Question

An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean equal to 809 hours and a standard deviation of 48 hours. Find the probability that a bulb burns between 728 and 848 hours

Answer 1

Let $X=$ a length of life: $X\sim N(\mu, \sigma^2).$

Given $\mu=809\ h, \sigma=48\ h.$

$$P(728<X<848)=P(X<848)-P(X\le 728)$$

$$=P(Z<\dfrac{848-809}{48})-P(Z\le \dfrac{728-809}{48})$$

$$= P(Z<0.8125)-P(X\le -1.6875)$$

$$\approx0.79175-0.04575= 0.7460$$