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.

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form