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lord [1]
3 years ago
13

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?
Mathematics
1 answer:
harina [27]3 years ago
4 0

Answer:

z=-1.28

And if we solve for a we got

a=122 -1.28*7.5=112.4

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

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

X \sim N(122,7.5)  

Where \mu=122 and \sigma=7.5

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

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

P(X   (a)

P(X>a) =0.9   (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:

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

P(X  

P(z

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

z=-1.28

And if we solve for a we got

a=122 -1.28*7.5=112.4

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

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