Answer:
86.65% probability that the mean monitor life would be greater than 74.4 months in a sample of 85 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
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes 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.
In this question:
Population: ![\mu = 75, \sigma = 5](https://tex.z-dn.net/?f=%5Cmu%20%3D%2075%2C%20%5Csigma%20%3D%205)
Sample of 85: ![n = 85, s = \frac{5}{\sqrt{85}} = 0.5423](https://tex.z-dn.net/?f=n%20%3D%2085%2C%20s%20%3D%20%5Cfrac%7B5%7D%7B%5Csqrt%7B85%7D%7D%20%3D%200.5423)
If the claim is true, what is the probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors
This is 1 subtracted by the pvalue of Z when X = 74.4. 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{74.4 - 75}{0.5423}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B74.4%20-%2075%7D%7B0.5423%7D)
![Z = -1.11](https://tex.z-dn.net/?f=Z%20%3D%20-1.11)
has a pvalue of 0.1335
1 - 0.1335 = 0.8665
86.65% probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors