Question

Suppose the round-trip airfare between Philadelphia and Los Angeles a month before the departure date follows the normal probability distribution with a mean of $387.20 and a standard deviation of $68.50. What is the probability that a randomly selected airfare between these two cities will be between $325 and $425?

Answer 1

Answer:

52.74% probability that a randomly selected airfare between these two cities will be between $325 and $425

Step-by-step explanation:

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.

In this question, we have that:

$\mu = 387.20, \sigma = 68.50$

What is the probability that a randomly selected airfare between these two cities will be between $325 and $425?

This is the pvalue of Z when X = 425 subtracted by the pvalue of Z when X = 325. So

X = 425

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

$Z = \frac{425 - 387.20}{68.50}$

$Z = 0.55$

$Z = 0.55$ has a pvalue of 0.7088

X = 325

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

$Z = \frac{325 - 387.20}{68.50}$

$Z = -0.91$

$Z = -0.91$ has a pvalue of 0.1814

0.7088 - 0.1814 = 0.5274

52.74% probability that a randomly selected airfare between these two cities will be between $325 and $425