EF 507
PS-Chapter 6
FALL 2008
QUESTIONS 1 THROUGH 4 ARE BASED ON THE FOLLOWING
INFORMATION:
The amount of time you have to wait at a particular stoplight is uniformly distributed
between zero and two minutes.
1.
What is the probability that you have to wait more than 30 seconds for the
light?
A) 0.25
B) 0.50
C) 0.75
D) 1.01
ANSWER: C
2.
What is the probability that you have to wait between 15 and 45 seconds for
the light?
A) 0.15
B) 0.25
C) 0.35
D) 0.45
ANSWER: B
3.
Eighty percent of the time, the light will change before you have to wait how
long?
A) 90 seconds
B) 24 seconds
C) 30 seconds
D) 96 seconds
ANSWER: D
4.
Sixty percent of the time, the light will change before you have to wait how
long?
A) 72 seconds
B) 60 seconds
C) 48 seconds
D) 36 seconds
ANSWER: A
QUESTIONS 5 THROUGH 7 ARE BASED ON THE FOLLOWING
INFORMATION:
Let the random variable Z follow a standard normal distribution.
5. What is P(Z > 1.2)?
A)
B)
C)
D)
0.1112
0.8849
0.1151
0.6112
ANSWER:
C
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6. What is P(Z > -0.21)?
A) 0.4207
B) 0.4168
C) 0.5793
D) 0.5832
ANSWER:
D
7. What is P(0.33 < Z < 0.45)?
A) 0.5443
B) 0.0443
C) 0.4557
D) 0.1515
ANSWER:
B
QUESTIONS 8 AND
INFORMATION:
10
ARE
BASED
ON
THE
FOLLOWING
Let the random variable X follow a normal distribution with a mean of 61.7 and a
standard deviation of 5.2.
8.
What is the value of k such that P(X > k) = 0.63?
A) 59.984
B) 66.446
C) 62.830
D) 67.576
ANSWER:
A
9. What is the value of k such that P(59 < X < k) = 0.54?
A) 65.8
B) 64.6
C) 63.7
D) 66.9
ANSWER:
D
10. Let the random variable Z follow a standard normal distribution. Find P(0 < Z <
0.57).
A) 0.2843
B) 0.7843
C) 0.2157
D) 0.7157
ANSWER:
C
11. Let the random variable Z follow a standard normal distribution. Find P(-2.21 < Z
< 0).
A)
B)
C)
D)
0.9864
0.4864
0.0136
0.5136
ANSWER:
B
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QUESTIONS 12 AND
INFORMATION:
13
ARE
BASED
ON
THE
FOLLOWING
During a professor’s office hours, students arrive, on average, every ten minutes.
Assume that the distribution of the time between arrivals follows an exponential
distribution. Suppose that a student has just left.
12. What is the probability that the professor has more than 20 minutes before the
next student shows up?
A) 0.8187
B) 0.8647
C) 0.1353
D) 0.1813
ANSWER: C
13. Suppose that a student has just left. Only 25% of the time will the professor have
to wait approximately how long or longer before the next student shows up?
A) 13.9 minutes
B) 14.6 minutes
C) 2.9 minutes
D) 15.3 minutes
ANSWER: A
14.
Let the random variable Z follow a standard normal distribution, and let z1 be
a possible value of Z that is representing the 10th percentile of the standard
normal distribution. Find the value of z1 .
A) 0.255
B) -0.255
C) 1.28
D) -1.28
ANSWER: D
15.
A very large logging operation has serious problems keeping their skidders
operating properly. The equipment fails at the rate of 3 breakdowns every 48
hours. Assume that x is time between breakdowns and is exponentially
distributed. The probability of two or less breakdowns in the next 48-hour
period is
A) 0.9672.
B) 0.4230.
C) 0.2231.
D) 0.7769.
ANSWER: B
16.
If the random variable x is exponentially distributed with parameter  = 1.5,
then the probability P(2  X  4) is equal to
A) 0.6667.
B) 0.5000.
C) 0.0473.
D) 0.2500.
ANSWER: C
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QUESTIONS 17 THROUGH 19 ARE BASED ON THE FOLLOWING
INFORMATION:
The amount of time you have to wait at a dentist office before you called in is
uniformly distributed between zero and twenty minutes.
17.
What is the probability that you have to wait more than 8 minutes?
ANSWER:
P(X > 8) = (20-8) / (20-0) = 12 / 20 = 0.60
18.
