Answer:
0.015 = 1.5% probability that the mean monitor life would be greater than 83.8 months in a sample of 71 monitors
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
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.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean life of 82 months with a standard deviation of 7 months.
This means that ![\mu = 82, \sigma = 7](https://tex.z-dn.net/?f=%5Cmu%20%3D%2082%2C%20%5Csigma%20%3D%207)
Sample of 71
This means that ![n = 71, s = \frac{7}{\sqrt{71}}](https://tex.z-dn.net/?f=n%20%3D%2071%2C%20s%20%3D%20%5Cfrac%7B7%7D%7B%5Csqrt%7B71%7D%7D)
What is the probability that the mean monitor life would be greater than 83.8 months?
1 subtracted by the p-value of Z when X = 83.8. 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)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{83.8 - 82}{\frac{7}{\sqrt{71}}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B83.8%20-%2082%7D%7B%5Cfrac%7B7%7D%7B%5Csqrt%7B71%7D%7D%7D)
![Z = 2.17](https://tex.z-dn.net/?f=Z%20%3D%202.17)
has a p-value of 0.985.
1 - 0.985 = 0.015
0.015 = 1.5% probability that the mean monitor life would be greater than 83.8 months in a sample of 71 monitors