Question

The amount of time a certain brand of light bulb lasts is normally distributed with a mean of 1600 hours and a standard deviation of 75 hours. What is the probability that a randomly chosen light bulb will last less than 1460 hours, to the nearest thousandth

Answer 1

Answer:

0.031 = 3.1% probability that a randomly chosen light bulb will last less than 1460 hours.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 1600 hours and a standard deviation of 75 hours.

This means that $\mu = 1600, \sigma = 75$

What is the probability that a randomly chosen light bulb will last less than 1460 hours?

This is the p-value of Z when X = 1460. So

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

$Z = \frac{1460 - 1600}{75}$

$Z = -1.87$

$Z = -1.87$ has a p-value of 0.031.

0.031 = 3.1% probability that a randomly chosen light bulb will last less than 1460 hours.