- What's the probability distribution for getting more than q questions
right, choosing each answer at random? I.e., let the random variable Q
denote the number of correct answers and calculate Pr(Q >
q) for q = 0, …, 79. Check your work on some
easy cases, e.g.:
- Pr(Q > 0) = 1
- Pr(Q > 79) = 1/480 ≈ 6.8 × 10-49
- and the median number of correct guesses is 1/4 * 80 = 20
Answer
55% of 80 questions = 44 questions answered correctly to pass.
Each question is a Bernouilli trial with success probability 25%. The total number right is therefore binomially distributed. The probability of getting more than a certain number right is given by the cumulative binomial distribution function.
In R,
pbinom()will compute that for us. Usinglower.tail = FALSEmeans we're measuring Pr(Q > q), i.e., strictly greater than. So 79 is the maximum reasonable for q.> pbinom(q = 0:79, size = 80, prob = 0.25, lower.tail = FALSE) [1] 1.000000e+00 1.000000e+00 1.000000e+00 9.999997e-01 9.999977e-01 [6] 9.999877e-01 9.999460e-01 9.997991e-01 9.993523e-01 9.981607e-01 [11] 9.953407e-01 9.893589e-01 9.778938e-01 9.579033e-01 9.260137e-01 [16] 8.792424e-01 8.159061e-01 7.364254e-01 6.436978e-01 5.428363e-01 [21] 4.402937e-01 3.426341e-01 2.553323e-01 1.819482e-01 1.238525e-01 [26] 8.047434e-02 4.988718e-02 2.949574e-02 1.662971e-02 8.939674e-03 [31] 4.581986e-03 2.239142e-03 1.043316e-03 4.635213e-04 1.963610e-04 [36] 7.931939e-05 3.055204e-05 1.122084e-05 3.929250e-06 1.311755e-06 [41] 4.174442e-07 1.266114e-07 3.659175e-08 1.007433e-08 2.641417e-09 [46] 6.593068e-10 1.565976e-10 3.537701e-11 7.597276e-12 1.549988e-12 [51] 3.002147e-13 5.516119e-14 9.606372e-15 1.584139e-15 2.471005e-16 [56] 3.641559e-17 5.063669e-18 6.633998e-19 8.175502e-20 9.460183e-21 [61] 1.025786e-21 1.039938e-22 9.832241e-24 8.644741e-25 7.045311e-26 [66] 5.302663e-27 3.670229e-28 2.324706e-29 1.339769e-30 6.978136e-32 [71] 3.258192e-33 1.350390e-34 4.907634e-36 1.539764e-37 4.086262e-39 [76] 8.919250e-41 1.537460e-42 1.962434e-44 1.648989e-46 6.842278e-49
Checks:
- Pr(Q > 0) is the first entry, which is 1.
- Pr(Q > 79) = Pr(Q = 80) = 1/480 ≈ 6.8 × 10-49 is the last entry.
- The median had better be 0.25 * 80 = 20; use
qbinom()to get the 50th percentile:> qbinom(p = 0.5, size = 80, prob = 0.25) [1] 20
Ok, we want Pr(Q > 43), which is the 44th element of the list above: guessing the answers gives a pass probability of 1.007433 × 10-08, or about 10 chances in a billion! If you administered this test to all 6 billion currently living humans and told them to guess, about 60 would pass. (It doesn't matter if they speak the language of the test; remember, they're guessing!)
- Comment on the reasonableness that one could achieve 55% correct at random.
(Hint: it was talk radio, and probably not NPR!)
Answer
Oh c'mon, do we even have to say anything? The probability of a successful pass from guessing is only 10 chances in a billion! Even Vegas gives better odds than that…
- How many times should you expect to have to take the test, if you answer each
question randomly? In more detail, what's the probability distribution for the
number of times you have to take the test? (Hint: you stop taking the
test after the first pass.)
Answer
This is a geometric distribution: you try k-1 times and fail, then succed on the kth attempt. (Think fertility treatment, where you stop at the first pregnancy!) The probability of success on any attempt is only 1.007433e-08. The probability we pass after k or fewer attempts is the cumulative geometric probability. The relevant R function is
pgeom(). Let's see what happens to a mathematically illiterate, but exceptionally stubborn student who's willing to go 50 times on the test:> pgeom(q = 1:50, prob = 1.007433e-08) [1] 2.014866e-08 3.022299e-08 4.029732e-08 5.037165e-08 6.044598e-08 [6] 7.052031e-08 8.059464e-08 9.066897e-08 1.007433e-07 1.108176e-07 [11] 1.208920e-07 1.309663e-07 1.410406e-07 1.511149e-07 1.611893e-07 [16] 1.712636e-07 1.813379e-07 1.914123e-07 2.014866e-07 2.115609e-07 [21] 2.216352e-07 2.317096e-07 2.417839e-07 2.518582e-07 2.619325e-07 [26] 2.720069e-07 2.820812e-07 2.921555e-07 3.022299e-07 3.123042e-07 [31] 3.223785e-07 3.324528e-07 3.425272e-07 3.526015e-07 3.626758e-07 [36] 3.727501e-07 3.828245e-07 3.928988e-07 4.029731e-07 4.130474e-07 [41] 4.231218e-07 4.331961e-07 4.432704e-07 4.533447e-07 4.634191e-07 [46] 4.734934e-07 4.835677e-07 4.936421e-07 5.037164e-07 5.137907e-07
Looks grim! How many times does he have to take it before he has a given chance of passing? That's the geometric quantile function
qgeom(); here we compute it for 0-90% chance of success, by 10% increments:> qgeom(p = seq(from = 0.0, to = 0.9, by = 0.1), prob = 1.007433e-08) [1] 0 10458314 22149716 35404333 50705666 68803302 90953018 [8] 119508969 159756321 228559624
So he needs to take it 68,803,302 times to get a 50% chance of passing. That's many, many attention spans of your average 18 year old.
