Question

An apple juice producer buys all his apples from a conglomerate of apple growers in one northwestern state. the amount of juice obtained from each of these apples is approximately normally distributed with a mean of 2.25 ounces and a standard deviation of .15 ounce. 77 percent of the apples will contain at least how many ounces of juice? select one: a. 2.12 b. 2.14 c. 2.36 d. 2.38

Answer 1

At least 77 percent of the apples will contain at least 2.36 ounces of juice, as calculated using the given mean, standard deviation, and the corresponding z-score for the 77th percentile.

Step-by-step explanation:

We are given that the amount of juice obtained from apples is normally distributed with a mean of 2.25 ounces and a standard deviation of 0.15 ounce.

To find the amount of juice that at least 77 percent of the apples will contain, we need to determine the z-score that corresponds to the 77th percentile of a standard normal distribution and then use the mean and standard deviation to find the specific value.

Firstly, using a standard normal distribution z-table or percentile calculator, we find the z-score that corresponds to the 77th percentile, which is approximately 0.74. Next, we use the formula for transforming a z-score into an X value:

X = μ + zσ

Where X is the value we want to find (the amount of juice), μ is the mean (2.25 ounces), and σ is the standard deviation (0.15 ounce).

Plugging in the values, we get:

X = 2.25 + 0.74(0.15)

X = 2.25 + 0.111 = 2.361

So, at least 77 percent of the apples will contain at least 2.36 ounces of juice. The closest answer choice is c. 2.36.