Question

An instructor in a large lecture class found, at the end of the semester, that the total point distribution in his class was approximately Normal, with a mean of 530 and a standard deviation of 80. If 10% of the class is to receive A’s (the instructor grades on a curve), what is the lowest number of points that a student can have and still earn an A? 632 610 428

Answer 1

Answer:

$a=530 +1.28*80=632.4 \approx 632$

So the value of height that separates the bottom 90% of data from the top 10% is approximately 632.

Step-by-step explanation:

For this case we can assume the following grade system:

A: 90-100

B: 80-89

C: 70-79

D: 60-69

F: 0-59

And we see that A represent the best scores.

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

$X \sim N(530,80)$  

Where $\mu=530$ and $\sigma=80$

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

$P(X>a)=0.10$   (a)

$P(X<a)=0.90$   (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 0.90 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)=0.90 and P(z>1.28)=0.1

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

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90$  

$P(z<\frac{a-\mu}{\sigma})=0.90$

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

$z=1.28=\frac{a-530}{80}$

And if we solve for a we got

$a=530 +1.28*80=632.4 \approx 632$

So the value of height that separates the bottom 90% of data from the top 10% is approximately 632.