Question

Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years.

a. Find the probability that a light bulb lasts less than one year.

b. Find the probability that a light bulb lasts between six and ten years.

c. Seventy percent of all light bulbs last at least how long?

d. A company decides to offer a warranty to give refunds to light bulbs whose lifetime is among the lowest two percent of all bulbs. To the nearest month, what should be the cutoff lifetime for the warranty to take place?

e. If a light bulb has lasted seven years, what is the probability that it fails within the $8th$year.

Short answer

a. $0.1175$

b. $0.1859$

c. $2.85years$

d. $0.1616$

e.$0.1175$

Step-by-step solution

introduction

a probability distribution is a numerical capacity that gives the probabilities of events of various potential results for an experiment.[1][2] It is a numerical portrayal of an irregular peculiarity as far as its example space and the probabilities of occasions (subsets of the example space)

Explanation (part a)

The longevity of a light bulb is exponential with a mean lifetime of eight years.

Let the life time of a light bulb. The decay parameter is , and . The cumulative distribution function is

Therefore, $P(T<1)=1-e^{-\frac{1}{8}}\approx 0.1175$

the probability that a light bulb lasts less than one year is$0.1175$

Explanation (part b)

We want to find $P(6<t<10).$

To do this, $P(6<t<10)–P(t<6)$

$=\left(1-e^{-\frac{1}{8}\ast 10}\right)-\left(1-e^{-\frac{1}{8}\ast 6}\right)\approx 0.7135-0.5276=0.1859$

the probability that a light bulb lasts between six and ten years is$0.1859$

Explanation (part c)

We want to find $0.70$$=P(T>t)=1-\left(1-e^{-\frac{t}{8}}\right)=e^{-\frac{t}{8}}$

Solving for $t$, localid="1651557906622" ${\mathrm{e}}^{-\frac{t}{8}}=0.70,\text{so}-\frac{t}{8}=\ln \left(0.70\right),\text{and}t=-8\ln \left(0.70\right)\approx 2.85\text{years.}$

or use localid="1651557912447" $t=\frac{\ln \left(\text{area\_to\_the\_right}\right)}{(-m)}=\frac{\ln \left(0.70\right)}{-\frac{1}{8}}\approx 2.85\text{years}.$

Seventy per cent of all light bulbs last at least$2.85$years

Explanation (part d)

We want to find $0.02=P(T<t)=1-e^{-\frac{t}{8}}$

Solving for $t,e^{-\frac{t}{8}}=0.98$so, $-\frac{t}{8}=\ln \left(0.98\right),\text{and}t=-8\ln \left(0.98\right)\approx 0.1616\text{years, or roughly two months.}$

The warranty should cover light bulbs that last less than 2 months.

Or use$\frac{\ln \left(\text{area\_to\_the\_right}\right)}{(-\mathrm{m})}=\frac{\ln (1-0.2)}{-\frac{1}{8}}=0.1616$

Explanation (part e)

We must find $P(T<8\mid T>7)$

Notice that by the rule of complement events, $P(T<8\mid T>7)=1-P(T>8\mid T>7)$

By the memory less property $\left(P\right(X>r+t\mid X>r)=P(X>t\left)\right)$

So $P(T>8\mid T>7)=P(T>1)=1-\left(1-e^{-\frac{1}{8}}\right)=e^{-\frac{1}{8}}\approx 0.8825$

Therefore, $P(T<8\mid T>7)=1-0.8825=0.1175$

the probability that it fails within the year is$0.1175$