For humans, gestation periods are normally distributed with a mean of \(266\) days and a standard deviation of \(16\) days.

a. Use the technology of your choice to simulate \(2000\) samples of nine human gestation periods each.

b. Find the sample mean of each of the \(2000\) samples

c. Obtain the mean, the standard deviation and a histogram of the \(2000\) sample means.

d. Theoretically, what are the mean, standard deviation and distribution of all possible sample means for sample of sizes \(9\)?

e. Compare your results from (c) and (d).

\(\mu=266\)

\(\sigma=16\)

\(n=9\)

The \(2000\) samples of nine human gestation periods each needs to be calculated

While using the R program, the following command can be used to create \(9\times2000=18000\) data values.

A<-rnorm(1800, 266, 16)\)

This will be result of the stated question.

Now, the sample mean of each of the sample calculated in part (a) needs to be calculated

While using the R program, the following command can be used to determine the mean of every sample.

for (i in 1:2000){

M[i]< -mean(a[9 \times (i-8)):(9 \times i)])

}

Next, it is needed to calculate the mean, standard deviation and histogram of the \(2000\) samples means calculated in part (b):

While using the R program, the following commands can be used to determine the required values:

mean(M)

sd(M)

hist(M)

After this, the mean comes out to be \(266.0248\) and the standard deviation comes out to be \(5.36694\). The graph is represented as per below:

\(\mu_{\bar{x}}=\mu=266\)

\(\sigma_{x}=\frac{\sigma}{\sqrt{n}}=\frac{16}{\sqrt{3}} \approx 5.3333\).

It can be noted that in part (c), the mean and standard deviation are little greater than the subsequent theoretical standard deviation and mean in the part (d),

However, both of values in both parts are pretty close to each other. The difference could ve because of sampling error.