Stat 61 - Solutions #3 (10/1/07)
1. Roll two standard dice. Let X = larger of the two numbers showing.
a. Construct a table showing both the probability mass function (pmf) pX and the cumulative density function (cdf) FX.
x / 1 / 2 / 3 / 4 / 5 / 6p(x) / 1/36 / 3/36 / 5/36 / 7/36 / 9/39 / 11/36
F(x) / 1/36 / 4/36 / 9/36 / 16/36 / 25/36 / 1
b. What is FX(7) ?(Note: “pmf” = what the text calls a “pdf” in the case of a discrete r.v.)
1
2. Let Y be a binomial random variable with parameters n = 6 and p = 1/2.
a. What is P ( Y = 3 ) ?
p(3) = (6-choose-3) (1/2)3 (1 – ½)3 = 20/64
b. What is P ( Y > 3 ) ? [Can you do part b with zero extra work?]
It would be p(4) + p(5) + p(6). But it’s easier to use symmetry:
P (Y > 3) = P(Y < 3) so these two events have to share equally
the probability that is left over from part (a):
P ( Y > 3 ) = (1 – 20/64) / 2 = 22/64.
3. Z is a random variable with the pmf
pZ (k) = C (1/10)k if k = 1, 2, 3, 4, …
(or zero if n has any other value)
(in particular, pZ(0) = 0)
…where C is a particular real number. What is C ?
If we erased the “C” then the sum of the series would be
1/10 + 1/100 + … = 1/9.
Since the sum of the probabilities must be 1, C must be 9.
( This isn’t strictly a geometric distribution as we have defined it, since the values start at 1 instead of 0. We could say “Z-1 has a geometric distribution with mean µ = 1/9” or “Z is a truncated geometric distribution, truncated below1.” We say µ = 1/9 because the “1/10” factor in the sequence is supposed to be ( µ / (µ+1) ), which forces µ = 1/9. )
4. A bar of Uranium contains n atoms. Each of them has (independent) probability p of decaying during a one-minute test. Let W be the random variable whose value is the number of atoms that decay during the one-minute test.
These are just binomial random variables, so the answers are all just
(n choose 2) (p)2 (1-p)n-2. The last one is a little harder to evaluate.
What is P ( W = 2 ) if…
a. n = 2 and p = 0.5
1/4
b. n = 10 and p = 0.1
(45)(0.01)(0.9)8 = 0.1937
c. n = 100 and p = 0.01
(495)(0.0001)(0.99)98 = 0.1849
d. n = 1023 and p = 10-23 (approximate if you must)
About 0.183939721.
If your calculator won’t do this directly (like Excel doesn’t) then you have three choices. Recognize this as some sort of limiting process and do the calculation with the largest n you can handle, and p = 1/n. Excel 2007 accepts n = 109, giving the probability 0.183939726, which turns out to be pretty good. Or, you could take the limit of (n - choose - 2) (1/n)2(1-1/n)n-2 as n approaches infinity; this is a moderately-hard L’Hopital problem and gives the answer 1/(2e) = 0.183939721. Or, you can recognize that the limit of binomials in this situation (that is, all with the same value of np) is a Poisson distribution with mean µ = np, and that the answer is therefore given by the formula for p(k) in the next problem with k = 2 and µ = np = 1.
5. A discrete random variable has a Poisson distribution with parameter and is called a Poisson random variable with parameter if its pmf is given by
p(k) = e- ( k / k! ) if k = 0, 1, 2, 3, …
or 0 for any other value of k.
Show that (as we would hope)
p(0) + p(1) + p(2) + … = 1.
(This is really a Math 25 problem. Think of a Taylor series.)
p(0) + p(1) + p(2) + …
= e-µ ( µ0/0!) + e-µ ( µ1/1!) + e-µ ( µ2/2!) + …
= e-µ [ µ0/0! + µ1/1! + µ2/2! + … ]
and now the series in the square brackets is just the Taylor series (about 0) for eµ, so the sum is
= e-µ [ eµ ] = 1.
6. (from text) A continuous random variable X has the following cdf:
FX (a) = 0 if a ≤ 0
a2 if 0 ≤ a ≤ 1
1 if a ≥ 1.
a. What is P ( 1/2 ≤ X ≤ 3/4 ) ?
F(3/4) – F(1/2) = (3/4)2-(1/2)2 = 5/16.
(Well, really you should use F(1/2 –); but since this particular F is a continuous function, F(1/2 –) is the same as F(1/2). )
b. What is the probability density function (pdf) for X ?
The pdf is just the derivative of FX.
So:
p(a) = 2a ( if 0 ≤ a ≤ 1 )
0 ( otherwise )
c. Express the answer to (a) as an integral involving the pdf.
P ( ½ ≤ X ≤ ¾ ) = integral from x=1/2 to 3/4 of (2x)dx = 5/16.
(end)