MCQ

MCQ 12 Marks

If $X$ is a random variable with probability mass function
$P (x)=k x \quad, \quad$ for $x=1,2,3$
$=0$. otherwise then $k=\ldots$

A
$\frac{1}{5}$
B
$\frac{1}{4}$
✓
$\frac{1}{6}$
D
$\frac{2}{3}$

Answer

Correct option: C.

$\frac{1}{6}$

$x$	$1$	$2$	$3$
$P(x)$	$k$	$2k$	$3k$

Since, the function is a $\text{p.m.f.}$
$\therefore ∑P(x_i) = 1$
$\therefore k + 2k + 3k = 1$
$\therefore k = 1/6$

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MCQ 22 Marks

If the $\text{p.m.f.}$ of a $\text{r.v. x}$ is
$P (x) =\frac{c}{x^3},$ for $x=1,2,3=0,$ otherwise,
then $E ( X )=.......$

A
$\frac{216}{251}$
✓
$\frac{294}{251}$
C
$\frac{297}{294}$
D
$\frac{294}{297}$

Answer

Correct option: B.

$\frac{294}{251}$

$\frac{294}{251}$
$P (x)=\frac{c}{x^3}$, for $x=1,2,3$
$\therefore P (x=1)=c, P (x=2)=\frac{c}{8}, P (x=3)=\frac{c}{27}$
We have $P (x=1)+ P (x=2)+ P (x=3)=1$
$\therefore c+\frac{c}{8}+\frac{c}{27} =1$
$ \therefore \frac{216 c+27 c+8 c}{27 \times 8} =1$
$ \therefore 251 c =216,$
$ \therefore c =\frac{216}{251}$

$x$	$1$	$2$	$3$
$P (x)$	$c$	$\frac{c}{8}$	$\frac{c}{27}$

$E(X)=(1)\left(\frac{216}{251}\right)+2\left(\frac{216}{8 \times 251}\right)+3\left(\frac{216}{27 \times 251}\right)$
$=\frac{216}{251}\left(1+\frac{1}{4}+\frac{1}{9}\right)$
$=\frac{216}{251} \times \frac{49}{36}$
$=\frac{294}{251}$
Hence option $(b)$

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MCQ 32 Marks

Let the $\text{p.m.f.}$ of a random variable $X$ be $-$
$P(x)=\frac{3-x}{10}, $ for $ x=-1,0,1,2$
$=0 $ otherwise Then $ E(X)$ is $..... $

A
$1$
B
$2$
✓
$0$
D
$-1$

Answer

Correct option: C.

$0$

$x$	$-1$	$0$	$1$	$2$
$p(x)$	$\frac{4}{10}$	$\frac{3}{10}$	$\frac{2}{10}$	$\frac{1}{10}$
$x.p(x)$	$\frac{-4}{10}$	$0$	$\frac{2}{10}$	$\frac{2}{10}$

$\sum x \cdot P(x)=0$

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MCQ 42 Marks

A random variable $X$ has the following probability distribution :

$X =x$	-2	-1	0	1	2	3
$P (x)$	0.1	0.1	0.2	0.2	0.3	0.1

Then $E (x)=$

A
0.8
B
0.9
C
0.7
D
1.1

Answer

coming soon

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MCQ 52 Marks

The integrating factor of linear differential equation $\frac{d y}{d x}+y \sec x=\tan x$ is

A
$\sec x-\tan x$
B
$\sec x \cdot \tan x$
C
$\sec x+\tan x$
D
$\sec x \cdot \cot x$

Answer

$\frac{d y}{d x}+P \cdot y=Q$
therefore P=secx
I. $f=e^{\int \sec x d x}=e^{\log |s e x+\tan x|}$
=secx+tanx

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MCQ 62 Marks

If the p.d.f. of a continuous random variable $X$ is given as
$
f(x)=\frac{x^2}{3},-1$
$=0$, otherwise then c.d.f. of $X$ is

A
$\frac{x^3}{9}+\frac{1}{9}$
B
$\frac{x^3}{9}-\frac{1}{9}$
C
$\frac{x^3}{4}+\frac{1}{4}$
D
$\frac{1}{9 x^3}+\frac{1}{9}$

Answer

coming soon

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