Tough probability question...any thoughts?

Question

A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If a random sample of size 75 is selected, the probability is ________ that between 30% and 50% of the students in the sample will be minority students.

Answer 1

a binomial distribution can probably be used to answer this accuratly.

i will approximate.

.45 minority
.55 not minority

lower bounds 75 * .3 = 23 minority students
upper bounds 75 * .5 = 38 minority students.

p(23 minority) = 75C23 * .45^23 * .55^52
p(38 minority) = 75C23 * .45^38 * .55^37

p(between 23 studenbts and 38 minority students) will be the sum of the probabilities from(23) up until p(38)

there are probably distribution tables that can also be used to get an immediate answer.

Answer 2

30% of 75 is 22.5, and 50% of 75 is 37.5. So you are asking what is the probability that between 23 and 37 students of the 75 are minorities (assuming we're not counting mixed ethnicities as fractions of people).
Since the sample is random, each student of the 75 has a .45 chance of being a minority.
So we add the probability that exactly 23 students will be minorities to the probability that exactly 24 will, to the probability that exactly 25 will, all the way up to 37.
The probability that exactly 23 students will be minorities is equal to the probability that any particular combination of students will be minorities (0.45^23 times 0.55^(75-23)) times the number of combinations of 23 students you can make out of 75 (75 choose 23). Add up this computation for each value from 23 to 37 and you get 0.808, or about 81 percent.

EDIT:
The previous two answers are close but incorrect. The probability of not being a minority is .55, not .65. Also, it's between 23 and 37, not 38, as 38 is not between 30% and 50% of 75.

Answer 3

30% of the students of size 75 is 22.5 (so 23)
50% of the students of size 75 is 37.5 (so 37).

so you want the probability that between 23 and 37 students are selected.

The probability that there are 23 students selected =
P(23) = [75choose23]*(.45)^23(.65)^52
Similarly, 24 students:
P(24) = [75choose24]*(.45)^24(.65)^53
....etc....
P(37) = [75choose37]*(.45)^37(.65)^38

To get the total probability, you need to add P(23)+P(24)+P(25)+...+P(37)

FYI - [75choose23] = 75! / 23!52!

Answer 4

Use H to represent the double-sided head coin, T for the double-sided tail, and N for the normal coin. Use x to represent the event that you see heads. x occurs if you draw H, (prob = 1/3) or if you draw N and it turns up showing heads (prob = (1/3)*(1/2)) Thus P(x) = (1/3)*1 + (1/3)*(1/2) = 1/2 as expected. Now use conditional probability: P(xH) = P(x given H)*P(H) or P(H given x)*P(x) The first of these gives P(xH) = 1/3, which is just common sense because P(H) = (1/3) and when H happens, x always does, so the probability of them happening together is 1/3. But applying the second expression to this gives P(H given x)*P(x) = 1/3, and since we already know that P(x) = 1/2, then P(H given x) = 2/3, so your second answer is correct. I suppose another way of looking at it is that there are three head faces altogether, of which two are on the double coin, so if you see a head the prob that it's on the double-head coin is 2/3, which is another way of saying what you said. What's wrong with the 50% argument? It assumes both coins are equally likely, which isn't the case if a head is already showing. All three head faces are equally likely.

Answer 5

Since you have a large sample size, and p is near 0.5, you can use the normal approximation to the binomial distribution.

The binomial's p value is 0.45, so (1-p) = 0.55.

For the normal approximation, the mean is equal to
n*p
= 75*0.45
= 33.75
The standard deviation is equal to
sqrt(n*p*(1-p))
= sqrt(75*0.45*0.55)
= sqrt(18.56)
= 4.308

Now, getting 30% minority students means that 0.3*75 = 22.5 students are minorities, and 50% means that 0.5*75 = 37.5 students are minorities. To obtain z-values for our normal distribution, we subtract the mean from each endpoint, then divide by the standard deviation.

(22.5 - 33.75) / 4.308
= -11.25 / 4.308
= -2.611

(37.5 - 33.75) / 4.308
= 3.75 / 4.308
= 0.8704

So, we integrate the basic normal distribution (with a mean of 0 and a stdev of 1) from -2.611 to 0.8704. Doing so, we obtain the answer:

integral[1/sqrt(2*pi) * e^(-x^2 / 2), x, -2.611, 0.8704]
= 0.8034

Answer 6

45%

Answer 7

I could really help you with this if I had my calculator... but if you have a TI-83 like me, it has a probability feature on it that can really help with problems like this