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Math 313
Exam 1
October 26, 2011
Name:
This is an untimed take-home exam. It is due Wednesday, October 26 at 2:30 pm. You may consult
any book, notes, computer program or resource that you want except you may not communicate with
another human being about this exam. Doing so is a violation of the Wartburg College Honor Code.
Answer 7 of the 11 questions on separate sheets of paper. Indicate clearly on each sheet which
question you are answering. Show all of your work, including any computer calculations.
1. For events A and B , suppose that A ⊂ B and P (A) > 0. Show that P (B|A) = 1 and explain why
this makes sense in terms of the original events.
Solution: P (B|A) = P (B ∩ A)/P (A) = P (A)/P (A) = 1 since B ∩ A = A. If we know that
event A happened, then an outcome that is also in B occured.
2. For two continuous random variables Y1 and Y2 with probability density functions f1 and f2 respectively, the random variable Y with probability density function
f (y) = α f1 (y) + (1 − α) f2 (y)
for some constant 0 ≤ α ≤ 1 is said to have a mixture distribution.
(a) Prove that f (y) is a probability density function.
Solution: There are two properties that we have to show. Because f1 and f2 are density
functions, f1 (y) ≥ 0 and f2 (y) ≥ 0 so f (y) = α f1 (y) + (1 − α) f2 (y) ≥ 0 because α ≥ 0
and 1 − α ≥ 0. For the integral
∫∞
∫∞
α f1 (y) + (1 − α) f2 (y) dy
f (y) dy =
−∞
−∞
∫∞
∫∞
=α
f1 (y) dy + (1 − α)
f2 (y) dy
−∞
−∞
= α + (1 − α) = 1.
(b) If E[Yi ] = µi and V (Yi ) = σi for i = 1, 2, find E[Y ] and V (Y ).
Math 313
Exam 1
Solution:
∫
∞
E[Y ] =
=
∫−∞
∞
October 26, 2011
y f (y) dy
y[α f1 (y) + (1 − α) f2 (y)] dy
∫∞
y f1 (y) dy + (1 − α)
y f2 (y) dy
−∞
∫∞
=α
−∞
−∞
= αµ1 + (1 − α)µ2 .
∫
∞
V (Y ) = E[Y ] − E[Y ] =
y 2 f (y) dy − (αµ1 + (1 − α)µ2 )2
−∞
∫∞
∫∞
2
=α
y f1 (y) dy + (1 − α)
y 2 f2 (y) dy − (αµ1 + (1 − α)µ2 )2
2
2
−∞
−∞
= ασ12 + (1 − α)σ22 − (αµ1 + (1 − α)µ2 )2
3. If a random variable X is normally distributed with mean µ and standard deviation σ , find the
X −µ
moment generating function of the random variable Z = σ . What distribution does Z have?
Solution: By Problem 4.137, if a random variable Y has moment generating function m(t )
then U = aY + b has moment generating
e b t m(at ). X has moment generating
function
function m(t ) = eµt +σ
2 t 2 /2
µ
so Z = σ1 X + − σ
mZ (t ) = e
=e
µ
has moment generating function
− σ t µ(t /σ)+σ 2 (t /σ)2 /2
e
t 2 /2
This is the moment generating function for a normal random variable with µ = 0 and σ 2 = 1,
so Z has a standard normal distribution.
4. The cost to repair a machine has a beta dstribution with α = 1 and β = 3, where Y is measured
in hundreds of dollars. How much money should be budgeted for repairs so that the actual cost
exceeds the budgeted amount with a probability not more than 10%?
Solution: We need to find a value x so∫ that P (Y ≤ x) = .9. The density function of Y is
x
f (y) = 3(1 − y)2 so we need to solve 0 3(1 − y)2 dy = .9 for x . Performing the integral,
Math 313
Exam 1
October 26, 2011
x
3(1 − y) dy = −(1 − y) = −(1 − x)3 + 1 = .9 and solving for x , (1 − x)3 = .1 ⇒ x =
0
0
p
3
1 − .1 = 0.5358. So we should budget $53.58 for repairs.
∫x
2
3
5. Two teams, let's call them the Rangers and the Cardinals, play a series of games until one team has
won four games. (This is often called a best-of-seven series.) If the probability that the Rangers win a
game is p , compute the probability that the series will take n games to complete for n = 4, 5, 6, 7.
Solution: For a four game series, the winning team has to win all four times so P (4) = p 4 +
(1 − p)4 where the first term is for the Rangers winning and the second for the Cardinals
winning. For 5 games, there are 41 ways to arrange the losing teams one loss since the winning
team always wins the last game. So, P (5) = 41 p 4 (1 − p) + 41 p(1 − p)4 . Similarly, P (6) =
5 4
5 2
6 4
6 3
2
4
3
p
(1
−
p)
+
p
(1
−
p)
and
P
(7)
=
p
(1
−
p)
+
p (1 − p)4 .
