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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 8, \sigma = 0.5[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8.64 - 8}{0.5}[/tex]
[tex]Z = 1.28[/tex]
[tex]Z = 1.28[/tex] has a pvalue of 0.8997
1 - 0.8997 = 0.1003
10.03% probability of getting a cup weighing more than 8.64oz
