Answer:
Area under the normal curve: 0.6915.
69.15% probability of putting less than 24 ounces in a cup.
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 problem, we have that:
[tex]\mu = 23, \sigma = 2[/tex]
You have been asked to calculate the probability of putting less than 24 ounces in a cup.
pvalue of Z when X = 24. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{24 - 23}{2}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a pvalue of 0.6915
Area under the normal curve: 0.6915.
69.15% probability of putting less than 24 ounces in a cup.