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 falls within the $8^{\text{th}}$ year.

Given data:

$\mu =8$,

$m=0.125$

The exponential distribution is defined as follows:

$\mathrm{P}(\mathrm{X}<\mathrm{x})=1-{\mathrm{e}}^{-\mathrm{mx}}$

The required probability can be estimated using the formula,

$\mathrm{P}(\mathrm{x}<1)=1-{\mathrm{e}}^{-0.125\times 1}$

$\mathrm{P}(\mathrm{x}<1)=1-0.8825$

$\mathrm{P}(\mathrm{x}<1)=0.1175$

$\mathrm{P}(\mathrm{x}<1)=0.1175$is the probability that a light bulb will last less than one year.

Given data:

$\mu =8$

$m=0.125$

The exponential distribution is defined as follows:

$\mathrm{P}(\mathrm{X}<\mathrm{x})=1-{\mathrm{e}}^{-\mathrm{mx}}$

The required probability can be estimated using the formula:

$\mathrm{P}(6<\mathrm{x}<10)=\mathrm{P}(\mathrm{x}<10)-\mathrm{P}(\mathrm{x}<6)$

$\mathrm{P}(6<\mathrm{x}<10)=1-{\mathrm{e}}^{-0.125\times 10}-\left(1-{\mathrm{e}}^{-0.125\times 6}\right)$$\mathrm{P}(6<\mathrm{x}<10)=0.4784-0.2865$

$\mathrm{P}(6<\mathrm{x}<10)=0.1859$

Given data:

$\mu =8$,

$m=0.125$

The exponential distribution is defined as follows:

$\mathrm{P}(X<x)=1-e^{-mx}$

$\mathrm{P}(x>k)=1-\left(1-e^{-0.125\times k}\right)$

localid="1653537354167" $0.70=e^{-0.125\times k}$

Taking log on both sides,

$In(0.70)=In\left({\mathrm{e}}^{-0.125\times \mathrm{k}}\right)$

$-0.125\mathrm{k}=-0.3567$

$\mathrm{k}=2.85$

Given data:

$\mu =8$

$m=0.125$

The given probability is $\mathrm{P}(x<k)$

$\mathrm{P}(\mathrm{x}<\mathrm{k})=1-{\mathrm{e}}^{-0.125\mathrm{k}}$

$0.02=1-{\mathrm{e}}^{-0.125\mathrm{k}}$

${\mathrm{e}}^{-0.125\mathrm{k}}=0.98$

Taking log on both sides,

${\mathrm{e}}^{-0.125\mathrm{k}}=0.98$

$-0.125\mathrm{k}=-0.0202$

$\mathrm{k}=0.1616$

The given probability is

$P(x<8\mid x>7)=1-P(x<8\mid x>7)$.

The memory less exponential distribution can be calculated as follows:

$\mathrm{P}(\mathrm{x}<\mathrm{t}+\mathrm{r}\mid \mathrm{x}>\mathrm{t})=\mathrm{P}(\mathrm{x}>\mathrm{r})$$\mathrm{P}(\mathrm{x}<8\mid \mathrm{x}>7)=1-\mathrm{P}(\mathrm{x}>7+1\mid \mathrm{x}>7)$$\mathrm{P}(\mathrm{x}<8\mid \mathrm{x}>7)=1-\mathrm{P}(\mathrm{x}>1)$$\mathrm{P}(\mathrm{x}<8\mid \mathrm{x}>7)=1-[1-\mathrm{P}(\mathrm{x}>1\left)\right]$

$\mathrm{P}(\mathrm{x}<8\mid \mathrm{x}>7)=\mathrm{P}(\mathrm{x}<1)$

$\mathrm{P}(\mathrm{x}<8\mid \mathrm{x}>7)=1-{\mathrm{e}}^{0.125\times 1}$

$\mathrm{P}(\mathrm{x}<8\mid \mathrm{x}>7)=1-0.8825$

$\mathrm{P}(\mathrm{x}<8\mid \mathrm{x}>7)=0.1175$