An orange juice producer buys oranges from a large orange grove that has one variety of orange.

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 4.20 ounces and a standard deviation of 0.24 ounce. Suppose that you select a sample of 36 oranges.

a. What is the probability that the sample mean amount of juice will be at least 4.14

​ounces?

b. The probability is 70​% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population​ mean? There is a 70​% probability that the sample mean amount of juice will be contained between ______​ounce(s) and _____ ​ounce(s) ​(Round to two decimal places as needed. Use ascending​ order.)

c. The probability is 79​% that the sample mean amount of juice will be greater than what​ value?

Answer 1

a. z = (4.14 - 4.2)/(0.24/sqrt(36) = -1.5

P(z > -1.5) = 1 - P(z < -1.5) = 1 - 0.0668 = 0.9332

b. 70% in the middle is between 15% and 85%.

z for 85% is 1.04

x is value in ounces

1.04 = (x - 4.2)/(0.04) = 4.24

z for 15% is -1.04

-1.04 = (x - 4.2)/0.04 = 4.16

sample mean will lie between 4.16 and 4.24

c. 79% of data is greater than 21st percentile.

z for 21% is -0.81

-0.81 = (x - 4.2)/0.04 = 4.17