Here as promised, word for word^{1}, is the second exam I gave my introductory statistics students. A prerequisite for the class was at least “Pre calculus”. I emphasize—in italics!—that I have taught this course before many times and have never had as much difficulty as I am having this time through.
This exam came after reviewing chapters three, four, and six for at least a week. These may be found in my typoridden class notes (a copy of which is linked on the far left of this page). These notes describe, in great detail, with plenty of inclass and outofclass exercises, all about the binomial and normal distributions (where the parameters are assumed to be known; i.e., no estimation: this is pure probability). We spent a great deal on the problems of chapter six, which are like those below, but which are much harder.
At the end of the exam, I include a list of “Potentially useful formulas”, such as the binomial, the normal approximation to the binomial, and each necessary combinatoric equation. In other words, a cheat sheet. The students did not have to memorize any equations.
I am embarrassed to show this because I find it is shockingly easy. I made it so to bring the grade of the class up from its onaverage failing (yes, failing) level. I do this because I am just a visitor and I do not want to cause a great disruption in the usual procedures.
Please, no sea lawyers arguing about ambiguity in the questions: the students (who came to class) have already had dozens of problems just like these and know what is expected of them. Like I said before, grades improved. The median score for this test was 76; the average was 70.
 Each floor in one of Behemoth University’s multistory dormitory houses 200 students. On Thursday nights, instead of remembering they are in college, and thus staying at home to study, history shows that about 70% of students “go out.”
 What is the probability that the third floor empties out on Thursday night?
 If the dormitory has 10 floors, what is the probability the entire building empties out?
 In order for a professor to have a completely successful Statistics 101 class, each student who registers should pass the course. 35 students are registered for this class. The probability that any individual student passes is 90%. What is the probability of not having a completely successful Statistics 101 class?

 Sketch the normal distribution x ~ N(25.0,12.5). Take your time and CLEARLY label your axes. Label all the usual points with their actual numbers, not their letters.^{2}
 For this same normal, what is the value of w such that Pr(x>wE_n)=0.025?
 What is the probability that Pr(x<25.0E_n)?
 The FCC designates AM radio stations with either three or four letters, with the restriction that the first letter must be a K (for stations West of the Mississippi) or W (East of the Mississippi). Examples are WOR, KSLT and so on. How many unique AM radio stations are possible?
 The amount of money that people spend on Christmas is of obvious interest to retailers. SMart’s topoftheline shotgun is the 12gauge doublebarreled Remington, retailing for $109.99. During the Christmas rush, each SMart branch sees about 5,000 paying customers. The chance any paying customer buys the Remington is 3%.
 What is the approximate probability any SMart branch sells more than 150 shotguns? Explain your answer for credit.
 We obviously do not know how much money an SMart store will make on selling Remingtons. But we can express our uncertainty in this number. Do so; and sketch a picture. Hint: If SMart sold just one Remington across all its stores, how much money would they make? If they sold just two? three? more?
The Answers are:

 0.7^{200}
 0.7^{2000}
 1 – 0.90^{35}

 The central parameter is 25, the two usual points are plus or minus 2 times the spread parameter^{3}, i.e. 50 and 0. These points demarcate the approximate interval in which we expect 95% of the observables to lie. This sounds much harder than it is.
 0
 50% or 0.5
 2 x 26^{2} x 27 (or any variation of this).

