Question

The Fight For Life emergency helicopter service is available for medical emergencies occurring 15 to 90 miles from the hospital. A long-term study of the service shows that the response time from receipt of the dispatch call to arrival at the scene of the emergency is normally distributed with standard deviation of 8 minutes. What is the mean response time (to the nearest whole minute) if only 6.7% of the calls require more than 54 minutes to respond? For a randomly received call, what is the probability that the response time will be less than 30 minutes? For a randomly received call, what is the probability that the response time will be within 1.5 standard deviations of the mean?

Answer 1

Answer:

The mean response time is 42 minutes.

There is a 6.68% probability that the response time will be less than 30 minutes.

There is an 86.62% probability that the response time will be within 1.5 standard deviations of the mean.

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:

A long-term study of the service shows that the response time from receipt of the dispatch call to arrival at the scene of the emergency is normally distributed with standard deviation of 8 minutes. This means that $\sigma = 8$.

What is the mean response time (to the nearest whole minute) if only 6.7% of the calls require more than 54 minutes to respond?

This means that Z when $X = 54$ has a pvalue of 1-0.067 = 0.933. This is $Z = 1.5$. So

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

$1.5 = \frac{54 - \mu}{8}$

$54 - \mu = 12$

$\mu = 42$

The mean response time is 42 minutes.

For a randomly received call, what is the probability that the response time will be less than 30 minutes?

This is the pvalue of Z when $X = 30$.

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

$Z = \frac{30 - 42}{8}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668.

This means that there is a 6.68% probability that the response time will be less than 30 minutes.

For a randomly received call, what is the probability that the response time will be within 1.5 standard deviations of the mean?

This is between 30 and 54 minutes.

Subtracing the pvalue of Z when $X = 54$ by the pvalue of Z when $X = 30$, we get that there is a 0.933 - 0.0668 = 0.8662 = 86.62% probability that the response time will be within 1.5 standard deviations of the mean.