An application of the sampling distribution of a proportion Of the 21.4 million U. S. firms without paid employees, 32% are female owned. [Data source: U. S. Census Bureau; data based on the 2007 Economic Census.]

A simple random sample of 503 firms is selected. Use the Distributions tool to help you answer the questions that follow.

Normal Distribution

Mean = 0.22

Standard Deviation = 0.025

The probability that the sample proportion is within ±.01 of the population proportion is .

Suppose the sample size is increased to 903. The probability that the sample proportion is within ±.01 of the population proportion is now .

(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:

*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:

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

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

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step-by-step explanation:

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