Question

According to the US Census Bureau's American Community Survey, 87, percent of Americans over the age of 25 have earned a high school diploma. Suppose we are going to take a random sample of 200 Americans in this age group and calculate what proportion of the sample has a high school diploma. a) What is the expected number of people in the sample with a high school diploma? b) What is the expected number of people in the sample without a high school diploma? c) Based on the answers in a) and b), Can the sampling distribution be approximated by a normal distribution? d) What is the mean of the sampling distribution ? e) What is the standard deviation of the sampling distribution ? f) What is the probability that the proportion of people in the sample with a high school diploma is less than 85%?

Answer 1

Answer:

B

Step-by-step explanation: pORBABLITY

Answer 2

Using the Central Limit Theorem, it is found that:

a) 174.

b) 26.

c) Since both $np \geq 10$ and $n(1-p) \geq 10$, the central limit theorem is applied, and the sampling distribution can be approximated by a normal distribution.

d) 0.87

e) 0.0238

f) 0.2005 = 20.05% probability that the proportion of people in the sample with a high school diploma is less than 85%.

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The Central Limit Theorem establishes that for a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$ , if $np \geq 10$ and $n(1-p) \geq 10$

In this problem:

• Sample of 200, thus $n = 200$.

• 87% have a diploma, thus $p = 0.87$.

Item a:

This is

$np = 200(0.87) = 174$

Item b:

This is:

$n(1-p) = 200(0.13) = 26$

Item c:

Since both $np \geq 10$ and $n(1-p) \geq 10$, the central limit theorem is applied, and the sampling distribution can be approximated by a normal distribution.

Item d:

The mean is:

$\mu = p = 0.87$

Item e:

The standard deviation is:

$s = \sqrt{\frac{0.87(0.13)}{200}} = 0.0238$

Item f:

Using z-scores, the probability is the p-value of Z when X = 0.85.

We have that:

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.85 - 0.87}{0.0238}$

$Z = -0.84$

$Z = -0.84$ has a p-value of 0.2005.

0.2005 = 20.05% probability that the proportion of people in the sample with a high school diploma is less than 85%.

A similar problem is given at https://brainly.com/question/15581844