 Use the normal approximation to the binomial formula, where the central parameter is found to be 5000 x 0.03 = 150; thus the probability of selling more than 150 guns is 50%.
 Sketch the appropriate normal approximation to the binomial from the first part and multiply all the usual points (like the 150) by $109.99. The actual calculation was not required: just writing “150 x $109.99” was sufficient.
—————————————————————
^{1}The school’s name, which was part of the exam, has been changed; so has the name of the course.
^{2}I.e., the “m” and “s” parameters of the normal. We had done dozens of these in and out of class by this point. Can you guess how many students wrote the letters and not numbers?
^{3}2 is close enough to 1.96 for anybody, especially in this class.
Forgive me for using you as a free educator…
Why isn’t this the answer to 4?
Possible two letter codes: 26**2
Possible three letter codes: 26**3
Possible two or three letter codes: (26**2 + 26**3)
Possible prefixes: 2
Every possible prefix combined with every possible code: 2*(26**2 + 26**3)
Onus,
Factor a 26^{2} from your final equation (inside the parentheses) and enlightenment shall be yours.
Yours is an approved variant.
Of course. Dim of me.
Thanks.
Shop smart! Shop Smart!
(I have nothing of value to add, sorry. It’s been more than 20 years since I did any statistics courses, and my work since then has not required that I use that particular toolset, with the result that it has almost completely rusted. I do intend to reeducate myself, somewhere between learning Latin, Arabic, and Flash programming. I expect to be back up to speed by about 2135 or so.)
If the answers were supplied with the test (as “word for word” implies) wouldn’t the answer to (2) be much closer to zero? How much closer is likely a function of the school’s dedication to filling stadium stands and the simple fact that, no matter how hard you try, there will be someone who just doesn’t Get It. Perhaps that’s why P(Passing)=90% vs 100%?
BTW: the answer to (3a) failed to CLEARLY label the axes. Was this the tricky part of the test?
Where I went to school, grading on the curve was common. That someone would fail was guaranteed.
Matt:
If it is not too much work, can you show a distribution of student answers and the nature of the errors made?
I’m stealing your questions for my quiz bank–my kids need the practice.
Regarding #1…
Now is NOT the time to use the binomial distribution. I estimate 0.7^2000 to be 10^300 (not using my calulator!). A blink of the eye in the history of the universe! Less! I say that the probability that the dorm is empty on a Thursday night is significantly greater than that.
What is the probabilty that the fire alarm goes of on a Thursday night?
Regarding 5b, would it be satisfatory to say that the standard devition of shotgun revenue is 109.99(0.03*0.97*150)^.5, and leave off the scetch?
I suppose this one–from today’s exam–would make your students’ heads explode (as it did for a few of mine):
MAKING A SAFE TURN: Cars are passing by your intersection at the rate of 10 per minute. You need an interval of at least 9 seconds before you can safely make a left turn into traffic. What’s the probability that the next interval between cars is at least 9 seconds long?
I don’t see the link to the notes, I hate to be a bother, but I really want to read them.
Mike,
Take anything you like. Yours is pretty good, too!
Bernie,
The errors were almost all of the type which indicated the students did not study. The formulas were there, but they didn’t know how to apply them. People would write various numbers down, but in no coherent order, hoping something would stick. Those that did well, did very well. Those that did poor, did very poorly.
I suspect that the bimodal distribution came as no surprise to you. It also supports the notion that the problem is not the test but the level of student effort – for if it was a matter of raw talent then I would expect a more normal type of distribution. Did it support any of your own hypotheses?
Alternative answers from the UEA creative writing course.
1a The third floor never empties on a Thursday night because thatâ€™s Chinese food night, unless 1b)
1b The probability the entire building empties out is proportionate to the number of days since the last time someone set off the fire alarm. 100% probability achieved after 8.6 days.
2 The probability of not having a completely successful Statistics 101 class. This question requires further information on price of alcohol in the student bar.
3 Sorry I forgot my pencil
4 Due to the derivative nature of entertainment media these days, there are no unique AM radio stations.
5a 50% in moose season, 78% in chocolate moose season.
5b My ruler broke.
‘The errors were almost all of the type which indicated the students did not study. The formulas were there, but they didnâ€™t know how to apply them. People would write various numbers down, but in no coherent order, hoping something would stick. Those that did well, did very well. Those that did poor, did very poorly.’
Yes, this mirrors my experience exactly (although I think my absolute fail rate is lower because I have fewer students who don’t study at all).
‘It also supports the notion that the problem is not the test but the level of student effort â€“ for if it was a matter of raw talent then I would expect a more normal type of distribution.’
There is really no reason to expect a ‘more normal type of distribution’ especially AFTER instruction. IMO, one of the largest mistaken ideas about assessment is that marks/grades should follow some approximately normal distribution.
Bernie,
Yes, that about half the students should not—at least not yet—be in college. Don’t forget that this test was exceedingly and, as I said, embarrassingly easy. I have “dumbed down” a regular test which should have seen every student get 100%, or close to it. I say this based on the lectures that I gave, the homework I assigned, and the problems we covered. Nobody received a 100; some did in the 90s.
Keith,
Right. It is impossible for grades to follow a normal distribution, but in a class we might expect a more symmetric distribution.
Keith & Matt:
My apologies, I guess I meant symmetric. Presumably most distributions of grades would approximate a gaussian with positive skew? However, I still think the bimodality is indicative of “effort”. Controlling for attendance and completion of homework assignments (which I would include as part of “effort”) should be revealing.
Matt:
How did the students react to the results of the tests?