To determine the percentage of light bulbs that will last longer than 880 hours, we need to calculate the z-score for 880 hours using the given mean and standard deviation. Then, we can use the z-score to find the corresponding probability from the standard normal distribution.
To determine the percentage of light bulbs that will last longer than 880 hours, we first calculate the z-score using the formula:
\[
z = \frac{x - \mu}{\sigma}
\]
where \( x = 880 \), \( \mu = 850 \), and \( \sigma = 50 \).
Substituting the values, we get:
\[
z = \frac{880 - 850}{50} = 0.6
\]
Next, we find the probability that a bulb lasts less than 880 hours using the cumulative distribution function (CDF) of the standard normal distribution:
\[
P(X < 880) = \Phi(z) = \Phi(0.6) \approx 0.7257
\]
To find the percentage of bulbs that last longer than 880 hours, we subtract the above probability from 1:
\[
P(X > 880) = 1 - \Phi(0.6) \approx 1 - 0.7257 = 0.2743
\]
Converting this probability to a percentage:
\[
\text{Percentage} = 0.2743 \times 100 \approx 27.43\%
\]