An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 801 hours and a standard deviation of 98 hours. Find the probability that a random samples of 19 bulbs will have an average life between 782 and 818 hours.

$\mu=801\\\sigma=98\\n=19$

Let the random variable $X$ represent the length of life of light bulbs. We are required to find the probability, $p(782\lt \bar x\lt 818)=p({782-\mu\over{\sigma\over\sqrt{n}}}\lt {\bar x-\mu\over{\sigma\over\sqrt{n}}}\lt{818-\mu\over{\sigma\over\sqrt{n}}})=p({782-801\over{98\over\sqrt{19}}}\lt Z\lt {{818-801\over{98\over\sqrt{19}}}})=p(-0.85\lt Z\lt0.76)=\phi(0.76)-\phi(-0.85)=0.7764-0.1977=0.5787$

Therefore, the probability that a random sample of 19 bulbs will have an average life between 782 and 818 hours is 0.5787