Question

Scores on a test are approximately normally distributed with a mean of 70 and a standard deviation of 9. The teacher wants to give A's to the top 10% of students, B's to the next 25%, C's to the next 40%, D's to the next 16%, and F's to the bottom 9%. What is the bottom cutoff for a D grade? Round your answer to the nearest whole number. a) 58b) 65c) 62d) 56

Answer 1

`z=-1.34<(a-70)/(9)`

And if we solve for a we got

`a=70 -1.34*9=57.95 \approx 58`

So the value of height that separates the bottom 9% of data from the top 91% is 58.

And the answer for this case would be:

a) 58

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

`X \sim N(70,9)`

Where
`\mu=70` and
`\sigma=9`

For this case the figure attached illustrate the situation for this case.

We know from the figure that the lower limti for D accumulates 9% or 0.09 of the area below and 0.91 or 91% of the area above.

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

`P(X>a)=0.91` (a)

`P(X<a)=0.09` (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 0.09 of the area on the left and 0.91 of the area on the right it's z=-1.34. On this case P(Z<-1.34)=0.09 and P(z>-1.34)=0.91

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.09`

`P(z<(a-\mu)/(\sigma))=0.09`

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

`z=-1.34<(a-70)/(9)`

And if we solve for a we got

`a=70 -1.34*9=57.95 \approx 58`

So the value of height that separates the bottom 9% of data from the top 91% is 58.

And the answer for this case would be:

a) 58