Respuesta :
Answer:
0.8919 = 89.19% probability of filling a cup between 33 and 35 ounces.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
A soft drink machine outputs a mean of 29 ounces per cup. The machine's output is normally distributed with a standard deviation of 4 ounces.
This means that [tex]\mu = 29, \sigma = 4[/tex]
What is the probability of filling a cup between 33 and 35 ounces?
This is the p-value of Z when X = 35 subtracted by the p-value of Z when X = 33.
X = 35
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35 - 29}{4}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a p-value of 0.9332.
X = 33
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{33 - 29}{4}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.0413.
0.9332 - 0.0413 = 0.8919
0.8919 = 89.19% probability of filling a cup between 33 and 35 ounces.
