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

Answer:

$z=-1.34<\frac{a-70}{9}$

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

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 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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.09$  

$P(z<\frac{a-\mu}{\sigma})=0.09$

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

$z=-1.34<\frac{a-70}{9}$

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