Question

At a national convention attended by many educators, about 30 percent of the attendees are from the northeast. Of all the attendees of the national convention, 25 will be selected at random to receive a free book. What are the mean and standard deviation of the sampling distribution of the proportion of attendees from the northeast for samples of size 25 ?

Answer 1

Answer:

The mean is 0.3 and the standard deviation is 0.0917.

Step-by-step explanation:

In this problem, we have that:

Proportion p = 0.3.

For a sample of size n, the mean and sandard deviation of a proportion are:

Mean: $E(X) = p$

Standard deviation:

$\sqrt{V(X)} = \sqrt{\frac{p(1-p)}{n}}$

What are the mean and standard deviation of the sampling distribution of the proportion of attendees from the northeast for samples of size 25 ?

n = 25. So

Mean:

E(X) = p = 0.3.

Standard deviation

$\sqrt{V(X)} = \sqrt{\frac{0.3*0.7}{25}} = 0.0917$

The mean is 0.3 and the standard deviation is 0.0917.

Answer 2

The mean of a sampling distribution is 0.3.

The standard deviation of a sampling distribution is 0.0917.

Given:

30 percent of the attendees are from the northeast.

So, Proportion (p) = 30% = 0.3

25 will be selected at random to receive a free book.

So, Sample size (n) = 25

To find mean:

$mean=E(X)=p\\E(X) = 0.3$

So, the mean is 0.3

To find Standard deviation

$SD=\sqrt{V(X)} =\sqrt{\frac{p(1-p)}{n} }\\\sqrt{V(X)} =\sqrt{\frac{0.3(1-0.3)}{25} }\\\sqrt{V(X)} =0.0917$

So, The standard deviation is 0.0917

Therefore, The mean of a sampling distribution is 0.3 and the standard deviation of a sampling distribution is 0.0917.

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