Aprofessor at a local community college noted that the grades of his students were normally distributed with a mean of 84 and a standard deviation of 6. the professor has informed us that 10 percent of his students received a's while only 2.5 percent of his students failed the course and received f's. a. what is the minimum score needed to make an a?
b. what is the maximum score among those who received an f?
c. if there were 5 students who did not pass the course, how many students took the course?

A. P(x>91.71)=0.10, so the minimum grade is 91.71

B. P(x<72.24)=0.025 so the maximum grade could be 72.24

C. By rule of three, 200 students took the course

Step-by-step explanation:

The problem says that the grades are normally distributed with mean 84 and STD 6, and we are asked some probabilities. We can´t find those probabilities directly only knowing the mean and STD (In that distribution), At first we need to transfer our problem to a Standard Normal Distribution and there is where we find those probabilities. We can do this by a process called "normalize".

P(x<a) = P( (x-μ)/σ < (a-μ)/σ ) = P(z<b)

Where x,a are data from the original normal distribution, μ is the mean, σ is the STD and z,b are data in the Standard Normal Distribution.

There´s almost no tools to calculate probabilities in other normal distributions. My favorite tool to find probabilities in a Standard Normal Distribution is a chart (attached to this answer) that works like this:

P(x<c=a.bd)=(a.b , d)

Where "a.b" are the whole part and the first decimal of "c" and "d" the second decimal of "c", (a.b,d) are the coordinates of the result in the table, we will be using this to answer these questions. Notice the table only works with the probability under a value (P(z>b) is not directly shown by the chart)

A. We are asked for the minimum value needed to make an "A", in other words, which value "a" give us the following:

P(x>a)=0.10

Knowing that 10% of the students are above that grade "a"

What we are doing to solve it, as I said before, is to transfer information from a Standard Normal Distribution to the distribution we are talking about. We are going to look for a value "b" that gives us 0.10, and then we "normalize backwards".

P(x>b)=0.10

Thus the chart only works with probabilities UNDER a value, we need to use this property of probabilities to help us out:

P(x>b)=1 - P(x<b)=0.10

P(x<b)=0.9

And now, we are able to look "b" in the chart.

P(x<1.28)=0.8997

If we take b=1.285

P(x<1.285)≈0.9

Then

P(x>1.285)≈0.1

Now that we know the value that works in the Standard Normal Distribution, we "normalize backwards" as follows:

P(x<a) = P( (x-μ)/σ < (a-μ)/σ ) = P(z<b)

If we take b=(a+μ)/σ, then a=σb+μ.

a=6(1.285)+84

a=91.71

And because P(x<a)=P(z<b), we have P(x>a)=P(z>b), and our answer will be 91.71 because:

P(x>91.71) = 0.1

B. We use the same trick looking for a value in the Standard Normal Distribution that gives us the probability that we want and then we "normalize backwards"

The maximum score among the students who failed, would be the value that fills:

P(x<a)=0.025

because those who failed were the 2.5% and they were under the grade "a".

We look for a value that gives us:

P(z<b)=0.025 (in the Standard Normal Distribution)

P(z<-1.96)=0.025

And now, we do the same as before

a=bσ+μ

a=6(-1.96)+84

a=72.24

So, we conclude that the maximum grade is 72.24 because

P(x<72.24)=0.025

C. if 5 students did not pass the course, then (Total)2.5%=5

So we have:

2.5%⇒5

100%⇒?

?=5*100/2.5

?=200

There were 200 students taking that course

y = 0.77(x)

step-by-step explanation:

the jacket retains this value after 23% cut or retains 77% of original value. plug original 87 jacket price for x and the remaining value should be y.