Answer to Question #292948 in Statistics and Probability for secret

Question

An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean equal to 817 hours and a standard deviation of 48 hours. Find the probability that a bulb burns between 736 and 823 hours.

Answer 1

$\mu=817\\\sigma=48$

Let the random variable $X$ denote the length of life of light bulbs then, $X\sim N(\mu=817,\sigma^2=48^2)$

We determine,

$p(736\lt x\lt 823)=p({736-\mu\over\sigma}\lt Z\lt{823-\mu\over\sigma})=p(({736-817\over48}\lt Z\lt{823-817\over48})=p(-1.69\lt Z\lt 0.13)=\phi(0.13)-\phi(-1.69)=0.5497382 -0.04575362=0.50398458$Therefore, the probability that a bulb burns between 736 and 823 hours is 0.50398458