STAT 331 Fall ’05: MIDTERM SOLUTIONS
1. Short Questions
a.
E[V] = Z4
0
4
3πr3f(r)dr
=Z4
0
π
6r4dr
=π
6r5
54
0
=512π
15 ≈107.
Some people computed E[V] = 4
3πE[R3], but keep in mind it is generally not true that
E[g(X)] = g(E[X]) unless gis linear.
b. Let Xbe the total number of errors in the sequence of ninformation bits, and let
Xi=1 if the ith information bit is decoded incorrectly
0 otherwise
so that X=Pn
i=1 Xi. Denote ˜p=P(Xi= 1), so that Xi∼Ber(˜p) and X∼bin(n, ˜p).
Then
P(an error) = 1 −P(no error)
= 1 −P(X= 0)
= 1 −(1 −˜p)n,
so it remains to find ˜p. Note that Xi= 1 if and only if 3 or more of the five repeated
bits are corrupted. Thus
˜p=P(3,4,or 5 flipped bits out of 5)
=5
3p3(1 −p)2+5
4p4(1 −p) + 5
5p5
= 0.0579
where I have plugged in p= 0.2. In conclusion, P(an error) = 1 −(0.94)n.
c. Let Xbe the point where the stick is broken. Then X∼unif[0,1]. The probability of
interest is
P(X∈[0,1/3] ∪[2/3,1]) = F(1/3) −F(0) + F(1) −F(2/3)
= 1/3 + 1/3
= 2/3
where F(x) = xis the CDF of X.Note: Many people looked at the random variable
Y= max(X, 1−X), the length of the longest piece, and computed P(Y≥2/3). This
is fine in principle. However, you cannot simply say that Y∼unif[1/2,1] without any
justifcation. While true, this must shown by, for example, finding the CDF of Y and
identifying it as that of a uniform.
