Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 inches and a standard deviation of 3.7 inches. Fill in the blanks. In 95.44% of the years, the precipitation in this city is between _ and _ inches. Round your answers to the nearest tenth as needed.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(288,3.7)[/tex]

Where [tex]\mu=288[/tex] and [tex]\sigma=3.7[/tex]

The confidence level is 95.44 and the signficance is [tex] 1-0.9544=0.0456[/tex] and the value of [tex]\alpha/2 =0.0228[/tex]. And the critical value for this case is [tex]z = \pm 1.99[/tex]

Using this condition we can find the limits

[tex]z=-1.99<\frac{a-288}{3.7}[/tex]

[tex]z=1.99<\frac{a-288}{3.7}[/tex]

And if we solve for a we got

[tex]a=288 -1.99*3.7=214.4[/tex]

[tex]a=288 +1.99*3.7=295.4[/tex]

And the limits for this case are: (214.4; 295.4)