Question

Assume that the empirical rule is a good model for the time it takes for all passengers to board an airplane. The mean time is 48 minutes with a standard deviation of 4 minutes. Which interval describes how long it takes for passengers to board the middle 95% of the time?

Answer 1

Answer:

The interval that describes how long it takes for passengers to board the middle 95% of the time is between 40.16 minutes and 55.84 minutes.

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:

$\mu = 48, \sigma = 4$.

Which interval describes how long it takes for passengers to board the middle 95% of the time?

This is between the 2.5th percentile and the 97.5th percentile.

So this interval is the value of X when Z has a a pvalue of 0.025 and the value of X when Z has a pvalue of 0.975

Lower Limit

Z has a pvalue of 0.025 when $Z = -1.96$. So

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

$-1.96 = \frac{X - 48}{4}$

$X - 48 = -1.96*4$

$X = 40.16$

Upper Limit

Z has a pvalue of 0.975 when $Z = 1.96$. So

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

$1.96 = \frac{X - 48}{4}$

$X - 48 = 1.96*4$

$X = 55.84$

The interval that describes how long it takes for passengers to board the middle 95% of the time is between 40.16 minutes and 55.84 minutes.