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?

Short answer

Part a. \(P(x<1)=0.1175\)

Part b. \(P(6<x<10)=0.1859\)

Part c. \(P(x>k)=2.85\)years

Step-by-step solution

Part a. Step 1. Explanation

It is given,

\(X=Exp\left ( \frac{1}{8} \right )\)

where,

\(\mu=8\)

\(m=\frac{1}{\mu}\)

\(m=\frac{1}{8}\)

\(m=0.125\)

Cumulative distribution function exponential distribution is given by,

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

Required probability can be calculated by,

\(P(x<1)=1-e^{-0.125\times1}\)

\(P(x<1)=1-0.8825\)

\(P(x<1)=0.1175\)

Therefore, the probability that a light bulb lasts less than one year is \(P(x<1)=0.1175\).

Part b. Step 1. Calculation

Cumulative distribution function exponential distribution is given by,

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

Required probability can be calculated by,

\(P(6<x<10)=P(x<10)-P(x<6)\)

\(P(6<x<10)=1-e^{-0.125\times10}-(1-e^{-0.125\times6}\)

\(P(6<x<10)=0.4784-0.2865\)

\(P(6<x<10)=0.1859\)

Therefore the probability that a light bulb lasts between six and ten years \(P(6<x<10)=0.1859\).

Part c. Step 1. Explanation 

Cumulative distribution function of exponential distribution is given by,

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

\P(x>k)=1-(1-e^{-0.125\timesk}\)

\(0.70=e^{-0.125\times k}\)

Taking log on both sides,

\(ln(0.70)=ln(e^{-0.125\times k})\)

\(-0.125k=-0.3567\)

\(k=2.85\)

Value of \(P(x>k)=2.85\)years