Answer to Question #307767 in Statistics and Probability for Italia

Question

The average precipitation for the first 7 months of the year is 19.32 inches with a standard deviation of 2.4 inches. Assume that the average precipitation is normally distributed.

a. What is the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months?

b. What is the average precipitation of 5 randomly selected years for the first 7 months?

c. What is the probability of 5 randomly selected years will have an average precipitation greater than 18 inches for the first 7 months?

Answer 1

We are given $\mu$ = 19.32 and $\sigma$ = 2.4

a.

P(X>18) = P(X-$\mu$ > 18-19.32) = P($\frac{X-\mu}{\sigma}$ > $\frac{18 - 19.32}{2.4}$)

We compute the z score using the formula:

z = $\frac{X-\mu}{\sigma}$

= $\frac{18-19.32}{2.4}$

= -0.55

P(X>18) = P(Z> -0.55)

where

P(Z> -0.55) = 1- P(Z<-0.55)

From the normal distribution table

P(Z<-0.55) = 0.2912

Therefore;

P(Z> -0.55) = 1- 0.2912

= 0.7088

Answer: 0.7088

b.

Answer:

We assume that the average precipitation of 5 randomly selected years for the first 7 months is the population mean $\mu$ = 19.32

c.

Since $\mu$ = 19.32 and $\sigma$ = 2.4

We are given that the sample size n = 5

Now, we will use the formula:

z = $\frac{X-\mu}{\sigma/\sqrt{n}}$

P(X>18) = P(X-$\mu$ > 18-19.32) = P($\frac{X-\mu}{\sigma/\sqrt{n}}$ > $\frac{18 - 19.32}{2.4/\sqrt{5}}$)

We compute the z score using the formula:

z = $\frac{X-\mu}{\sigma/\sqrt{n}}$

= $\frac{18-19.32}{2.4/\sqrt{5}}$

= -1.23

P(X>18) = P(Z> -1.23)

where

P(Z> -1.23) = 1- P(Z<-1.23)

From the normal distribution table

P(Z<-1.23) = 0.10935

Therefore;

P(Z> -1.23) = 1- 0.10935

= 0.89065

Answer: 0.8907