A DVD rental store wants to know what proportion of its customers are under age 21. A simple random sample of 500 customers was taken, and 375 of them were under age 21. Presume that the true population proportion of customers under age 21 is 0.68.
Find the mean and standard deviation of P̂
What is the probability that the sample proportion P̂ is within 0.03 of the true proportion of customers who are under age 21?
.21
To find the mean and standard deviation of the sample proportion P̂, you can use the formulas:
Mean (μᵖ̂) = p
Standard Deviation (σᵖ̂) = √(p(1-p)/n)
where:
p = true population proportion (0.68 in this case)
n = sample size (500)
Substituting the values, we get:
Mean (μᵖ̂) = 0.68
Standard Deviation (σᵖ̂) = √(0.68(1-0.68)/500)
Calculating the standard deviation, we have:
Standard Deviation (σᵖ̂) = √(0.68(0.32)/500) ≈ 0.016
To find the probability that the sample proportion P̂ is within 0.03 of the true proportion of customers who are under age 21, we need to calculate the z-scores and use them to find the probability.
First, calculate the z-scores for the upper and lower limits:
Lower limit z-score = (0.68 - 0.03 - 0.375) / 0.016
Upper limit z-score = (0.68 + 0.03 - 0.375) / 0.016
Next, use a standard normal distribution table or a calculator to find the probabilities associated with the z-scores.
Probability = P(Z ≤ upper limit z-score) - P(Z ≤ lower limit z-score)
Using the z-table or calculator, find the probabilities associated with the z-scores and subtract the lower probability from the upper probability to get the final probability.