A certain manufacturer makes $400 \mathrm{~W}$ light bulbs. Assume that these light bulbs have lifetimes that are normally distributed with a mean lifetime of 480 hours and a standard deviation of 50 hours.

Find the percentage of light bulbs with lifetimes shorter than 384 hours. For your intermediate computations, use four or more decimal places. Give your final answer to two decimal places (for example $98.23 \%$ ).

Step 1 ：We are given that the manufacturer makes 400W light bulbs with a mean lifetime of 480 hours and a standard deviation of 50 hours. We are asked to find the percentage of light bulbs with lifetimes shorter than 384 hours.

Step 2 ：This is a problem of finding the probability of a normally distributed random variable being less than a certain value. We can use the z-score formula to find the z-score for 384 hours.

Step 3 ：The z-score formula is \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value we are interested in (in this case, 384 hours), \(\mu\) is the mean (in this case, 480 hours), and \(\sigma\) is the standard deviation (in this case, 50 hours).

Step 4 ：Substituting the given values into the z-score formula, we get \(z = \frac{384 - 480}{50} = -1.92\).

Step 5 ：Once we have the z-score, we can use a z-table or a function to find the probability associated with that z-score. This will give us the percentage of light bulbs with lifetimes shorter than 384 hours.

Step 6 ：The probability associated with a z-score of -1.92 is approximately 0.0274.

Step 7 ：To convert this probability to a percentage, we multiply by 100, giving us approximately 2.74%.

Step 8 ：Final Answer: The percentage of light bulbs with lifetimes shorter than 384 hours is approximately \(\boxed{2.74\%}\).