Question

A state administered standardized reading exam is given to eighth grade students. The scores on this exam for all students statewide have a normal distribution with a mean of 502 and a standard deviation of 81. A local Junior High principal has decided to give an award to any student who scores in the top 10% of statewide scores. How high should a student score be to win this award?

Answer 1

Answer:

$a=502 +1.28*81=605.68$

So the value of height that separates the bottom 90% of data from the top 10% is 605.68.  

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:

$X \sim N(502,81)$  

Where $\mu=502$ and $\sigma=81$

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

$P(X>a)=0.1$   (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 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

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

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9$  

$P(z<\frac{a-\mu}{\sigma})=0.9$

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

$z=1.28<\frac{a-502}{81}$

And if we solve for a we got

$a=502 +1.28*81=605.68$

So the value of height that separates the bottom 90% of data from the top 10% is 605.68.