Question

At NC State University, 16.4% of the undergraduate classes have more than 50 students. If a random sample of 200 undergraduate classes was taken, which of the following would accurately describe the sampling distribution? Select all that apply:
Select one or more:

The sampling distribution will be approximately normal.

The sampling distribution will be skewed right.

The sampling distribution will be skewed left.

The mean of the sampling distribution will be 16.4%.

The mean of the sampling distribution will be equal to 50%.

We can not determine the mean of the sampling distribution from the given information.

The standard deviation of the sampling distribution will be 0.0262.

The standard deviation of the sampling distribution will be 0.0007.

Answer 1

Answer:

We need to check the conditions in order to use the normal approximation.

$np=200*0.164=32.8 \geq 10$

$n(1-p)=200*(1-0.164)=167.2 \geq 10$

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The mean is given by:

$p =0.164$

And the deviation is given by:

$\sigma_{p}= \sqrt{\frac{0.164*(1-0.164)}{200}}= 0.0262$

So then the correct options for this case are:

The sampling distribution will be approximately normal.

The mean of the sampling distribution will be 16.4%.

The standard deviation of the sampling distribution will be 0.0262.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=200, p=0.164)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We need to check the conditions in order to use the normal approximation.

$np=200*0.164=32.8 \geq 10$

$n(1-p)=200*(1-0.164)=167.2 \geq 10$

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The mean is given by:

$p =0.164$

And the deviation is given by:

$\sigma_{p}= \sqrt{\frac{0.164*(1-0.164)}{200}}= 0.0262$

So then the correct options for this case are:

The sampling distribution will be approximately normal.

The mean of the sampling distribution will be 16.4%.

The standard deviation of the sampling distribution will be 0.0262.