Question

According to Runzheimer International, a typical business traveler spends an average of $281 per day in Chicago. This cost includes hotel, meals, car rental, and incidentals. A survey of 60 randomly selected business travelers who have been to Chicago on business recently is taken. For the population mean of $281 per day, what is the probability of getting a sample average of more than $266 per day if the population standard deviation is $47?

Answer 1

Answer:

99.32% probability of getting a sample average of more than $266 per day if the population standard deviation is $47

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 normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 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 $\mu$ 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 $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 281, \sigma = 47, n = 60, s = \frac{47}{\sqrt{60}} = 6.07$

For the population mean of $281 per day, what is the probability of getting a sample average of more than $266 per day if the population standard deviation is $47?

This is 1 subtracted by the pvalue of Z when X = 266. Then

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{266 - 281}{6.07}$

$Z = -2.47$

$Z = -2.47$ has a pvalue of 0.0068

1 - 0.0068 = 0.9932

99.32% probability of getting a sample average of more than $266 per day if the population standard deviation is $47