Question

The amount of time a certain brand of light bulb lasts is normally distributed with a mean of 1200 hours and a standard deviation of 60 hours. What percentage light bulbs last less than 1160 hours, to the nearest tenth?

Answer 1

To determine the percentage of light bulbs that last less than 1160 hours, we first need to convert the raw score (1160 hours) into a z-score using the formula:

\[
z = \frac{(X - \mu)}{\sigma}
\]

where:
- \(X\) is the value we are interested in (1160 hours),
- \(\mu\) is the mean (1200 hours),
- \(\sigma\) is the standard deviation (60 hours).

Plugging in the values, we get:

\[
z = \frac{(1160 - 1200)}{60} = \frac{-40}{60} = -\frac{2}{3} \approx -0.6667
\]

Next, we use the z-score to find the corresponding percentile. We can look up the z-score of -0.6667 in the standard normal distribution table, or use a calculator or software that provides cumulative probabilities for the standard normal distribution.

Using a z-table or calculator:
- The cumulative probability for \(z = -0.6667\) is approximately 0.2525 (or 25.25%).

Thus, about 25.3% of the light bulbs last less than 1160 hours when rounded to the nearest tenth.

So, the final answer is:
\[
\text{Approximately } 25.3\%
\]