Distribution of equal probability?

Question

This involves statistical methods, and is for my homework.  I have so far been very good with my homework and being able to do it myself, but I am just STUCK on this one, and I think it's largely because of the terminology.  It's entirely possible that this question is poorly worded.  The answers to this question, and the method by which you solve it, influences the answers you get in the next few questions as well, so it's a nice little potential "chain of failure" if I get this one wrong.

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Twenty-five percent of the students at Midwestern are taking 6 credit hours, 25% are taking 9 credit hours, 25% are taking 12 credit hours, 25% are taking 15 credit hours. Use EXCEL to do parts a through e.

a. Give the distribution of X = number of credit hours taken by students at Midwestern.
b. Draw a scatter plot of the distribution.
c. Find the mean of X.
d. Find the variance of X.
e. Find the standard deviation of X.

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Now here's my problem with this question... Asking for the "distribution" seems kinda silly, since the distribution seems to have been given already, as 25% for each!

If we're talking a "normal distribution", and asking "what's the chance of a random student being in any of these 4 particular groups", the answer would be... 25% chance, right?

Now if' we're talking about "the chance a student is taking AT LEAST 12 credit hours", it would be the cumulative chance... but even then it would be additive, or 50% chance (25% chance of being 12, 25% chance of being 15).

Please help!

Stochastic Seeker · Accepted Answer

I have only taken one course in probability theory, but let me try this out:

a. I assume that they want the c.d.f. (the p.d.f. should be obvious from the question) so we can just define the piece-wise function:

P(X<=x) = F_X(x) ={

0 if x < 6

0.25 if 6 <= x < 9

0.5 if 9 <= x < 12

0.75 if 12 <= x < 15

1 if x > 15

b. I assume you can do this part on your own.

c. The mean is the expectation and is defined by:

E[X] = sum [x in range(x)] of x*f_X(x) = 0.25*6 + 0.25*9 + 0.25*12 + 0.25*15 = 10.5

d. The variance is defined as:

Var[X] = E[X^2] - [E[X]]^2

Now E[X^2] = sum [x in range(x)] of (x^2)*f_X(x) = 0.25*36 + 0.25*81 + 0.25*144 + 0.25*225 = 121.5

So Var[X] = 121.5 - (10.5)^2 = 11.25

e. The standard deviation is the square root of variance:

SD[X] = sqrt(11.25) ~= 3.3541

In regards to your other comments:

- If X was distributed normally, then it is assuming that X is a continuous random variable, which clearly it is not (# of credits is discrete). If we assume that we can take on real values for the number of credits, then X~ N(m,v^2) where m = E[X] = 10.5 and v^2 = Var[X] = 11.25.

For normal distributions, you cannot ask what is the probability that someone will be in this particular group, but instead will have to ask what is the probability that someone will lie in this range of values.

To do this, you standardize your normal distribution and use the standardized normal distribution tables.

- A student is taking at least 12 credits: P(X>=12) = 1 - P(X<12) = 1- P(X<=11) = 1 - 0.5 = 0.5 as you have stated.

~ Hope this is helpful.

Distribution of equal probability?

1 Answer