Question

Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights of oranges are also normally distributed with a mean of 131 grams and a standard deviation of 20 grams. Amy has an apple that weighs 90 grams and an orange that weighs 155 grams.
Required:
a. Find the probability a randomly chosen apple exceeds 100 g in weight.
b. What weight do 80% of the apples exceed?

Answer 1

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that $\mu = 85, \sigma = 8$

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{100 - 85}{8}$

$Z = 1.875$

$Z = 1.875$ has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

$Z = \frac{X - \mu}{\sigma}$

$-0.84 = \frac{X- 85}{8}$

$X - 85 = -0.84*8$

$X = 78.28$

The weight that 80% of the apples exceed is of 78.28g.