Question

An application of the sampling distribution of a proportion

Answer 1

Answer:

(1) 0.3108

(2) 0.5223

Step-by-step explanation:

The information provided is:

Mean = p = 0.22

Standard Deviation = 0.025

n = 503

Compute the probability that the sample proportion is within ±0.01 of the population proportion as follows:

$P(-0.01$

                                    $=P(-0.40$

*Use a z-table.

Thus, the probability that the sample proportion is within ±0.01 of the population proportion is 0.3108.

The new sample size is, n = 903.

Compute the standard deviation as follows:

$\sigma_{p}=\sqrt{\farc{p(1-p)}{n}}=\sqrt{\frac{0.22(1-0.22)}{903}}=0.014$

Compute the probability that the sample proportion is within ±.01 of the population proportion as follows:

$P(-0.01$

                                    $=P(-0.71$

Thus, the probability that the sample proportion is within ±.01 of the population proportion is 0.5223.