Using the normal distribution, it is found that the 10.56% of fire station response times are under 3 minutes.
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
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])
then it can be converted to a standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
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:
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z = \dfrac{3 - 4.5}{1.2}, \\\\Z = -1.25[/tex]
-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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