Respuesta :
Answer:
a) [tex]P(X>8)=P(\frac{X-\mu}{\sigma}>\frac{8-\mu}{\sigma})=P(Z>\frac{8-7.6}{0.4})=P(Z>1)[/tex]
And we can find this probability using the complement rule and the standard normal table or excel:
[tex]P(Z>1)=1-P(Z<1)= 1-0.841=0.159[/tex]
b)[tex]P(X<8)=P(\frac{X-\mu}{\sigma}<\frac{8-\mu}{\sigma})=P(Z<\frac{8-7.6}{0.4})=P(Z<1)[/tex]
And we can find this probability using the standard normal table or excel:
[tex]P(Z<1)=0.841[/tex]
c) For this case we have a total of 868 cups and we know that the fraction of cups that will be overflow is 0.159, so then the expected number of cups overflow would be:
[tex] N = 868*0.159= 138.012 \approx 138[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the amount of soft frink of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(7.6,0.4)[/tex]
Where [tex]\mu=7.6[/tex] and [tex]\sigma=0.4[/tex]
We are interested on this probability
[tex]P(X>8)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>8)=P(\frac{X-\mu}{\sigma}>\frac{8-\mu}{\sigma})=P(Z>\frac{8-7.6}{0.4})=P(Z>1)[/tex]
And we can find this probability using the complement rule and the standard normal table or excel:
[tex]P(Z>1)=1-P(Z<1)= 1-0.841=0.159[/tex]
Part b
We are interested on this probability
[tex]P(X<8)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<8)=P(\frac{X-\mu}{\sigma}<\frac{8-\mu}{\sigma})=P(Z<\frac{8-7.6}{0.4})=P(Z<1)[/tex]
And we can find this probability using the standard normal table or excel:
[tex]P(Z<1)=0.841[/tex]
Part c
For this case we have a total of 868 cups and we know that the fraction of cups that will be overflow is 0.159, so then the expected number of cups overflow would be:
[tex] N = 868*0.159= 138.012 \approx 138[/tex]
