Using the normal distribution, it is found that the 10.56% of fire station response times are under 3 minutes.
<h3>How to get the z scores?</h3>
If we've got a normal distribution, then we can convert it to a standard normal distribution and its values will give us the z score.
If we have
![X \sim N(\mu, \sigma)](https://tex.z-dn.net/?f=X%20%5Csim%20N%28%5Cmu%2C%20%5Csigma%29)
(X is following a normal distribution with mean
and standard deviation
)
then it can be converted to a standard normal distribution as
![Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)](https://tex.z-dn.net/?f=Z%20%3D%20%5Cdfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D%2C%20%5C%5C%5C%5CZ%20%5Csim%20N%280%2C1%29)
The mean is of 4.5 minutes,
The standard deviation is of 1.2 minutes,
The proportion of fire station response times are under 3 minutes is the p-value of Z when X = 3, hence:
![Z = \dfrac{X - \mu}{\sigma}, \\\\Z = \dfrac{3 - 4.5}{1.2}, \\\\Z = -1.25](https://tex.z-dn.net/?f=Z%20%3D%20%5Cdfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D%2C%20%5C%5C%5C%5CZ%20%3D%20%5Cdfrac%7B3%20-%204.5%7D%7B1.2%7D%2C%20%5C%5C%5C%5CZ%20%3D%20-1.25)
-1.25 has a p-value of 0.1056.
0.1056 x 100% = 10.56%
10.56% of fire station response times are under 3 minutes.
To learn more about the normal distribution, brainly.com/question/24663213
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