High School

The Statistical Abstract of the United States, published by the U.S. Census Bureau, reports that the average annual consumption of fresh fruit per person is 99.9 pounds. The standard deviation of fresh fruit consumption is about 30 pounds. Suppose a researcher took a random sample of 38 people and had them keep a record of the fresh fruit they ate for one year.

a. What is the probability that the sample average would be less than 90 pounds?

b. What is the probability that the sample average would be between 98 and 105 pounds?

c. What is the probability that the sample average would be less than 112 pounds?

d. What is the probability that the sample average would be between 93 and 96 pounds?

Answer :

Final answer:

Using the Central Limit Theorem, we can calculate the probabilities for various sample averages of fresh fruit consumption.

Explanation:

To calculate the probability of sample averages, we need to use the Central Limit Theorem. Given that the average annual consumption of fresh fruit per person is 99.9 pounds with a standard deviation of 30 pounds, we can use the formula: Z = (X - μ) / (σ / √n), where Z is the Z-score, X is the sample average, μ is the population mean, σ is the population standard deviation, and n is the sample size.

a. To find the probability that the sample average would be less than 90 pounds, we calculate the Z-score: Z = (90 - 99.9) / (30 / √38) = -2.41. Using a Z-table, we find that the probability is approximately 0.008, or 0.8%.

b. To find the probability that the sample average would be between 98 and 105 pounds, we calculate the Z-scores for both values: Z1 = (98 - 99.9) / (30 / √38) = -0.38 and Z2 = (105 - 99.9) / (30 / √38) = 0.78. Using a Z-table, we find the probability for Z2 and subtract the probability for Z1 to get approximately 0.353, or 35.3%.

c. To find the probability that the sample average would be less than 112 pounds, we calculate the Z-score: Z = (112 - 99.9) / (30 / √38) = 1.89. Using a Z-table, we find that the probability is approximately 0.970, or 97.0%.

d. To find the probability that the sample average would be between 93 and 96 pounds, we calculate the Z-scores for both values: Z1 = (93 - 99.9) / (30 / √38) = -1.14 and Z2 = (96 - 99.9) / (30 / √38) = -0.72. Using a Z-table, we find the probability for Z2 and subtract the probability for Z1 to get approximately 0.338, or 33.8%.

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