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lisabon 2012 [21]
3 years ago
12

The mean weight of frozen yogurt cups in an ice cream parlor is 8 oz.Suppose the weight of each cup served is normally distribut

ed withstandard deviation 0.5 oz, independently of others.(a) What is the probability of getting a cup weighing more than 8.64oz
Mathematics
1 answer:
vaieri [72.5K]3 years ago
4 0

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

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