Final answer:
To find the probability that the mean price for a sample of federal income tax returns is within $16 of the population mean, we calculate the standard error of the mean and use z-scores. The probabilities are then found using the standard normal distribution table. For part (d), none of the sample sizes (20, 60, 121) will give us a probability higher than 0.95.
Step-by-step explanation:
Part (a):
To find the probability that the mean price for a sample of 20 federal income tax returns is within $16 of the population mean, we can use the normal distribution since the sample size is greater than 30. Given that the population mean is $302 and the population standard deviation is $100, we can calculate the standard error of the mean using the formula: SE = σ/√n, where σ is the population standard deviation and n is the sample size. Plugging in the values, we get SE = 100/√20 = 22.36. Now we can calculate the z-scores for $16 above and below the population mean using the formula: Z = (X - μ) / SE, where X is the value we want to find the probability for, μ is the population mean, and SE is the standard error of the mean. The z-score for $16 above the mean is (302 + 16 - 302) / 22.36 ≈ 0.67, and the z-score for $16 below the mean is (302 - 16 - 302) / 22.36 ≈ -0.67. Looking up these z-scores in the standard normal distribution table, we find the probabilities: 0.7486 (for the z-score of 0.67) and 0.2514 (for the z-score of -0.67). To find the probability that the mean price falls within $16 of the population mean, we subtract the probability of being $16 above the mean from the probability of being $16 below the mean: 0.7486 - 0.2514 = 0.4972.
Part (b):
To find the probability that the mean price for a sample of 60 federal income tax returns is within $16 of the population mean, we follow the same steps as in part (a), but with different values. The standard error of the mean is now SE = 100/√60 ≈ 12.91. The z-score for $16 above the mean is (302 + 16 - 302) / 12.91 ≈ 1.24, and the z-score for $16 below the mean is (302 - 16 - 302) / 12.91 ≈ -1.24. Looking up these z-scores in the table, we find the probabilities: 0.8921 (for the z-score of 1.24) and 0.1079 (for the z-score of -1.24). Subtracting the probability of being $16 above the mean from the probability of being $16 below the mean gives us: 0.1079 - 0.8921 = -0.7842.
Part (c):
To find the probability that the mean price for a sample of 121 federal income tax returns is within $16 of the population mean, we use the same steps as in parts (a) and (b), but with different values. The standard error of the mean is now SE = 100/√121 = 9.13. The z-score for $16 above the mean is (302 + 16 - 302) / 9.13 ≈ 1.75, and the z-score for $16 below the mean is (302 - 16 - 302) / 9.13 ≈ -1.75. Looking up these z-scores in the table, we find the probabilities: 0.9599 (for the z-score of 1.75) and 0.0401 (for the z-score of -1.75). Subtracting the probability of being $16 above the mean from the probability of being $16 below the mean gives us: 0.0401 - 0.9599 = -0.9198.
Part (d):
To ensure at least a 0.95 probability that the sample mean is within $16 of the population mean, we need to choose a sample size that gives us a higher probability than 0.95. From the calculations in parts (a), (b), and (c), we can see that the probability decreases as the sample size increases. None of the sample sizes (20, 60, 121) will give us a probability higher than 0.95, so we would recommend choosing none of the above.