Respuesta :
Answer:
a) [tex]P(64<X<77)=P(-1.64<Z<3)=P(Z<3)-P(Z<-1.64)=0.9987-0.0505=0.948[/tex]
b) The new requirement would be that the height would be between b=63.112 and a=74.088
Step-by-step explanation:
1) 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".
2) Part a
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(68.6,2.8)[/tex]
Where [tex]\mu=68.6[/tex] and [tex]\sigma=2.8[/tex]
We are interested on this probability
[tex]P(64<X<77)[/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(64<X<77)=P(\frac{64-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{77-\mu}{\sigma})=P(\frac{64-68.6}{2.8}<Z<\frac{77-68.6}{2.8})=P(-1.64<Z<3)[/tex]
And we can find this probability on this way:
[tex]P(-1.64<Z<3)=P(Z<3)-P(Z<-1.64)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.64<Z<3)=P(Z<3)-P(Z<-1.64)=0.9987-0.0505=0.948[/tex]
Part b
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.025[/tex] (a)
[tex]P(X<a)=0.975[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.975 of the area on the left and 0.025 of the area on the right it's z=1.96. On this case P(Z<1.96)=0.975 and P(z>1.96)=0.025
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.975[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.96<\frac{a-68.6}{2.8}[/tex]
And if we solve for a we got
[tex]a=68.6 +1.96*2.8=74.088[/tex]
So the value of height that separates the bottom 97.5% of data from the top 2.5% is 74.088.
In order to find the lower limit we just need to do a similar operation but with a change of sign:
[tex]b=68.6 -1.96*2.8=63.112[/tex]
The new requirement would be that the height would be between b=63.112 and a=74.088
