Stat 100

[Alex]

I'm not sure why you're putting the 0.8 x 0.2 in the Part B answer. All you need is the second part.

But otherwise that's right.

[BarTopDancer]

Thanks Alex!

I'm not sure why I put that either. I'm sure at the time I had a good (incorrect) reason.

[Ghoulish Delight]

Originally Posted by BarTopDancer

-----

New question:

636 students answered the survey.

0.47 said they consumed more alcohol then usual

636 x .47 = 299 students who consumed more alcohol than usual

337 students did not consume more alcohol than usual. This is 53%.

(0.47 + 0.53 = 1)

So, I think the caculation would be

.47 x .53 x .47 = .12

Close, but I think this is incorrect. Notice they said that the order matters.

Remember the "what assumptions are being made" part of question b? Don't make those same assumptions for this one.

[BarTopDancer]

Originally Posted by Ghoulish Delight

Close, but I think this is incorrect. Notice they said that the order matters.

Remember the "what assumptions are being made" part of question b? Don't make those same assumptions for this one.

Confused.

.47 drank more than usual (1st and 3rd students)
.53 did not consume more than usual (2nd)

If you remove either the 1st or the third student the probability doesn't change.

If I do .47 x .53 = .25

I'm so lost.

[Alex]

Sorry, BTD, I suspect this is going to confuse you more but now GD and I hash it out and I don't know any other way without getting explicit.


GD: I thought about that, but the resulting difference doesn't make a huge difference until 1/100th of a percent even if pool depletion is taken into account.

299/636 x 337/635 x 298/634 = 0.1172727 = 11.72727%

Not taking pool depletion into account it is:

299/636 x 337/636 x 299/636 = 0.1171119 = 11.71119%

Using BTD two-digit rounding results in:

.47 x .53 x .47 = 0.117077 = 11.7077% (ignoring the issue of significant digits).

However, the question, as set up does not draw the three students from the pool of survey respondents so it isn't actually possible to take pool depletion into account. The survey is a 636 student subset. The three students in question are drawn from the entire pool of Penn State students and the question does not give the information on that pool size.

But that is the hard part of seeing questions out of context of the class. I don't know what level of imprecision is acceptable. So if I were doing this homework I'd probably present the second equation above. But I don't think the first one (which is what I think GD is going for) is justified. Though it would be if the three students were drawn from the population of survey respondents.

[BarTopDancer]

You two can hash it out all you want. I've gotta turn it in now, so hopefully I'll understand the teacher's corrections.

Thanks for your help!

[BarTopDancer]

I got number 3 wrong.

Suppose studies have shown that approximately 20% (.2) of Americans regularly engage in exercise (at least 3 times per week).

a. If one American is randomly selected, what is the probability that he or she does not regularly engage in exercise? Set up calculation.

1 - .20 = .80

The probability that a randomly selected American does not engage in regular exercise is 80%

b. If three Americans are randomly selected, what is the probability that all three regularly engage in exercise? Set up the calculation.

.8 x .8 x .8 = .51

(-1)
(.2) x (.2) x (.2) = .008

c. What assumption is being made in part b above?

The assumption being made in part B is that if you remove one person from the equation it will not change the outcome of the probability.

(-1) sure the probability will change!!!

The answer is that we are assuming independence of the events. Otherwise, we could not just multiply the individual event probabilities together to arrive at the “joint probability” of all 3 events happening

d. If two Americans are randomly selected, what is the probability that the first American regularly engages in exercise while the second American does not regularly engage in exercise? Set up the calculation.

.80 x .20 = .16
1 - .20 = .80

(-.5)
Correct order is important and will be important on the exam. The answer is :
(.20) x (.80) = .16



Make sure that the values are in the correct order.

[Alex]

Crap. I'm partly to blame for b being wrong. GD saw that mistake on your part of using the odds on the wrong side but then I flipped them around again in helping.

Part C is the difficulty of trying to help without the class context. Odds are that was a specific point in the chapter, though I'd argue the answer you gave is a subset of the answer he was looking for. GD's answer is still mostly correct (though adding the word "significantly" to "will not change" might make it better). It just isn't the assumption being sought.

Part D was just a continuation of the same mistake in Part B of having flipped the odds of the two outcomes.

[BarTopDancer]

It's ok Alex. I still ended up with a C on the assignment. I did get the last question correct (the 2nd one I posted about).

I just wanted to understand the discrepancies.

Only 2 more assignments, and 2 tests. If I can hold this C I will be super happy. If I get a D in the class I'll still be super happy, cause I passed!

[BarTopDancer]

I need a total of 365 points to pass. I have 255 points. There are 220 points left in the class. I need to get 110 of them.

2 assignments left, one 50 points, one 70 points and one test worth 100 points.

I'm starting to freak out.