Question

If np is greater than or equal to 15 and n(1-p) is greater than or equal to 15, what is the approximate shape of the sampling distribution of the sample proportion?

A. p-hat
B. true proportion p
C. x-bar
D. sqrt(p*(1-p)/n)
E. Normal
F. sigma/sqrt(n)
G. Binomial
H. Bimodal

Answer 1

Answer:

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

$np \geq 15$

$n(1-p) \geq 15$

If we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np$

$\sigma=\sqrt{np(1-p)}$

$X \sim N (\mu = np, \sigma=\sqrt{np(1-p)})$

So then the correct answer for this case would be:

E. Normal

Step-by-step explanation:

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 assume that:

$X \sim Binom(n, p)$

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.

$np \geq 15$

$n(1-p) \geq 15$

If we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np$

$\sigma=\sqrt{np(1-p)}$

$X \sim N (\mu = np, \sigma=\sqrt{np(1-p)})$

So then the correct answer for this case would be:

E. Normal

Answer 2

The approximate shape of the sampling distribution of the sample will be normal.

Option E is correct.

Sampling distribution:

It is a statistic that determines the probability of an event based on data from a small group within a large population.

Given that, $np$ is greater than or equal to 15 and $n(1-p)$ is greater than or equal to 15.

     $np\geq 15$  and $n(1-p)\geq 15$

So that the new mean and standard deviation will be,

           $mean=\mu=np\\\\Deviation=\sigma=\sqrt{np(1-p)}$

Thus, the approximate shape of the sampling distribution of the sample will be normal.

Learn more about the sampling distribution here:

https://brainly.com/question/12892403