Answer to Question #292067 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 796 hours and a standard deviation of 55 hours. Find the probability that a random samples of 40 bulbs will have an average life between 788 and 810 hours.

Answer 1

$\mu=796\\\sigma=55\\n=40$

We find the probability,

$p(788\lt \bar x\lt 810)$

We standardize this probability to a standard normal distribution as below.

$p(788\lt \bar x\lt 810)=p({788-\mu\over{\sigma\over \sqrt {n}}}\lt {\bar x-\mu\over{\sigma\over\sqrt{n}}}\lt {810-\mu\over{\sigma\over \sqrt {n}}})$. Substituting for the values of $\mu,\sigma$ and $n$ given above, we have,

$p({788-796\over{55\over \sqrt {40}}}\lt Z\lt {810-796\over{55\over \sqrt {40}}})=P(-0.92\lt Z\lt 1.61)=\phi(1.61)-\phi(-0.92)=0.9463-0.1788=0.7675$

Therefore, the probability that a random sample of 40 bulbs will have an average life between 788 and 810 hours is 0.7675.