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
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%203.5%2C%20%5Csigma%20%3D%200.75)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{2 - 3.5}{0.75}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B2%20-%203.5%7D%7B0.75%7D)
![Z = -2](https://tex.z-dn.net/?f=Z%20%3D%20-2)
has a p-value of 0.0228
0.0228 = 2.28% probability one of these light bulbs burns out in under two years.