A soft drink machine outputs a mean of 24 ounces per cup. The machine's output is normally distributed with a standard deviation of 3 ounces. What is the probability of filling a cup between 21 and 28 ounces? Round your answer to four decimal places.

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Answer:

[tex]P(21<X<28)=P(\frac{21-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{28-\mu}{\sigma})=P(\frac{21-24}{3}<Z<\frac{28-24}{3})=P(-1<z<1.33)[/tex]

And we can find the probability with this difference

[tex]P(-1<z<1.33)=P(z<1.33)-P(z<-1)[/tex]

And using the normal standard distribution or excel we got:

[tex]P(-1<z<1.33)=P(z<1.33)-P(z<-1)=0.908-0.159=0.749[/tex]

Step-by-step explanation:

Let X the random variable that represent the soft drink machine outputs of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(24,3)[/tex]  

Where [tex]\mu=24[/tex] and [tex]\sigma=3[/tex]

We want to find this probability:

[tex]P(21<X<28)[/tex]

The z score is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Using this formula we got:

[tex]P(21<X<28)=P(\frac{21-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{28-\mu}{\sigma})=P(\frac{21-24}{3}<Z<\frac{28-24}{3})=P(-1<z<1.33)[/tex]

And we can find the probability with this difference

[tex]P(-1<z<1.33)=P(z<1.33)-P(z<-1)[/tex]

And using the normal standard distribution or excel we got:

[tex]P(-1<z<1.33)=P(z<1.33)-P(z<-1)=0.908-0.159=0.749[/tex]

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