Question

a certain manufacturer makes 400w light bulbs. assume that these light bulbs have lifetimes that are normally distributed with a mean lifetime of 680 hours and a standard deviation of 125 hours. use this table or the aleks calculator to find the percentage of light bulbs with lifetimes shorter than 720 hours. for your intermediate computations, use four or more decimal places. give your final answer to two decimal places (for example 98.23%).

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 720$ (the value we are interested in), $\mu = 680$ (the mean), and $\sigma=125$ (the standard deviation).
$z=\frac{720 - 680}{125}=\frac{40}{125}=0.3200$

Step2: Find the cumulative probability

We look up the z - score of $0.32$ in the standard normal distribution table. The cumulative probability $P(Z < 0.32)$ corresponds to the percentage of light - bulbs with lifetimes shorter than 720 hours. From the standard normal table, $P(Z < 0.32)=0.6255$

Answer:

$62.55\%$