2
2
3
3
6. Consider the problem of trying to predict frost on a given night in October in Waverly, IA. Historically, there is frost (defined as air temperatures below 32◦ C) on 15% of nights in October. On
nights when frost did occur, there was also frost on the previous night 65% of the time. On nights
without frost, there was frost on the previous night only 12% of the time. Using these historical
probabilities, compute the probability of frost, given that there was frost on the previous night.
Solution: Let F be the event of frost and N the event of not having frost. The historical data
shows P (F ) = 0.15 and P (N ) = 0.85. Let F p be the event of frost on the previous night so
that P (F p |F ) = 0.65 and P (F p |N ) = 0.12. Using Bayes' Rule,
P (F |F p ) =
=
P (F p |F )P (F )
P (F p |F )P (F ) + P (F p |N )P (N )
(0.65)(0.15)
(0.65)(0.15) + (0.12)(0.85)
= 0.4887
7. A shipment of 20 transistors contains 3 that are defective. If 6 transistors from the shipment are
randomly selected to be used in a product, what is the probability that none of them are defective?
How many transistors could we use from the shipment so that P (none defective) < 0.2?
Math 313
Exam 1
October 26, 2011
Solution: The number of defective transistors out of the six selected is a hypergeometric
random variable with N = 20, n = 6, r = 3. Using the probability distribution,
p(0) =
r N −r y
n−y
N
n
=
3 17
0
6
20
6
= 0.3193
By trial and error, using n = 8 gives p(0) = 0.1930 < 0.2.
8. To qualify for a police academy, applicants are given a test of physical fitness. The scores are
normally distributed with a mean of 64 and a standard deviation of 9.
(a) Find the probability of scoring a 70 or above.
Solution: P (Y ≥ 70) = 0.2525 by WolframAlpha.
(b) If only the top 20% of applicants are selected, find the cutoff score.
Solution: We need to find z0 such that P (Z > z0 ) = 0.2. Looking at a table (or some
such) we find z0 = 0.842. Un-standardizing that value, (y0 −64)/9 = 0.842 ⇒ y0 = 71.6.
9. Dr. Birgen claims that 80% of students cry when taking his tests. To test his claim, we select 12
students at random and find that 8 of them cry when taking a Birgen test.
(a) If Dr. Birgen's claim is correct, what is the probability that 8 out of 12 randomly selected
students cry when taking his tests?
Solution: The number of students who cry is binomial
with n = 12 and p = 0.8. Using
12
8
the binomial probability distribution, p(8) = 8 (0.8) (0.2)4 = 0.1329.
(b) If Dr. Birgen's claim is correct, what is the probability that 8 or less out of 12 randomly
selected students cry when taking his tests?
Solution: Either by brute force or using the cumulative distribution function, P (Y ≤
8) = p(0) + · · · + p(8) = .2054.
(c) Based on your above probabilities, do you think that our experiment refutes Dr. Birgen's
claim?
Solution: No. There is at least a one in five chance that a sample of size twelve would
have eight or fewer students who cry. Thus our experiment is no reason to believe that
Dr. Birgen is not telling the truth.
Math 313
Exam 1
October 26, 2011
10. 80% of underclassmen thought Outfly came too late this year, while only 60% of upperclassmen
thought so. Enrollment data from this fall shows that 52.8% of students at Wartburg are underclassmen. What is the probability that a randomly selected Wartburg student thinks that Outfly
came too late?
Solution: Let U be the event that the randomly selected student is an upperclassman and
L be the event that that student is a lowerclassman. Then P (L) = 0.528 and P (U ) = 1 −
0.528 = 0.472. The other information is conditional, so if O is the event that the randomly
selected student thinks Outfly came too late then P (O|U ) = 0.6 and P (O|L) = 0.8. Using
the law of total probability P (O) = P (O ∩ U )+ P (O ∩ L) = P (O|U )P (U )+ P (O|L)P (L) =
(0.6)(0.472) + (0.8)(0.528) = 0.7056.
11. (a) A fair die is tossed six times, and the number on the uppermost face is recorded each time.
What is the probability that the numbers recorded are 1, 2, 3, 4, 5, and 6 in any order?
Solution: Using the sample point method, there are 66 possible rolls of a fair die six
times. Of these, there are 6 · 5 . . . 1 = 6! ways to arrange the numbers 1 through 6 among
the six rolls. So P = 66!6 = 5/324 ≈ 0.0154.
(b) Suppose now that the die has been altered so that the faces are 1, 2, 3, 4, 5, and 5. If the die
is tossed five times, what is the probability that the numbers recorded are 1, 2, 3, 4, and 5 in
any order?
Solution: Assuming all of the faces were still distinct (perhaps one of the fives was
painted yellow), there are 65 ways to roll the die five times. There are 5! ways to get
1, 2, 3, 4 and the first five and 5! ways to get 1, 2, 3, 4, and the second five. So P =
2 · 5!/65 = 5/162 ≈ 0.0309 or twice the previous answer.