Question

An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean equal to 778 hours and a standard deviation of 40 hours. Find the probability that a bulb burns between 710 and 841 hours.

Answer 1

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

Given $\mu=778\ h, \sigma=40\ h.$

$$P(710<X<841)=P(X<841)-P(X\le 710)$$

$$=P(Z<\dfrac{841-778}{40})-P(Z\le \dfrac{710-778}{40})$$

$$= P(Z<1.575)-P(X\le -1.7)$$

$$\approx0.94237-0.04457= 0.8978$$