Question

The mean weight of frozen yogurt cups in an ice cream parlor is 8 oz.Suppose the weight of each cup served is normally distributed withstandard deviation 0.5 oz, independently of others.(a) What is the probability of getting a cup weighing more than 8.64oz

Answer 1

Answer:

10.03% probability of getting a cup weighing more than 8.64oz

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 8, \sigma = 0.5$

What is the probability of getting a cup weighing more than 8.64oz

This is the 1 subtracted by the pvalue of Z when X = 8.64. So

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

$Z = \frac{8.64 - 8}{0.5}$

$Z = 1.28$

$Z = 1.28$ has a pvalue of 0.8997

1 - 0.8997 = 0.1003

10.03% probability of getting a cup weighing more than 8.64oz