Answer:
a) P(7.99 ≤ X ≤ 8.01) = 0.8164
b) P(X ≥ 8.01) = 0.0475.
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples can be solved using 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 random variable X, with mean
and standard deviation
, the sample means with size n can be approximated to a normal distribution with mean
In this problem, we have that:
![\mu = 8, \sigma = 0.03](https://tex.z-dn.net/?f=%5Cmu%20%3D%208%2C%20%5Csigma%20%3D%200.03)
(a) Calculate P(7.99 ≤ X ≤ 8.01) when n = 16.
n = 16, so ![s = \frac{0.03}{4} = 0.0075](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B0.03%7D%7B4%7D%20%3D%200.0075)
This probability is the pvalue of Z when X = 8.01 subtracted by the pvalue of Z when X = 7.99. So
X = 8.01
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
Applying 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{8.01 - 8}{0.0075}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B8.01%20-%208%7D%7B0.0075%7D)
![Z = 1.33](https://tex.z-dn.net/?f=Z%20%3D%201.33)
has a pvalue of 0.9082
X = 7.99
![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{7.99 - 8}{0.0075}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B7.99%20-%208%7D%7B0.0075%7D)
![Z = -1.33](https://tex.z-dn.net/?f=Z%20%3D%20-1.33)
has a pvalue of 0.0918
0.9082 - 0.0918 = 0.8164
P(7.99 ≤ X ≤ 8.01) = 0.8164
(b) How likely is it that the sample mean diameter exceeds 8.01 when n = 25? P(X ≥ 8.01) =
n = 25, so ![s = \frac{0.03}{5} = 0.006](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B0.03%7D%7B5%7D%20%3D%200.006)
This is 1 subtracted by the pvalue of Z when X = 8.01. So
![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{8.01 - 8}{0.006}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B8.01%20-%208%7D%7B0.006%7D)
![Z = 1.67](https://tex.z-dn.net/?f=Z%20%3D%201.67)
has a pvalue of 0.9525
1 - 0.9525 = 0.0475
P(X ≥ 8.01) = 0.0475.