# Solved: The useful life of an electrical component is exponentially distributed with a mean of 2,500 hours

#### ByDr. Raju Chaudhari

May 31, 2021

The useful life of an electrical component is exponentially distributed with a mean of 2,500 hours.

a. What is the probability the circuit will last more than 3,000 hours?
b. What is the probability the circuit will last between 2,500 and 2,750 hours?
c. What is the probability the circuit will fail within the first 2,000 hours?

#### Solution

Let $X$ be useful life of an electrical component . $X\sim \exp(2500)$.

The pdf of $X$ is

 \begin{aligned} f(x) & = \frac{1}{2500}e^{-x/2500}, x>0 \end{aligned}

The distribution function of $X$ is

 \begin{aligned} F(x) & = 1-e^{-x/2500} \end{aligned}

a. The probability the circuit will last more than 3,000 hours is

 \begin{aligned} P(X > 3000) & = 1-P(X\leq 3000)\\ &= 1-F(3000)\\ &= 1- (1 - e^{-3000/2500})\\ & = e^{-3000/2500} \\ &= 0.3012 \end{aligned}

b. The probability that the circuit will last between 2,500 and 2,750 hours is

 \begin{aligned} P(2500 < X < 2750) & = F(2750)-F(2500)\\ &= (1 - e^{-2750/2500})-(1 - e^{-2500/2500})\\ & = e^{-2500/2500} -e^{-2750/2500}\\ &= 0.3679-0.3329\\ &= 0.035 \end{aligned}

c. The probability that the circuit will fail within the first 2,000 hours is

 \begin{aligned} P(0 < X < 2000) & = P(X < 2000)\\ &= F(2000)\\ &= (1 - e^{-2000/2500})\\ & = 1-e^{-2000/2500} \\ &= 1-0.4493\\ &= 0.5507 \end{aligned}