Question

The lifetimes of a certain brand of light bulb are normally distributed, with a mean of 3.5 years and a standard deviation of .75 years. What is the probability one of these light bulbs burns out in under two years

Answer 1

Answer:

0.0228 = 2.28% probability one of these light bulbs burns out in under two years.

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 3.5 years and a standard deviation of .75 years.

This means that $\mu = 3.5, \sigma = 0.75$

What is the probability one of these light bulbs burns out in under two years?

This is the p-value of Z when X = 2. So

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

$Z = \frac{2 - 3.5}{0.75}$

$Z = -2$

$Z = -2$ has a p-value of 0.0228

0.0228 = 2.28% probability one of these light bulbs burns out in under two years.