Using the normal distribution, it is found that 10.56% of fire station response times are under 3 minutes.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem:
The proportion of fire station response times are under 3 minutes is the p-value of Z when X = 3, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3 - 4.5}{1.2}[/tex]
[tex]Z = -1.25[/tex]
[tex]Z = -1.25[/tex] 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, you can take a look at https://brainly.com/question/24663213