Suppose compact fluorescent light bulbs last, on average, 11,500 hours. The distribution is normal and the standard deviation is 400 hours. What percent of light bulbs burn out within 12,300 hours?

2.275%

Step-by-step explanation:

The first thing to do here is to calculate the z-score

Mathematically;

z-score = (x-mean)/SD

from the question, x = 12,300 hours , mean = 11,500 hours while Standard deviation(SD) = 400 hours

Plugging the values we have;

z-score = (12,300-11,500)/400 = 800/400 = 2

Now, we want to calculate P(z ≤ 2)

This is so because we are calculating within a particular value

To calculate this, we use the z-score table.

Mathematically;

P(z ≤ 2) = 1 - P(z > 2) = 1 - 0.97725 = 0.02275

To percentage = 2.275%

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step-by-step explanation: