MTH 4451 Test #2 - Solutions Download

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MTH 4451 Test #2 - Solutions
Spring 2009
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers.
1. A large jar contains US coins. In this jar, there are 350 pennies ($0.01), 300 nickels
($0.05), 250 dimes ($0.10), and 300 quarters ($0.25) - each type of coin is evenly
distributed throughout the jar. I take one coin from the jar.
Let the random variable Y be the value of the coin that I take.
⎧
⎪
⎪
⎪
⎨
when
when
when
when
$0.01
$0.05
Then Y = ⎪
$0.10
⎪
⎪
⎩
$0.25
the
the
the
the
coin is
coin is
coin is
coin is
a
a
a
a
penny
nickel
dime
quarter
Furthermore, the probability function for Y is
⎧ 350
⎪
⎪
1200
⎪
⎨ 300
1200
250
P (Y = y) = ⎪
⎪
⎪
⎩ 1200
300
1200
= 0.29167
= 0.25
= 0.20833
= 0.25
when
when
when
when
y
y
y
y
= $0.01
= $0.05
= $0.10
= $0.25
(a) What is the expected value of the coin?
E (Y ) =
P
Y =y
y·P (Y = y) = ($0.01) (0.29167)+($0.05) (0.25)+($0.10) (0.20833)
+ ($0.25) (0.25) = $0.09875
i.e., E (Y ) = $0.09875
(b) If I perform this experiment a number of times, what is the variance?
V (Y ) = E (Y 2 ) − [E (Y )]2
We need to find the value of E (Y 2 )
E (Y 2 ) =
P
Y =y
y 2 ·P (Y = y) = ($0.01)2 (0.29167)+($0.05)2 (0.25)+($0.10)2 (0.20833)
+ ($0.25)2 (0.25) = $2 0.018362
i.e., E (Y 2 ) = $2 0.018362
Thus the variance is:
V (Y ) = E (Y 2 ) − [E (Y )]2 = $2 0.018362 − (0.09875)2 = $2 0.0086104
i.e., V (Y ) = $2 0.0086104
2. A group of 6 software packages available to solve a linear programming problem has
been ranked from 1 to 6 (best to worst). An engineering team, unaware of the rankings,
randomly selected and then purchased two of the packages. Let Y denote the number
of packages purchased (by the team) that are ranked 3, 4, 5, 6.
(a) Give the probability distribution for Y.
In this experiment, we have two identical trials, and each trial has exactly two
outcomes - Success (Package is ranked 3, 4, 5, 6) or Failure (Package is ranked 1,
2).
Because the sample size (n = 2) is large compared to the population size (N = 6) ,
the trials are NOT independent. This is a hypergeometric distribution
N = 6
n = 2
M = 4
P (Y = k) =
N −M
(M
(4)( 2 )
K )( n−k )
= k 62−k for k = 0, 1, 2
N
(n)
(2)
(b) Compute P (1) (i.e., compute P (Y = 1))
P (Y = 1) =
2
(41)(2−1
)
=
6
(2)
8
15
= 0.53333
3. A recent survey indicated that a record low number (41%) of adult American citizens
expressed “a great deal of confidence” in the US Supreme Court. If you conduct your
own random poll on the same issue,
(a) find the probability distribution for Y, the number people who must be polled in
order to find a person who does express “a great deal of confidence” in the US
Supreme Court.
Each time a person is polled, there is a 41% probability that they will express
“a great deal of confidence” in the US Supreme Court. So each time a person is
polled, it constitutes a trail in which there are two outcomes - Success (a person
DOES express a great deal of confidence) or Failure (a person does NOTexpress
a great deal of confidence).
Furthermore each trial is independent of all other trials. Since we are looking
for the number of trials necessary to acheive the first success, the experiment is
geometric.
p = 0.41
q = 0.59
Y = the number of the trial on which the first success is acheived.
2
P (Y = y) = q y−1 p = (0.59)y−1 (0.41)
i.e., P (Y = y) = (0.59)y−1 (0.41)
(b) Compute P (5) (i.e., compute P (Y = 5))
P (Y = 5) = (0.59)5−1 (0.41) = 0.049681
i.e., P (Y = 5) = 0.049681
4. A multiple choice examination has 15 questions, each with 5 possible answers, of which
only one answer is correct. Suppose that one of the students who takes the examination answers each of the questions with an independent, random guess. What is the
probability that he answers at least 10 and no more than 12 questions correctly?
