Download MATH 3160, SPRING 2013 HOMEWORK #8

MATH 3160, SPRING 2013
HOMEWORK #8—SOLUTIONS
JOHANNA FRANKLIN
This assignment will be due on Wednesday, April 10 at the beginning of class. Remember to
show your reasoning and name the classmates you worked with. Answers without work shown will
receive almost minimal credit.
When calculating probabilities for normal random variables, use the table on p. 201 of your
textbook.
(1) (Chapter 5, #17) A man aiming at a target receives 10 points if his shot is within 1 inch of
the target, 5 points if it is between 1 and 3 inches of the target, and 3 points if it is between
3 and 5 inches of the target. Find the expected number of points scored if the distance from
the shot to the target is uniformly distributed between 0 and 10 inches.
Solution. The distribution function for the distance X from the shot to the target is
(
1
0 < x < 10
f (x) = 10
0 otherwise
R1 1
1
2
2
Therefore, P (0 < X < 1) = 0 10
= 10
dx, P (1 < X < 3) = 10
, and P (3 < X < 5) = 10
.
Since these probabilities correspond to the only nonzero numbers of points P the man can
receive, the expected number of points is
2
2
6
1
+5·
+3·
=2 .
E[P ] = 10 ·
10
10
10
10
(2) Suppose that X is a normally distributed random variable with µ = 10 and σ = 3.
(a) Find the probability that X is between 7 and 16.
Solution. Let Z be a standard normal random variable. Then Z = X−µ
σ =
7 − 10
X − 10
16 − 10
P (7 < X < 16) = P
<
<
= P (−1 < Z < 2)
3
3
3
= Φ(2) − (1 − Φ(1)) = .9772 − (1 − .8413) = .8185.
X−10
3 ,
(b) Find the probability that X is greater than 12.
Solution. We convert our question about X to a question about Z as above:
12 − 10
X − 10
P (12 < X) = P
<
= P (.67 < Z) = 1 − Φ(.67) = 1 − .7486 = .2514.
3
3
1
and
2
FRANKLIN
(3) A crack in the world has opened due to a mad scientist’s attempt to harness the energy from
the earth’s magma layer.1 To save the earth’s crust, a Pico-Omega device must be built.2
The rotors for this device have normally distributed masses with parameters µ = 5 kg and
σ = .04 kg. If only widgets with masses within .05 kg of µ are acceptable, what fraction of
the rotors produced will be acceptable for the Pico-Omega device?
Solution. Let X be a random variable representing the mass of a rotor. We want P (4.95 <
X < 5.05). We convert this problem to a question about the standard normal random
variable Z:
4.95 − 5
5.05 − 5
P (4.95 < X < 5.05) = P
= P (−1.25 < Z < 1.25)
<Z<
.04
.04
= Φ(1.25) − (1 − Φ(1.25)) = .8944 − (1 − .8944) = .7888
(4) Suppose that we roll 2 dice 180 times. Let E be the event that we roll two fives no more
than once.
(a) Find the exact probability of E.
Solution. The probability of rolling two fives in a particular roll is
bility that we roll two fives no more than once in 180 rolls is
180 35
180
180 1 35 179
p=
+
≈ .0386.
0
36
1 36 36
1
36 ,
so the proba-
(b) Approximate P (E) using the normal distribution.
Solution. We want the number of successes to be 0 or 1, so we want P (0 ≤ S180 ≤ 1).
Since the binomial is integer-valued, we apply the continuity correction and calculate
P (−.5 ≤ S180 ≤ 1.5) instead. We calculate
that the expected value is µ = 180 · p = 5
p
and the standard deviation is σ = 180p(1 − p) ≈ 2.205. Now, as always, we convert
this question to a question about the standard normal random variable Z:
1.5 − 5
−.5 − 5
<Z<
= P (−2.49 < Z < −1.59)
P (−.5 < S180 < 1.5) = P
2.205
2.205
= (1 − Φ(1.59)) − (1 − Φ(2.49)) = (1 − .9441) − (1 − .9936) = .0495
(c) Approximate P (E) using the Poisson distribution.
Solution. We use λ = np = 5 (note that we calculated this already in (b)!). Now we
see that
50
51
P (E) ≈ e−5 + e−5 ≈ .0404.
0!
1!
(5) Suppose that the time required to replace a car’s windshield can be represented by an
exponentially distributed random variable with parameter λ = 12 .
(a) What is the probability that it will take at least 3 hours to replace a windshield?
1“Crack in the World,” 1965
2Clearly, this is not the actual device. I’m just trying to avoid spoilers.
MATH 3160, SPRING 2013
HOMEWORK #8—SOLUTIONS
3
Solution. Let X be the exponentially distributed random variable representing the
time required to replace a windshield. We recall the pdf for such a variable and see
that
Z 3
h
i
x 3
3
3
1 −x
P (3 < X) = 1 − P (0 < X < 3) = 1 −
e 2 dx = 1 − −e− 2 = 1 + e− 2 − 1 = e− 2 ≈ .2231.
0
0 2
(b) What is the probability that it will take at least 5 hours to replace a windshield given
that it hasn’t been finished after 2 hours?
Solution. There are two ways to do this. The longer one is to calculate P (X > 5|X >
2). The shorter one is to remember that the exponential distribution is memoryless
and to observe that P (X > t + 3|X > t) = P (X > 3), so the answer is the same as the
answer to (a).
Suggested problems: Chapter 5 Problems: 10, 13, 15, 16, 18, 21, 22, 32, 33, 34;
Chapter 5 Theoretical Exercises: 9, 10, 13