The right formula for finding prob. that three get above C

[mhb22079]

Hoping someone can provide the correct formula for the following question:
Given grades A,B,C,D,F, and 5 students randomly selected, what is the probability of 3 of them getting above a C, and the other two below a C.

I'm totally lost here and have no clue. If someone could please post the formula and/or steps to solving this I'd greatly appreciate it.

 

[pka]

Re: The right formula, should be easy I don't get it though.

Given grades A,B,C,D,F, and 5 students randomly selected, what is the probability of 3 of them getting above a C, and the other two below a C.
I have no clue.

I am not at all surprise by that, because the way you have stated the problem there is no answer. Please post the exact wording of the problem. If the above is the exact question, just say there is no way of knowing.

 

[mhb22079]

Sorry I tried to simplify the question by using alpha grades in liu of values. What is provided is:
A = 100
B = 90
C = 80
D = 70
E = 60

If 5 students are randomly selected, find the probability that 3 of them received a score greater than the mean (which is obviously 80) and the other two received a score less than the mean.

Thanks!

 

[pka]

What is provided is:
A = 100
B = 90
C = 80
D = 70
E = 60
If 5 students are randomly selected, find the probability that 3 of them received a score greater than the mean (which is obviously 80) and the other two received a score less than the mean. WHY IS TRAT TRUE??

Still not enough information!
What assumptions are you to make about the nature of the distribution?
Is it normal? Uniform? What?
What about variance?

 

[mhb22079]

I honestly don't know. I am only in the third chapter of the course and those topics have not been covered. I know the answer to be 5/16 but am not sure how it was arrived at.

My feeling is that the 16 is determined in the same way that 2 students randomly selected would both have scores between 70 & 80 as:

P(A)*P(B/A) = 1/4 * 1/4 = 1/16.

 

[pka]

Were you told how many of each grade was given?
That is, so many A’s, so many B’s etc.
There must be more to it than you have stated.
Otherwise, as I said to begin with, there is nothing to do.

 

[mhb22079]

No, the way I understand it is that a students could have a score within that range of 60-100. The mean is determined by the average of the range of values which is 80. For one of the other questions

"If 1 student was randomly selected find the probaboility that their score was between 60 and 70 I got 1/4 which was correct according to the text.

The other question was If a student was selected the probability that their score would be below 70 or above 80 I got 3/4 which was correct.

No sure what else I can provide.

 

[pka]

I know the answer to be 5/16 but am not sure how it was arrived at. Students could have a score within that range of 60-100. The mean is determined by the average of the range of values which is 80. For one of the other questions "If 1 student was randomly selected find the probability that their score was between 60 and 70 I got 1/4 which was correct according to the text.

O.K. It appears that your text material is using a uniform distribution on the interval [60,100]. The mean in that case is 80. The probability that anyone would be in the [80,100] range is (1/2).

Now it appears that your text switches to a binomial distribution.
The probability that three of five independent selections are above 80 is
\(\displaystyle {5 \choose 3} \left( {\frac{1}{2}} \right)^3 \left( {\frac{1}{2}} \right)^2 = \frac{{5!}}{{3!2!}}\left( {\frac{1}{2}} \right)^5 = \frac{5}{{16}}\).

 

[mhb22079]

Ok, I'm about 1/2 (no pun intented) way there. I see that 1/2 is the probability that one student will be over / under the mean. So 1/2 cubed = the three that are over, and 1/2 squared are the other two under. What then is the 5/3?

Thanks again for everything.

 

[galactus]

\(\displaystyle C(5,3)=10\). That is the coefficient in the binomial expansion for

\(\displaystyle (p+q)^{5}=p^{5}+5p^{4}q+10p^{3}q^{2}+10p^{2}q^{3}+5pq^{4}+q^{5}\)

\(\displaystyle p=1/2\ and\ q=1/2\)

Since they're asking for 3 students above the mean, you use \(\displaystyle 10p^{3}q^{2}\)

\(\displaystyle 10(1/2)^{3}(1/2)^{2}=\frac{5}{16}\)

This is one way of looking at it.

 

[pka]

You really need to lookup binominal distribution in your text.
That is a very important concept and you will have to use it a good bit therefore you need to study it until you understand it.
The symbol you asked about is the combination N things taken k at a time:
\(\displaystyle \L
{N \choose k} = \frac{{N!}}{{k!\left( {N - k} \right)!}}\). That is the number of ways of choosing k items from N.
We are given 5 students, we want to know how ways to choose 3.

The binominal distribution is a series of independent trials with the probability of success p. If we do N trials the probability of exactly k successes is
\(\displaystyle \L
{N \choose k}\left( p \right)^k (1 - p)^{N - k}\) .

 

[mhb22079]

OK, I think I'm nearly there. The "binomial" stuff is not covered until the next chapter but they do show the "permutations & combinations rule" formulas.

Thank you both so much for your help on this! The Prof advised that this would be the hardest chapter and I believe he is right.

 

[pka]

The "binomial" stuff is not covered until the next chapter but they do show the "permutations & combinations rule" formulas.

It seems to me as if your textbook may have topical order a bit askew.
I would like to know how it expected you to work that problem otherwise.

 

[mhb22079]

I suppose they know that there are great websites like this to get help. I think I finally got it with the formulas you guys posted. What I was not getting was the "permutations rule".

Thanks again.

 

[pka]

What I was not getting was the "permutations rule".

Not quite sure what you mean by your statement?
Maybe you are confused as to the distinction between combinations and permutations.
Here is a rule of thumb that I have given students for years.
Combinations are content driven.
Permutations are order driven.

Combinations are about “how many ways we can form subsets from a set?”.
Permutations are about “how many ways can we form a queue of distinct elements?”.

 

[galactus]

You'll learn about Pascal's triangle. That comes in to play here.

                                     1
                                1         1
                             1      2        1
                           1    3       3       1
                        1    4      6       4     1
                      1   5     10     10     5     1

See the bottom row?. That's the row that corresponds to the coefficients in your expansion.

Each number is derived by adding the two numbers directly above it.

See?. Pretty cool, huh?. You can keep going and going.

The 5th row down(not counting the 1 at the top of the pyramid)

You have \(\displaystyle (p+q)^{5}\). Notice the coiefficients in the expansion I gave in my last post?. Notice the numbers in the fifth row of the pyramid?.

They're the same, ain't they?.

You'll learn about it in class, though. I don't wanna spoil all the fun.