Question

According to a recent poll, 64% of adults who use the Internet have paid to download music. In a random sample of size 675, adults who use the Internet, let [^(p)] represent the sample proportion who have paid to download music.

Find the mean of the sampling distribution of [^(p)]. Answer to 2 decimal places.

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Find the standard deviation of the sampling distribution of [^(p)]. Answer to 4 decimal places.

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What does the Central Limit Theorem say about the shape of the sampling distribution of [^(p)]?
The Central Limit Theorem says that the shape of the sampling distribution of [^(p)] is completely determined by the population parameter p.
The Central Limit Theorem says that as the sample size grows large, the sampling distribution of [^(p)] will approach the population distribution.
The Central Limit Theorem says that as the sample size grows large, the sampling distribution of [^(p)] will become approximately normal.
The Central Limit Theorem does not say anything about the sampling distribution of [^(p)] because it is not a population mean.

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Compute the probability that [^(p)] is less than 0.598. Answer to 4 decimal places.

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Find a such that P(0.64 − a < [^(p)] < 0.64 +a ) = 0.59. Answer to 3 decimal places.

Answer 1

p = 0.64, n = 675

Mean of the sampling distribution, μ_p̂ =  0.64

Standard deviation of the sampling distribution, σ_p̂ = √(p*(1-p)/n) = √(0.64 * 0.36 / 675) = 0.0185

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The Central Limit Theorem says that as the sample size grows large, the sampling distribution of p̂ will become approximately normal.

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P(p̂ < 0.598)

= P((p̂ - μ_p̂)/σ_p̂ < (0.598 - 0.64)/0.0185)

= P(z < -2.2733)

Using excel function:

= NORM.S.DIST(-2.2733, 1)

= 0.0115

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P(0.64-a < p̂ < 0.64+a) = 0.59

= P(p̂ < 0.64+a) - P(p̂ < 0.64-a) = 0.59

= P(p̂ < 0.64+a) - [1 - P(p̂ < 0.64+a)] = 0.59

= 2P(p̂ < 0.64+a) - 1 = 0.59

= P(p̂ < 0.64+a) = 0.795

Z score at p = 0.795 using excel = NORM.S.INV(0.795) = 0.8239

Value of p = µ_p̂ + z*σ_p̂ = 0.64 + (0.8239)*0.0185 = 0.655

a = 0.655 - 0.64 = 0.015