An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 806 hours and a standard deviation of 90 hours. Find the probability that a random samples of 15 bulbs will have an average life of greater than 788 hours

Given that,

$\mu=806\\\sigma=90\\n=15$

We are required to determine the probability, $p(\bar x\gt 788)=p(Z\gt{788-806\over {90\over\sqrt{15}}})=p(Z\gt-0.77)=1-p(Z\lt -0.77)=1-0.2206=0.7794$

Therefore, the probability that a random sample of 15 bulbs will have an average life greater than 788 hours is 0.7794.