Statistics Class

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 0.40 ounce. Suppose that you select a sample of 25 oranges.

a. What is the probability that the sample mean amount of juice will be at least 4.36

​ounces?

b. The probability is 72% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population​ mean? There is a 72% 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 75% that the sample mean amount of juice will be greater than what​ value?

Answer 1

This is a normal distribution with µ = 5.0 ounces and σ = 0.40 ounces. The sample size is 25 oranges.

a.) Use the z-score formula to find the probability. You will need to have the z-score table either from the textbook or online.

z = (X - μ) / (σ / √n) = (4.36 - 5) / (0.40 / √25) = -0.64/0.08 = -8.00

P(z ≥ -8.00) = 1 - P(z < -8.00) = 1 - 0.0001 = 0.9999

The p-value for the z-score will equal to virtually zero since it is outside of the original table. Therefore, the probability that the sample mean amount of juice will be at least 4.36 ounces is about 0.9999.

b.) The probability of 72% is 0.72. The z-score for the p-value of 0.72 is about 0.5284. This means, -0.5284 and +0.5284 gives a probability (0.72) or confidence level of 72%.

We need to find the X values (raw scores) using the z-score formula.

-0.5284 = (X - 5) / 0.08 ⇒ -0.042272 = X - 5 ⇒ X ≈ 4.96

0.5284 = (X - 5) / 0.08 ⇒ 0.042272 = X - 5 ⇒ X ≈ 5.04

There is a 72% probability that the sample mean amount of juice will be contained between 4.96 ounces and 5.04 ounces.

c.) The probability of 75% is 0.75. The z-score for the p-value of 0.75 is about 0.675. We need to find the X value (raw score) using the z-score formula.

0.675 = (X - 5) / 0.08 ⇒ 0.054 = X - 5 ⇒ X ≈ 5.054

There is a 75% probability that the sample mean amount of juice will be greater than 5.054 ounces.