Question

The distribution of grade point averages​ (GPAs) for medical school applicants of a certain year were approximately​ Normal, with a mean of 3.48 and a standard deviation of 0.36. Suppose a medical school will only consider candidates with GPAs in the top 20​% of the applicant pool. An applicant has a GPA of 3.75. Does this GPA fall in the top 20​% of the applicant​ pool?

Answer 1

Answer:

$a=3.48 +0.842*0.36=3.783$

So the score that separates the bottom 80% of data from the top 20% is 3.783. Since the value obtained by the applicant is 3.75 <3.783 not falls on the top 20%.  

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 grade point averages​ (GPAs) for medical school applicants of a certain year of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.48,0.36)$  

Where $\mu=3.48$ and $\sigma=0.36$

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

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

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

$P(X>a)=0.20$   (a)

$P(X<a)=0.80$   (b)

Both conditions are equivalent on this case. We can use the z score in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. We can found this with the following excel code:"=NORM.INV(0.8,0,1)". On this case P(Z<0.842)=0.80 and P(Z>0.842)=0.2

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

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.80$  

$P(Z<\frac{a-\mu}{\sigma})=0.80$

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

$z=0.842<\frac{a-3.48}{0.36}$

And if we solve for a we got

$a=3.48 +0.842*0.36=3.783$

So the score that separates the bottom 80% of data from the top 20% is 3.783. Since the value obtained by the applicant is 3.75 <3.783 not falls on the top 20%.  

Answer 2

Z-table is also known as the standard normal distribution table. The student is among the top 20%.

What is a Z-table?

A z-table also known as the standard normal distribution table, helps us to know the percentage of values that are below (or to the left of the Distribution) a z-score in the standard normal distribution.

Since it is given that the mean GPA of the students is 3.48, while the standard deviation is 0.36, therefore, the GPA which is in the top 20% can be given as:

$P(Z\leq z)=20\% =0.2\\\\\text{From z-tables}\\z=\dfrac{x-\mu}{\sigma} = 0.5793\\\\\dfrac{x-3.48}{0.36}=0.573\\\\x = 3.69$

Thus, the minimum GPA a student needs to score to be in 20% of the class is 3.69.

As the score of the student is 3.75, therefore, the student is among the top 20%.

Learn more about Z-table:

https://brainly.com/question/6096474