AMU MCA Probability Distribution Previous Year Questions (PYQs) – Page 1 of 4

For exponential distribution:

Mean $= \frac{1}{\lambda}$

Variance $= \frac{1}{\lambda^2}$

Given mean $=3 \Rightarrow \lambda=\frac{1}{3}$

Variance $=\frac{1}{(1/3)^2}=9$

Even values: $2,4,6,\dots$

$P(\text{even})=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\dots$

$=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots$

This is geometric series:

$=\frac{1/4}{1-1/4}=\frac{1/4}{3/4}=\frac{1}{3}$

Mean of uniform integers $1$ to $8$:

$=\frac{1+8}{2}=4.5$

Let $p=P(X=1)$

Then $P(X=0)=1-p$

$E(X)=p$

$V(X)=p(1-p)$

Given $p=3p(1-p)$

$1=3(1-p)$

$1=3-3p$

$3p=2$

$p=\frac{2}{3}$

Therefore $P(X=0)=1-p=\frac{1}{3}$

$Z=\frac{X-\mu}{\sigma} \sim N(0,1)$

Square of standard normal variable follows $\chi^2$ distribution with 1 d.f.

For normal curve, points of inflexion are at

$\mu \pm \sigma$

Thus $\mu=\mu_0$ with $\sigma^2$ known is simple.