6$ and $\operatorname{Var}(X)=4 \cdot 2$ therefore $n p q=4 \cdot 2$
$
\begin{aligned}
& \frac{n p q}{n p}=\frac{4 \cdot 2}{6} \quad \therefore \quad q=0.7 \\
& \therefore p=1-q=1-0.7 \quad \therefore \quad p=0.3 \\
& n p=6 \\
& \therefore n \times 0 \cdot 3=6 \quad \therefore \quad n=\frac{6}{0 \cdot 3}=20 \\
\end{aligned}
$

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Question 41 Mark

Let the p.m.f. of r.v. $X$ be
$
P(X=x)={ }^4 C_x\left(\frac{5}{9}\right)^x \times\left(\frac{4}{9}\right)^{4-x} \text {, for } x=0,1,2,3,4 .
$
then find $E(X)$ and $\operatorname{Var}(X)$.

Answer

$P(X=x)$ is binomial distribution with $n=4, p=\frac{5}{9}$ and $q=\frac{4}{9}$
$
\begin{aligned}
E(X) & =n p \\
& =4 \times\left(\frac{5}{9}\right)=\frac{20}{9} \\
\operatorname{Var}(X) & =n p q \\
& =4 \times\left(\frac{5}{9}\right) \times\left(\frac{4}{9}\right)=\frac{80}{81}
\end{aligned}
$

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Question 51 Mark

Given $X \sim B(n, P)$ : If $n=10, E(X)=8$, find $\operatorname{Var}(X)$.

Answer

Given: $n=10, E(X)=8$
$ E(X)=n p$
$8=10 p$
$p=\frac{8}{10}=\frac{4}{5}$
$q=1-p=1-\frac{4}{5}=\frac{1}{5}$
$\operatorname{Var}(X)=n p q=10\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)=\frac{8}{5} $
Hence, $\operatorname{Var}(X)=\frac{8}{5}$.

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Question 61 Mark

Given $X \sim B(n, P)$ : If $n=25, E(X)=10$, find $p$ and $\operatorname{SD}(X)$.

Answer

Given: $n=25, E(X)=10$
$ E(X)=n p$
$10=25 p$
$p=\frac{10}{25}=\frac{2}{5}$
$\therefore q=1-p=1-\frac{2}{5}=\frac{3}{5}$
$\operatorname{Var}(X)=n p q=25 \times \frac{2}{5} \times \frac{3}{5}=6$
$\therefore S D(X)=\sqrt{ } \operatorname{Var}(X)=\sqrt{ } 6$
$\text { Hence, } p=\frac{2}{5} \text { and } S \cdot D \cdot(X)=\sqrt{ } 6 . $

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Question 71 Mark

Given $X \sim B(n, P)$ : If $p=0.6$ and $E(X)=6$, find $n$ and $\operatorname{Var}(X)$.

Answer

Given: $p=0.6, E(X)=6$
$ E(X)=n p$
$6=n(0.6)$
$n=\frac{6}{0.6}=10 $
Now, $q=1-p=1-0.6=0.4$
$\therefore \operatorname{Var}(X)=n p q=10(0.6)(0.4)=2.4$
Hence, $\mathrm{n}=10$ and $\operatorname{Var}(X)=2.4$.

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Question 81 Mark

Given $X \sim B(n, P)$ : If $n=10$ and $p=0.4$, find $E(X)$ and $\operatorname{Var}(X)$.

Answer

Given: $\mathrm{n}=10$ and $p=0.4$
$ \therefore q=1-p=1-0.4=0.6$
$\therefore E(X)=n p=10(0.4)=4$
$\operatorname{Var}(X)=n p q=10(0.4)(0.6)=2.4$
$\text { Hence, } E(X)=4, \operatorname{Var}(X)=2.4 $

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Maths Answer the following questions in short.

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