Answer: 0.6065
Step-by-step explanation:
Given : The machine's output is normally distributed with
[tex]\mu=27\text{ ounces per cup}[/tex]
[tex]\sigma=3\text{ ounces per cup}[/tex]
Let x be the random variable that represents the output of machine .
z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 21 ounces
[tex]z=\dfrac{21-27}{3}\approx-2[/tex]
For x= 28 ounces
[tex]z=\dfrac{28-27}{3}\approx0.33[/tex]
Using the standard normal distribution table , we have
The p-value : [tex]P(21<x<28)=P(-2<z<0.33)[/tex]
[tex]P(z<0.33)-P(z<-2)=0.6293-0.0227501=0.6065499\approx0.6065[/tex]
Hence, the probability of filling a cup between 21 and 28 ounces= 0.6065
