Question

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.

Answer 1

`z=-1.99<(a-288)/(3.7)`

`z=1.99<(a-288)/(3.7)`

And if we solve for a we got

`a=288 -1.99*3.7=214.4`

`a=288 +1.99*3.7=295.4`

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

Explanation:

Previous concepts

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:

`X \sim N(288,3.7)`

Where
`\mu=288` and
`\sigma=3.7`

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

Using this condition we can find the limits

`z=-1.99<(a-288)/(3.7)`

`z=1.99<(a-288)/(3.7)`

And if we solve for a we got

`a=288 -1.99*3.7=214.4`

`a=288 +1.99*3.7=295.4`

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