Answer:
0.0524 = 5.24% probability that the sample mean would differ from the population mean by more than 2 millimeters.
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 diameter of 144 millimeters, and a variance of 49.
This means that ![\mu = 144, \sigma = \sqrt{49} = 7](https://tex.z-dn.net/?f=%5Cmu%20%3D%20144%2C%20%5Csigma%20%3D%20%5Csqrt%7B49%7D%20%3D%207)
Sample of 46:
This means that ![n = 46, s = \frac{7}{\sqrt{46}}](https://tex.z-dn.net/?f=n%20%3D%2046%2C%20s%20%3D%20%5Cfrac%7B7%7D%7B%5Csqrt%7B46%7D%7D)
Wat is the probability that the sample mean would differ from the population mean by more than 2 millimeters?
Above 144 + 2 = 146 or below 144 - 2 = 142. Since the normal distribution is symmetric, these probabilities are equal, which means that we find one of them and multiply by two.
Probability the sample mean is below 142:
p-value of Z when X = 142, 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{142 - 144}{\frac{7}{\sqrt{46}}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B142%20-%20144%7D%7B%5Cfrac%7B7%7D%7B%5Csqrt%7B46%7D%7D%7D)
![Z = -1.94](https://tex.z-dn.net/?f=Z%20%3D%20-1.94)
has a p-value of 0.0262
2*0.0262 = 0.0524
0.0524 = 5.24% probability that the sample mean would differ from the population mean by more than 2 millimeters.