- Comment on the reasonableness of the hypothesis that retaking the test is being
"too forgiving" of students who know no math.
Answer
68,803,302 times for a 50% pass probability? Are you high?! Easier to learn some math! (Which, of course, is kind of the point…)
- Extra credit: A rhesus monkey lives about 30 years in captivity (though
only 4 years
in the wild). Assume your rhesus monkey knows how to use a typewriter, and is
cooperative enough to do so. He sleeps 8hr/day but devotes his waking life to
taking the exam (in captivity, of course -- in the wild, monkeys are far too busy for
math tests). Using your result from (2a), how much time can he take for each
iteration of the exam?
Answer
Obviously I'm starting to yank your chain here, but...
Sleeping 8hr/day means we get about 2/3 of his existence for exam taking:
30 yr * 3.16e7 sec/yr * 2/3 = 6.31e8 sec
So he has that much "monkey time" to take 68 million exams:
6.31e8 sec / 68,803,302 exams = 9.2 sec/exam
We can therefore be reasonably confident that a rhesus monkey would not pass the California math exit exam in his lifetime. Actually, being a monkey, the results won't improve much even if you give him more time.
(It's more fun to contemplate what a rhesus monkey would actually do with a keyboard… maybe Web programming?)
- If you took that much time for the exam, could you pass it?
Answer
9 seconds for 80 questions? Not a chance. I might be able to fill out the form that fast, given sufficient practice… like 68 million times worth of practice. Given sufficiently little time on the exam, you can make monkeys of us all.
- This procedure results in an elderly monkey who passes the exam and then
instantly dies. Comment on the usefulness of mathematics skills to dead monkeys.
Answer
What? Are you expecting a serious answer? Look, we left reality far behind the very instant we took the talk radio host seriously. Absurd questions like this is where that kind of thinking leads!
- (Semi-serious) Suppose you know just barely enough math to answer k
questions correctly (where k < 44 is not enough to pass). Then
you guess randomly at the rest. What's the minimum value of k such
that you pass with 50% probability?
Answer
This is a surprisingly reasonable question, and it's a real pity that the talk radio host didn't ask it.
We get k questions right because we actually know some math, then guess on the remaining (80 - k) questions and get some fraction of those right by chance. If the total right is exceeds 44, we pass. If we want to pass with 50% probability, then the median number we guess correctly should just put us over the threshold to pass:
k + [median # guesed right out of (80 - k)] = 44
So what's that median expression? It's the 50% quantile of the binomial distribution with probability of success 25% and number of trials 80 - k.
Again, this is a 1-liner in R (no, I did not grow up programming in APL!) using our old friend
qbinom():> 0:80 + qbinom(p = 0.5, size = 80 - 0:80, prob = 0.25) [1] 20 21 21 22 23 24 24 25 26 27 27 28 29 30 30 31 32 33 33 34 35 36 36 37 38 [26] 39 39 40 41 42 42 43 44 45 45 46 47 48 48 49 50 51 51 52 53 54 54 55 56 57 [51] 57 58 59 60 60 61 62 63 63 64 65 66 66 67 68 69 69 70 71 72 72 73 74 75 75 [76] 76 77 78 78 79 80
(The first term is a vector of values for k, the number you get right by actually knowing the answer. The second term is a vector of medians of the number you get right by guessing on the remainder. Add 'em up, look for the magic number 44.)
So it looks like you can pass by knowing k=32 questions, and guessing on the rest!
Here's a simpler argument: the median number guessed right is going to be about 1/4 of the number guessed. Thus our pass condition is:
k + 1/4 * (80 - k) ≥ 44 3/4 k + 20 ≥ 44 3/4 k ≥ 24 k ≥ 24 * 4/3 k ≥ 32
Satisfying that we get the same answer of k=32 questions, both ways.
Now, people say the test measures math at the 8th grade level (to graduate 12th grade, but that's another issue). If we assume grade-level performance is proportional to number of questions right (8th grade = 44 correct), then 32 questions is 32/44 * 8 = 5.8th grade. I.e., you could pass with a 6th grade knowledge of math and random guessing on the rest. (How someone with a 6th grade math education could figure that out, we leave as an exercise to the reader... :-)