Respuesta :
Answer:
a) [tex]P(X<72)=P(\frac{X-\mu}{\sigma}<\frac{72-\mu}{\sigma})=P(Z<\frac{72-75}{10})=P(z<-0.3)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-0.3)=0.382[/tex]
b) [tex]P(X>77)=P(\frac{X-\mu}{\sigma}>\frac{77-\mu}{\sigma})=P(Z>\frac{77-75}{10})=P(z>0.2)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel:
[tex]P(z>0.2)=1-P(Z<0.2) = 1-0.579=0.421 [/tex]
c) [tex] z = \frac{70-75}{\frac{10}{\sqrt{16}}}= -2[/tex]
[tex] z = \frac{80-75}{\frac{10}{\sqrt{16}}}= 2[/tex]
And we can find this probability with this difference:
[tex] P(-2<z<2) = P(Z<2) -P(z<-2)= 0.97725-0.02275= 0.9545[/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".
Solution to the problem
Let X the random variable that represent the life of a particular brand of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(75,10)[/tex]
Where [tex]\mu=75[/tex] and [tex]\sigma=10[/tex]
Part a
We are interested on this probability
[tex]P(X<72)[/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<72)=P(\frac{X-\mu}{\sigma}<\frac{72-\mu}{\sigma})=P(Z<\frac{72-75}{10})=P(z<-0.3)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-0.3)=0.382[/tex]
Part b
[tex]P(X>77)=P(\frac{X-\mu}{\sigma}>\frac{77-\mu}{\sigma})=P(Z>\frac{77-75}{10})=P(z>0.2)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel:
[tex]P(z>0.2)=1-P(Z<0.2) = 1-0.579=0.421 [/tex]
Part c
For this case we select a sample size of n =16. and we want this probability:
[tex] P(70 < \bar X <80)[/tex]
We know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we can find the z scores for the limits like this:
[tex] z = \frac{70-75}{\frac{10}{\sqrt{16}}}= -2[/tex]
[tex] z = \frac{80-75}{\frac{10}{\sqrt{16}}}= 2[/tex]
And we can find this probability with this difference:
[tex] P(-2<z<2) = P(Z<2) -P(z<-2)= 0.97725-0.02275= 0.9545[/tex]
