What is the probability that the sample proportion is between 0.22 and 0.49?

Question

Based on historical data, your manager believes that 45% of the company's orders come from first-time customers. A random sample of 97 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.22 and 0.49?

(Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Answer: 0.7852

Explanation:

Because we have a random sample of 97, which quite a bit larger than 30, we can assume the Central Limit Theorem applies and the distribution of sample proportions is approximately normal.

We know the mean of the distribution phat = 0.45. Therefore, the standard deviation is sqrt(phat(1-phat)/n). Plugging in phat=0.45 and n=97, we get sqrt(0.45*0.55/97) = 0.0505128339 for the standard deviation.

Next, we use the standard deviation and the phat to find the z-scores of p1=0.22 and p2=0.49, so

Z(p1=0.49) = 0.49-0.45/0.0505128339 = 0.7919

Z(p2=0.22) = 0.22-0.45/0.0505128339 = -4.5533

Finally, we can use the z-scores and a Z Table to calculate the desired probability:

P(p2 < p < p1) = P(p<=p1) - P(p<=p2) = 0.7852 - 0 = 0.7852.

Because the z-score for p2=0.22 is -4.5533, we can assume P(p<p2) is essentially 0 as approximately 99.7% of all values fall within three standard deviations of the mean for a normal distribution using the empirical rule.