Question

The amount of time a certain brand of light bulb lasts is normally distributed with a mean of 1400 hours and a standard deviation of 75 hours. What is the probability that a randomly chosen light bulb will last less than 1380 hours, to the nearest thousandth?

Answer 1

Answer:

0.394 = 39.4% probability that a randomly chosen light bulb will last less than 1380 hours

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 1400, \sigma = 75$

What is the probability that a randomly chosen light bulb will last less than 1380 hours, to the nearest thousandth?

This is the pvalue of Z when X = 1380. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1380 - 1400}{75}$

$Z = -0.27$

$Z = -0.27$ has a pvalue of 0.394

0.394 = 39.4% probability that a randomly chosen light bulb will last less than 1380 hours