Question

The average ticket price for a Washington Redskins football game was $81.89 for the season. With the additional costs of parking, food. drinks, and souvenirs, the average cost for a family of four to attend a game totaled $442.54. Assume the normal distribution applies and that the standard deviation is $65. What is the probability that a family of four will spend more than $400?

Answer 1

Answer:

74.22% probability that a family of four will spend more than $400.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 442.54, sigma = 65$

What is the probability that a family of four will spend more than $400?

This is 1 subtracted by the pvalue of Z when X = 400. So

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

$Z = \frac{400 - 442.54}{65}$

$Z = -0.65$

$Z = -0.65$ has a pvalue of 0.2578.

So there is a 1-0.2578 = 0.7422 = 74.22% probability that a family of four will spend more than $400.