Answer to Question #185811 in Statistics and Probability for Betty

Question

1.Suppose on average, Nepal experiences 6 earthquakes per year.

a. What is the mean number of earthquakes in Nepal in the first four month of a year?

b. What is the probability that there’ll be 7 earthquakes in Nepal in the next two years?

c. What is the probability that there’ll be at least 9 earthquakes in Nepal in 2021?

2.A TV manufacturing company is planning to launch a new type of product recently. To check the lifespan status of their previous products they conduct a study. And they found that the lifetime of their previous products (plasma TV sets) follows an exponential distribution with a mean of 100,000 hours. Compute the probability a television set:

a. Fails in less than 10,000 hours.

b. Lasts more than 120,000 hours.

c. Fails between 60,000 and 100,000 hours of use.

d. Find the 90th percentile. So, 10 percent of the TV sets last more than what length of time?

Answer 1

1.Average number of earthquake nepal experiences per year=6

Average number of earthquake nepal experiences per month$=\dfrac{6}{12}=0.5,\lambda=0.5$

Let X denote the number of months-

(a) Mean number of earthquakes in Nepal in the first four month of a year-

$P(X=4)=\dfrac{e^{-\lambda}\lambda^{4}}{4!}=\dfrac{e^{-0.5}(0.5)^4}{24}=0.001579$

(b) probability that there’ll be 7 earthquakes in Nepal in the next two yearsi.e. in next 24 months

$P(X=7) =24\dfrac{e^{-0.5}(0.5)^7}{7!}$

$=24\times 0.000940=0.02256$

(C) probability that there’ll be at least 9 earthquakes in Nepal in 2021

$P(X\ge 9) =12\dfrac{e^{-0.5}(0.5)^9}{9!}$

$=12\times 0.0000391 =0.0047009$

2.Mean lifetime $\mu=100000$

This implies $\lambda=\dfrac{1}{\mu}=\dfrac{1}{100000}=10^{-5}$

Let X be the failing time-

(a) $P(X<10000)=1-e^{-\lambda(10000)}=1-e^{-10^{-5}(10^4)}=1-e^{-0.1}=1-0.905=0.0095$

(b) $P(X>120000)=e^{-\lambda(120000)}=e^{-10^{-5}(120000)}=e^{-1.2}=0.3012$

(c)$P(60000<X<100000)=1-e^{-\lambda(100000)}-(1-e^{-\lambda(60000)})$

$=-e^{-10^{-5}(10^5)}+e^{-10^{-5}(60000)}$

$=-e^{-1}+e^{-0.6}$

$=-0.3679+0.4720=0186$

(d) $P(X>X_1)=1-P(X<X_1)$

$P(X>X_1)=1-(1-e^{-\lambda X_1})$

$\Rightarrow 0.1=e^{-\lambda X_1}\\ \Rightarrow 0.1=e^{-X_1(10^{-5}}$

Taking Log on Both sides-

$ln0.1=-X_1(10^{-5})lne$

On solving we get-

$X_1=0.2302$

Hence The $90^{th}$ Percentile is 0.2302