Question

The American Automobile Association reported that families planning to travel over the Labor Day weekend would spend an average of $735. Assume that the amount spent is normally distributed with a standard deviation of $230. What is the probability of family expenses for the weekend being: a. more than $580

Answer 1

Answer:

0.7486 = 74.86% probability of family expenses for the weekend being more than $580.

Step-by-step explanation:

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 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.

The American Automobile Association reported that families planning to travel over the Labor Day weekend would spend an average of $735.

This means that $\mu = 735$

Assume that the amount spent is normally distributed with a standard deviation of $230.

This means that $\sigma = 230$

What is the probability of family expenses for the weekend being: a. more than $580

This is 1 subtracted by the pvalue of Z when X = 580. So

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

$Z = \frac{580 - 735}{230}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514

1 - 0.2514 = 0.7486

0.7486 = 74.86% probability of family expenses for the weekend being more than $580.