Question

Individual taxi fares in Metro City are normally distributed with a population mean equals to $10 and a population standard deviation of $4. Assume drawing from an infinite population, a random sample of 64 fares is obtained. Based on these information, what is the probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10

Answer 1

0.3413 = 34.13% probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10.

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
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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`

Sample of 64

This means that
`n = 8, s = (4)/(√(64)) = 0.5`

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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (10 - 10)/(0.5)`

`Z = 0`

`Z = 0` has a pvalue of 0.5

X = 9.5

`Z = (X - \mu)/(s)`

`Z = (9.5 - 10)/(0.5)`

`Z = -1`

`Z = -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.