Question

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally​ distributed, with a mean of 5.0 ounces and a standard deviation of 0.16 ounce. Suppose that you select a sample of 16 oranges. a. What is the probability that the sample mean amount of juice will be at least 4.68 ​ounces? b. The probability is 76​% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population​ mean? c. The probability is 73​% that the sample mean amount of juice will be greater than what​ value? Question content area bottom Part 1 a. The probability is enter your response here . ​(Round to three decimal places as​ needed.) Part 2 b. There is a 76​% probability that the sample mean amount of juice will be contained between enter your response here​ounce(s) and enter your response here ​ounce(s). ​(Round to two decimal places as needed. Use ascending​ order.) Part 3 c. There is a 73​% probability that the sample mean amount of juice will be greater than enter your response here ​ounce(s). ​(Round to two decimal places as​ needed.)

Answer 1

a. The probability is 1.000. b. There is a 76% probability that the sample mean amount of juice will be contained between 4.979 and 5.021 ounces. c. There is a 73% probability that the sample mean amount of juice will be greater than 5.013 ounces.

Step-by-step explanation:

a. To find the probability that the sample mean amount of juice will be at least 4.68 ounces, we need to find the z-score corresponding to that value and then find the probability of getting a z-score greater than or equal to that value. The z-score is calculated as: (4.68 - 5.0) / (0.16 / sqrt(16)) = -4.09. Using a standard normal distribution table or a calculator, we find that the probability of getting a z-score greater than or equal to -4.09 is approximately 1.000. Therefore, the probability that the sample mean amount of juice will be at least 4.68 ounces is 1.000.

b. To find the values symmetrically distributed around the population mean with a probability of 76%, we need to find the z-scores corresponding to the upper and lower percentiles. The upper percentile is (1 + 0.76) / 2 = 0.88. Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a percentile of 0.88 is approximately 1.170. The lower percentile will be (-1.170). The sample mean amount of juice will be contained between 5.0 - 1.170 * (0.16 / sqrt(16)) and 5.0 + 1.170 * (0.16 / sqrt(16)), which is approximately 4.979 and 5.021 ounces, respectively.

c. To find the value that corresponds to a probability of 73%, we need to find the z-score corresponding to that probability. The z-score can be found by calculating the inverse of the cumulative distribution function for the standard normal distribution. Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a probability of 73% is approximately 0.663. The sample mean amount of juice will be greater than 5.0 + 0.663 * (0.16 / sqrt(16)), which is approximately 5.013 ounces.