Answer:
[tex]z=-1.28<\frac{a-66}{1.5}[/tex]
And if we solve for a we got
[tex]a=66 -1.28*1.5=64.08[/tex]
So the value of height that separates the bottom 10% of data from the top 90% is 64.08.
The graph attached illustrate the situation.
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 heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(66,1.5)[/tex]
Where [tex]\mu=66[/tex] and [tex]\sigma=1.5[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.90[/tex] (a)
[tex]P(X<a)=0.10[/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 0.10 of the area on the left and 0.90 of the area on the right it's z=-1.28. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.10[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.10[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.28<\frac{a-66}{1.5}[/tex]
And if we solve for a we got
[tex]a=66 -1.28*1.5=64.08[/tex]
So the value of height that separates the bottom 10% of data from the top 90% is 64.08.
The graph attached illustrate the situation.
