Assume that blood pressure readings are normally distributed with μ = 122 and σ = 7.5. A researcher wishes to select people for a study but wants to exclude the top and bottom 10 percent. What would be the upper and lower readings to qualify people to participate in the study?

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 blood pressures of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(122,7.5)[/tex]

Where [tex]\mu=122[/tex] and [tex]\sigma=7.5[/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]

For this part we want to find a value a, such that we satisfy this condition:

[tex] P(X<a) =0.1[/tex] (a)

[tex] P(X>a) =0.9[/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 01 of the area on the left and 0.90 of the area on the right it's z=-11.28. On this case P(Z<-1.28)=0.10 and P(z>-1.28)=0.9

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.1[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.1[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-1.28<\frac{a-122}{7.5}[/tex]

And if we solve for a we got

[tex]a=122 -1.28*7.5=112.4[/tex]

So the value of height that separates the bottom 10% of data from the top 90% is 112.4.