Answer:
0.3413 = 34.13% probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10.
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 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.
Mean equals to $10 and a population standard deviation of $4
This means that ![\mu = 10, \sigma = 4](https://tex.z-dn.net/?f=%5Cmu%20%3D%2010%2C%20%5Csigma%20%3D%204)
Sample of 64
This means that ![n = 8, s = \frac{4}{\sqrt{64}} = 0.5](https://tex.z-dn.net/?f=n%20%3D%208%2C%20s%20%3D%20%5Cfrac%7B4%7D%7B%5Csqrt%7B64%7D%7D%20%3D%200.5)
What is the probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10?
This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 9.5.
X = 10
![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{10 - 10}{0.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B10%20-%2010%7D%7B0.5%7D)
![Z = 0](https://tex.z-dn.net/?f=Z%20%3D%200)
has a pvalue of 0.5
X = 9.5
![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{9.5 - 10}{0.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B9.5%20-%2010%7D%7B0.5%7D)
![Z = -1](https://tex.z-dn.net/?f=Z%20%3D%20-1)
has a pvalue of 0.1587
0.5 - 0.1587 = 0.3413
0.3413 = 34.13% probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10.