What is the probability that you have to wait between 10 and 15 minutes?
ANSWER:
P(10 < X < 15) = (15-10) / (20-0) = 5 / 20 = 0.25
19.
Seventy percent of the time, you will be called in before you have to wait
how long?
ANSWER: The range is a total of 20 minutes. Seventy percent of the time,
you should be called in before you wait 14 minutes
20.
Suppose that 24% of all sales in a grocery store are for amounts greater than
$100. In a random sample of 50 invoices, what is the probability that more
than ten of the invoices are for over $100? Use the normal approximation to
the binomial distribution, both with and without the continuity correction.
ANSWER:
9.12,  = 3.02
P = 0.23, n = 50,  = E(X) = nP = 12,  2 = Var(X) = nP(1-P) =
Without continuity correction P(X>10) = P(Z>-0.66) = 0.7454.
21. It has been found that 62.1% of all unsolicited third class mail delivered to
households goes unread. If, over the course of a month, a household receives
150 pieces of unsolicited mail, what is the probability that the household
discards more than 80 pieces of the mail without reading it? Use the normal
approximation to the binomial distribution without the continuity correction.
ANSWER: P = 0.621, n = 150,  =E(X) = 93.15,  2 =Var(X) = nP(1-P) =
35.3039,  X = 5.942
P(X > 80) = P(Z > -2.21) = 0.9864.
22. You are the Webmaster for your firm’s Website. From your records, you know
that the probability that a visitor will buy something from your firm is 0.23. If
the number of visitors in one day is 952, what is the probability that less than
200 of them will buy something from your firm? Use the normal
approximation to the binomial without the continuity correction.
ANSWER: n = 952, P = 0.23,  =E(X) = 218.96,  2 =Var(X) = nP(1-P) =
168.5992,  = 12.985
P(X<200) =P(Z<-1.46) = 0.0721
QUESTIONS 23 AND 24 ARE BASED ON THE FOLLOWING
INFORMATION:
During hours, students arrive, on average, every ten minutes. Assume that the
distribution of the time between arrivals follows an exponential distribution. Suppose
that a customer has just left the gas station.
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23. What is the probability that the cashier at the gas station has more than 15 minutes
before the next customer arrive?
ANSWER: Exponential with mean 1/  =10, then,  = 0.10, and P(T>15)
= e15  e1.5  0.223
24. The probability is 0.30 that the cashier has to wait at least how long or longer
before the next customer arrives?
ANSWER: P(T > t) = e t  e 0.10t = 0.30  -0.10t = -1.204  t =12.04 minutes
25.
The total cost for a production process is equal to $1,200 plus 2.5 times the
number of units produced. The mean and variance for the number of units
produced are 520 and 840, respectively. Find the mean and variance of the
total cost.
ANSWER: Let X = number of units produced, and T = total cost for the
production process.
Since T = 1200 + 2.5X,  X = 520 and  X2 = 840, then
T  1200  2.5 X  1200 + (2.5)(520) = 2500, and  T2  (2.5)2  X2  (6.25)(840)
= 5250.
26.
A random variable X is normally distributed with mean of 50 and variance of
50, and a random variable Y is normally distributed with mean of 100 and variance of
200. Given the random variables X and Y have a correlation coefficient equal to 0.50,
find the mean and variance of the random variable W = 4X+3Y.
ANSWER:
X  50,  X2  50, Y  100,  Y2  200, Corr( X ,Y )  0.50
W  4 X  3Y  (4)(50) + (3)(100) = 500
W2  (4)2 X2  (3)2 Y2  2(4)(3)Corr( X , Y ) X  Y
= (16)(50) +(9)(200) + (24)(0.50)(7.071)(14.142) = 3,799.977
27.
A normal random variable x has an unknown mean  and standard deviation
  2.5 If the probability that x exceeds 7.5 is 0.8289, find  .
ANSWER:
It is given that x is normally distributed with  = 2.5 but with unknown
mean  , and that P (x > 7.5) = 0.8289. In terms of the standard normal
random variable z, we can write
P( X  7.5)  P[ Z  (7.5   ) / 2.5]  0.8289
Since the area to the right of (7.5   ) / 2.5 is greater than 0.5,
then (7.5   ) / 2.5 must be negative, and that P[ (7.5   ) / 2.5 < z < 0] =
0.3289. Hence,
(7.5   ) / 2.5 = -.095. This implies  = 9.875.
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