Question

A random sample of n = 83 measurements is drawn from a binomial population with probability of success 0.4. Complete parts a through d below.
a. Give the mean and standard deviation of the sampling distribution of the sample proportion, p. The mean of the sampling distribution of p is The standard deviation of the sampling distribution of p is (Round to four decimal places as needed.)
b. Describe the shape of the sampling distribution of p. 0 A The shape of the sampling distribution of p is approximately normal because the sample size is small. The shape of the sampling distribution of p is approximately normal because the sample size is large The shape of the sampling distribution of p is approximately uniform because the sample size is smal ○ C. O D. The shape of the sampling distribution of p is approximately uniform because the sample size is large.
c. Calculate the standard normal z-score corresponding to a value of p=0.41. The standard normal z-score corresponding to a value of p: 041 is . Round to two decimal places as needed.) Finn)

Answer 1

Answer:

a) The mean of the sampling distribution of p is 0.4.

The standard deviation of the sampling distribution of p is 0.0538.

b) The shape of the sampling distribution of p is approximately normal because the sample size is large.

c) z=0.19

Step-by-step explanation:

We have a random sample of size n=83, drawn from a binomial population with proabiliity p=0.4. We have to compute the characteristic of the sampling distribution of the sample proportion.

a) The mean of the sampling distribution is equal to the mean of the distribution p:

$\mu_p=p=0.4$

The standard deviation of the sampling distribution is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.4*0.6}{83}}=\sqrt{0.0029}=0.0538$

b) The shape of sampling distribution with enough sample size tend to be approximately normal. In this case, n=83 is big enough for a binomial distribution.

c) The z-score fof p-0.41 can be calculated as:

$z=\dfrac{p-\mu_p}{\sigma_p}=\dfrac{0.41-0.4}{0.0538}=\dfrac{0.01}{0.0538}=0.19$

Answer 2

The mean of the sampling distribution is 0.4 and the standard deviation is 0.0538.

How to solve the sampling distribution?

From the information given, the sample is drawn from a binomial population with probability of success 0.4. Therefore, the mean is 0.4.

The standard deviation will be:

= [(✓0.4 × ✓0.6) / ✓83)]

= 0.0538

Furthermore, the shape of the sampling distribution of p is approximately normal because the sample size is large.

The standard normal z-score will be:

= (0.41 - 0.4) / 0.0538

= 0.10/0.0538

= 0.19.

Therefore, the standard normal z-score is 0.19.

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