Question

According to records, the amount of precipitation in a certain city on a November day has a mean of 0.10 inches; with a standard deviation of 0.05m inches.


What is the probability that the mean daily precipitation will be 0.09 inches or less for a random sample of 40 November days (taken over many years)?


Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

Answer 1

Answer:

Probability that the mean daily precipitation will be 0.09 inches or less for a random sample of 40 November days is 0.103.

Step-by-step explanation:

We are given that according to records, the amount of precipitation in a certain city on a November day has a mean of 0.10 inches; with a standard deviation of 0.05 inches.

A random sample of 40 November days is taken(taken over many years).

Let $\bar X$ = sample mean daily precipitation

The z score probability distribution for sample mean is given by;

                                 Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\mu$ = population mean = 0.10 inches

           $\sigma$ = population standard deviation = 0.05 inches

           n = sample of days = 40

Now, the probability that the mean daily precipitation will be 0.09 inches or less is given by = P( $\bar X$ $\leq$ 0.09 inches)

 P($\bar X$ $\leq$ 0.09 inches) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{0.09-0.10}{\frac{0.05}{\sqrt{40} } }$ ) = P(Z $\leq$ -1.265) = 1 - P(Z < 1.265)

                                                                    = 1 - 0.8971 = 0.1029

The above probability is calculated by looking at the value of x = 1.265 from the z table which will lie between x = 1.26 and x = 1.27 which has an area of 0.89617 and 0.89796 respectively.

Hence, the probability that the mean daily precipitation will be 0.09 inches or less for a random sample of 40 November days is 0.103.