Answer to Question #296065 in Statistics and Probability for secret

Question

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 808 hours and a standard deviation of 93 hours. Find the probability that a random samples of 19 bulbs will have an average life of greater than 799 hours.

Answer 1

Let $X=$ an average life: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=808\ h, \sigma=93\ h, n=19.$

$$P(X>799)=1-P(X\le 799)$$

$$=1-P(Z\le \dfrac{799-808}{93/\sqrt{19}})$$

$$\approx 1-P(Z\le-0.421829)$$

$$\approx0.663425$$