Question

[img]https://upload.cc/i1/2020/04/21/sDqOnK.jpg[/img]


question in photo

Answer 1

X=life time of the a relationship for couple.

$\mu=5months\\ \lambda=\frac{1}{\mu}=\frac{1}{5}=0.2\\ X\thicksim exp(0.2)$

$f(x)= \begin{cases} 0.2e^{(-0.2x)} &\text{if } x\ge0 \\ 0 &\text{if } x<0 \end{cases}$

and

$F(x)= \begin{cases} 1-e^{(-0.2x)} &\text{if } x\ge0 \\ 0 &\text{if } x<0 \end{cases}$

i)

$P(X>12)=1-P(X\le12)\\ P(X>12)=1-F(12)\\ P(X>12)=1-(1-e^{(0.2*12)})\\ P(X>12)=0.0907$

Probability that a couple in hong kong can last longer that 1 year =0.0907

ii)Median of a exponential distribution is given by,

$median=\frac{ln(2)}{\lambda}\\ median=\frac{ln(2)}{0.2}\\ \bold{median=3.4657}$

iii) Probability(P₁) that a couple in hong kong can last longer that 1 year =0.0907

Probability(P)that exact three out of eight couples in hong kong can last longer than 1 year,

$P=\binom{8}{3} P_1^3(1-P_1)^{8-3}\\ P=\frac{3!}{8!5!}*0.0907^3*(1-0.0907)^5\\ P=5.7518*10^{-10}$

Probability that exact three out of eight couples in hong kong can last longer than 1 year=$\bold{5.7518*10^{-10}}$