Question

The summer monsoon rains in a country follow approximately a Normal distribution with mean 857 millimeters (mm) of rainfall and standard deviation 85 mm. (Round your answers to two decimal places.)

(a) In a drought year, 699 mm of rain fell. In what percent of all years will this country have 699 mm or less of monsoon rain?

Answer 1

`P(X<699)=P((X-\mu)/(\sigma)<(699-\mu)/(\sigma))=P(Z<(699-857)/(85))=P(z<-1.86)`

And we can find this probability using Excel or the normal standard table:

`P(z<-1.86)=0.0314`

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 summer monsoon rains of a population, and for this case we know the distribution for X is given by:

`X \sim N(857,85)`

Where
`\mu=857` and
`\sigma=85`

We are interested on this probability

`P(X<699)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<699)=P((X-\mu)/(\sigma)<(699-\mu)/(\sigma))=P(Z<(699-857)/(85))=P(z<-1.86)`

And we can find this probability using Excel or the normal standard table:

`P(z<-1.86)=0.0314`