Stat 330 (Spring 2015): Homework 6
Due: February 28, 2015
Show all of your work, and please staple your assignment if you use more than one sheet. Write your name,
the course number and the section on every sheet. Problems marked with * will be graded and one additional
randomly chosen problem will be graded. Must show work for full credit.
1. *
(a) What are the two properties of a probability density function f (x)?
Answer: See Slide Set #12 page 3:
(i) fX (x) ≥ 0 for all x,
R∞
(ii) −∞ f (x)dx = 1.
(b) Which of the following are valid probability density functions? Explain why or why not, a yes or
no is not a sufficient answer.
(
x2 + 2x, for − 1 ≤ x ≤ 1
f (x) =
0,
otherwise.
(
2x, for 0 ≤ x ≤ 1
g(x) =
0,
otherwise.
Answer:
We need to check if the functions satisfy the two criteria in (a).
Because f (−1) = −1 < −0, f (x) fails to satisfy property (i). Therefore, f (x) is not a valid p.d.f.
On the other hand, g(x) ≥ 0 for all x ∈ (−∞, ∞). Also
Z ∞
Z 1
1
g(x)dx =
2xdx = x2 0 = 1.
−∞
0
Since g(x) satisfies the two properties, g(x) is a valid p.d.f. By a result stated in class, there is a
r.v. with p.d.f. g(x).
(c) Let X be a random variable with probability density function h(x):
(
C(4x − 2x2 ), for 0 < x < 2
h(x) =
0,
otherwise
i. What is the value of C?
Answer:
Z 2
C(4x − 2x2 )dx
0
2 2x3 2
8C set
= C(2x2 −
= 1 gives C = 3/8
)=
3 0
3
0
ii. Determine the values of E[X] and V ar(X).
Answer:
Z ∞
Z
E[X] =
xh(x)dx =
−∞
∞
E[X 2 ]
Z
=
0
x2 h(x)dx =
−∞
V ar(X)
2
Z
0
3x(4x − 2x2 )
=1
8
2
3x2 (4x − 2x2 )
= 6/5
8
= E[X 2 ] − (E[X])2 = 6/5 − 1 = 1/5
iii. P ( 12 < X < 23 ) =?
Answer:
1
3
P( < X < )
2
2
=
=
1
3
2
3(4x − 2x2 )
dx
1
8
2
3x2 32
x3 32
1 − 1 ) = 11/16
4 2
4 2
Z
Stat 330 (Spring 2015): Homework 6
Due: February 28, 2015
2. A web page is accessed at an average of 10 times an hour. Assume that waiting time until the next hit
has an exponential distribution and that the hits are independent of each other.
(a.) Determine the rate parameter λ of the distribution of the time until the first hit?
(b.) What is the expected time between hits?
(c.) What is the distribution of the time until the second hit? (Give the name of the distribution and
the value(s) of parameter(s).)
(d.) What is the probability that the next hit is within 10 minutes?
(e.) Describe the distribution of the total waiting time for 10 hits? (Give the name of the distribution
and the value(s) of parameter(s).)
(f.) What is the expected total waiting time for 10 hits on the web page?
(g.) What is the probability that there will be less than 10 hits in the first hour?
Answer:
Let X be the time until the next hit. It is the same as the time between hits.
(a) Let X be the time until the next hit. By the the description, the rate parameter (number of hits per
hour ) λ = 10. Alternatively, the expected waiting time between hits, E[X] = 1/10 = 1/λ giving
λ = 10.
(b) Since X ∼ Exp(10), we have E[X] = 1/λ = 1/10 = .1 (hours)
(c) The time until the second hit is Y = X1 + X2 where X1 ∼ Exp(10) and X2 ∼ Exp(10) and X1 and
X2 are independent. Thus Y ∼ Gamma(2, 10) (or Erlang(2, 10)).
(d) Need P (X ≤ 10/60) where X ∼ Exp(10) Using the cdf for the exponential distribution Exp10 (.16667) =
1 − e−.16667×10 = 1 − .18888 = .8111
P10
(e) The waiting time for 10 hits is W = i=1 Xi ; so W ∼ Gamma(10, 10) (or Erlang(10, 10)).
(f) We need E[W ] which is given by k/λ = 10/10 = 1 hour, as one might expect!
(g) Let N be the number of hits in the first hour and assume N ∼ P oi(10 × 1). The we need P (N <
10) = P (N ≤ 9) = 0.45793.
3. Suppose that the lifetimes of batteries used in a tablet computer follow independent Exponential distributions with mean 1.5 years. If a battery fails, it is immediately replaced by an identical battery. You
buy a new tablet today. Find the probability that it will not need the third battery replacement within
the next 45 months.
Answer: The lifetime distribution of each battery is Exp(λ). Since E(Y ) = λ1 = 1.5 years ⇒ λ = 23
Let T represent time to failure of two replacement batteries. Sum of two sequential exponential distributions, T , has a Gamma(2, 32 ) distribution.
Thus, we need to compute:
P (T >
2 45
45
) = P (X < 2) where X ∼ P o( ×
= 2.5) using the Gamma-Poisson formula
12
3 12
= P (X ≤ 1) = .287 from Poisson cdf Table A3 with λ = 2.5
4. * (Baron’s book): 4.8
Answer: Given X ∼ Gamma(2, 2). Thus:
P (X <
1
)
2
=
=
=
P (X ≤
1
)
2
1
where Y ∼ P o(2 · ) from Gamma-Poisson formula
2
1 − P (Y ≤ 1) = 1 − 0.736 = 0.264 from Poisson cdf Table A3 with λ = 1
P (Y ≥ 2)
2
Stat 330 (Spring 2015): Homework 6
Due: February 28, 2015
5. (Baron’s book): 4.9
Answer:
3