Each question constitutes a trial which has exactly two outcomes - Success (Correct)
or Failure (incorrect).
The probability of getting a success is constant from one problem to the next and each
trial is independent of all others.
This is binomial.
Y = the number of successes (number of questions answered correctly)
n = 15
p = 0.2
q = 0.8
We want P [(Y = 10) ∪ (Y = 11) ∪ (Y = 12)] = P (Y = 10)+P (Y = 11)+P (Y = 12)
because the trials are independent. Therefore,
P [(Y = 10) ∪ (Y = 11) ∪ (Y = 12)] = P (Y = 10) + P (Y = 11) + P (Y = 12)
=
=
³ ´
15
10
p10 q15−10 +
³ ´
15
10
³ ´
15
11
p11 q15−11 +
(0.2)10 (0.8)15−10 +
³ ´
15
11
³ ´
15
12
p12 q 15−12
(0.2)11 (0.8)15−11 +
³ ´
15
12
(0.2)12 (0.8)15−12 = 0.00011317
i.e., P [(Y = 10) ∪ (Y = 11) ∪ (Y = 12)] = 0.00011317
3
5. The mean number of cars entering a narrow mountain tunnel during a 5 minute period
is 6.
(a) Compute the probability that during the next 5 minutes exactly 4 cars enter the
tunnel.
This is a Poisson Distribution with
(i.e., µ =
µ=6
6 cars
)
5 minutes
Y = number of cars entering the tunnel
We want P (Y = 4) =
µy e−µ
y!
=
64 e−6
4!
= 0.13385
i.e., P (Y = 4) = 0.13385
(b) Compute the probability that during the next 15 minutes at least 18 and no more
than 20 cars enter the tunnel.
We must “re-parameterize” µ, since the time period is different.
µ=
6 cars
5 minutes
=
i.e., µ = 18
6 cars
5 minutes
·
3
3
(i.e., µ =
=
18 cars
15 minutes
18 cars
)
15 minutes
We want P [(Y = 18) ∪ (Y = 19) ∪ (Y = 20)] = P (Y = 18)+P (Y = 19)+P (Y = 20)
because the trials are independent. Therefore,
P [(Y = 18) ∪ (Y = 19) ∪ (Y = 20)] = P (Y = 18) + P (Y = 19) + P (Y = 20)
=
µ18 e−µ
18!
+
µ19 e−µ
19!
+
µ20 e−µ
20!
=
1818 e−18
18!
+
1819 e−18
19!
+
1820 e−18
20!
i.e., P [(Y = 18) ∪ (Y = 19) ∪ (Y = 20)] = 0.26207
4
= 0.26207
6. A student answers a multiple choice question that offers five possible answers. Suppose
the probability that the student knows the answer to the question is 0.7 and the
probability that the student will guess is 0.3. Assume that if the student guesses, the
probability of guessing the correct answer is 0.2. If the student correctly answers the
question, what is the probability that the student really knew the correct answer?
Let’s name the events as follows:
B1 − Student knows the correct answer
B2 − Student guesses the answer (does not know the correct answer)
A1 − Student correctly answers the question
A2 − Student answers the question incorrectly
Then:
P (B1 ) = 0.7
P (B2 ) = 0.3
P (A1 |B2 ) = 0.2
Also, note that we can justify assuming that the following must be true:
P (A1 |B1 ) = 1.0
Note that:
(a) B1 ∩ B2 = ∅
(mutually exclusive)
and
(b) B1 ∪ B2 = S
(collectively exhaustive)
Thus, Bayes’ Rule applies.
We want P (B1 |A1 ) =
P (B1 )P (A1 |B1 )
P (B1 )P (A1 |B1 )+P (B2 )P (A1 |B2 )
i.e., P (B1 |A1 ) = 0.92105
5
=
(0.7)(1.0)
(0.7)(1.0)+(0.3)(0.2)
= 0.92105