ch06
Short Description
Download ch06...
Description
CHAPTER SIX Continuous Distributions D
1. Which of the following is NOT a continuous distribution?
E Term
D E Term
A. B. C. D. 2.
normal distribution exponential distribution uniform distribution binomial distribution
The uniform distribution is _______________. A. B. C. D.
bimodal skewed to the right skewed to the left symmetric
173
174
Test Bank A
3.
E Term
A. B. C. D. 4.
rectangular distribution gamma distribution beta distribution Erlang distribution
The distribution in the following graph is a ________ distribution.
f(X)
D
The uniform distribution is also known as the __________.
0.06 0.05 0.04 0.03 0.02 0.01 0.00 35
E Term
A
E Term
A. B. C. D. 5.
40
45
50
55
60 x 65
normal gamma exponential uniform
The distribution in the following graph is a ________ distribution.
A. B. C. D.
normal gamma exponential uniform
Chapter 6: Continuous Distributions 175 C
6.
E Term
B
A. B. C. D. 7.
E Term
A
8.
M Calc
1/8 1/4 1/12 1/20
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then the mean() of this distribution is __________________. A. B. C. D.
9.
normal gamma exponential uniform
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then the height of this distribution, f(x), is __________________. A. B. C. D.
M Calc
C
The distribution in the following graph is a ________ distribution.
10 20 5 incalculable
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then the standard deviation () of this distribution is __________________. A. B. C. D.
4 1.33 1.15 2
176
Test Bank B
10.
M Calc
C
A. B. C. D. 11.
M Calc
D
12.
M Calc
A
M Calc
0.250 0.500 0.375 0.000
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then P(X < 7) is __________________. A. B. C. D.
14.
0.250 0.333 0.375 0.000
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then the P(13 X 15) is __________________. A. B. C. D.
13.
0.250 0.500 0.333 1.000
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then the P(10.0 X 11.5) is __________________. A. B. C. D.
M Calc
B
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then the P(9 X 11) is __________________.
0.500 0.000 0.375 0.250
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then P(X 11) is __________________. A. B. C. D.
0.750 0.000 0.333 0.500
Chapter 6: Continuous Distributions 177 D
15.
E Calc
A
A. B. C. D. 16.
E Calc
B
17.
M Calc
C
M Calc
50 25 10 5
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then the standard deviation () of this distribution is __________________. A. B. C. D.
19.
1/10 1/20 1/30 1/50
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then the mean () of this distribution is __________________. A. B. C. D.
18.
0.750 0.000 0.333 0.500
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then the height of this distribution, f(x), is __________________. A. B. C. D.
M Calc
D
If X is uniformly distributed over the interval 8 to 12, inclusively (8 X 12), then P(X 10) is __________________.
incalculable 8.33 0.833 2.89
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then P(25 X 28) is __________________. A. B. C. D.
0.250 0.500 0.300 1.000
178
Test Bank A
20.
M Calc
B
A. B. C. D. 21.
M Calc
C
22.
M Calc
D
M Calc
0.500 0.300 0.000 0.250
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then P(X 22) is __________________. A. B. C. D.
24.
0.500 0.000 0.375 0.200
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then P(X < 17) is __________________. A. B. C. D.
23.
0.250 0.333 0.375 0.000
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then P(33 X 35) is __________________. A. B. C. D.
M Calc
A
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then P(21.75 X 24.25) is __________________.
0.200 0.300 0.000 0.250
If X is uniformly distributed over the interval 20 to 30, inclusively (20 X 30), then P(X 24) is __________________. A. B. C. D.
0.100 0.000 0.333 0.600
Chapter 6: Continuous Distributions 179 C
25.
Helen Casner, a labor relations arbitrator, feels that the amount of time needed to arbitrate a labor dispute is uniformly distributed over the interval 4 to 24 hours, inclusively (4 X 24). Accordingly, the mean (average) time needed to arbitrate a labor dispute is ____________.
M BCalc
A. B. C. D.
D
Helen Casner, a labor relations arbitrator, feels that the amount of time needed to arbitrate a labor dispute is uniformly distributed over the interval 4 to 24 hours, inclusively (4 X 24). Accordingly, the probability that a labor dispute will be arbitrated in 8 hours or less is ____________.
26.
20 hours 16 hours 14 hours 12 hours
M BCalc
A. B. C. D.
C
Helen Casner, a labor relations arbitrator, feels that the amount of time needed to arbitrate a labor dispute is uniformly distributed over the interval 4 to 24 hours, inclusively (4 X 24). Accordingly, the probability that a labor dispute will require between 8 and 16 hours, inclusively, for arbitration is ____________.
27.
0.3333 0.6667 0.0000 0.2000
M BCalc
A. B. C. D.
B
The normal distribution is an example of _______.
28.
E Term
A. B. C. D.
0.3333 0.6667 0.4000 0.2000
a discrete distribution a continuous distribution a bimodal distribution an exponential distribution
180
Test Bank B
29.
The total area underneath any normal curve is _______.
E Term
A. B. C. D.
D
The area to the left of the mean in any normal distribution is _______.
30.
equal to the mean equal to 1 equal to the variance equal to the coefficient of variation
E Term
A. B. C. D.
B
For any normal distribution, any value less than the mean would have a _______.
31.
equal to the mean equal to 1 equal to the variance equal to 0.5
E Term
A. B. C. D.
D
A standardized normal distribution has the following characteristics:
32.
positive Z-score negative Z-score negative variance negative probability of occurring
E Term
A. B. C. D.
C
If X is a normal random variable with mean 80 and standard deviation 5, calculate the Z score if X=88.
33.
E Calc
D
E Calc
A. B. C. D. 34.
the mean and variance are both equal to 1 the mean and variance are both equal to 0 the mean is equal to the variance the mean is equal to 0 and the variance is equal to 1
1.8 -1.8 1.6 -1.6
If X is a normal random variable with mean 80 and standard deviation 5, calculate the Z score if X=72. A. B. C. D.
1.8 -1.8 1.6 -1.6
Chapter 6: Continuous Distributions 181 D
35.
E Calc
C
A. B. C. D. 36.
E Calc
C
37.
M Calc
B
E Calc
63.4 56.6 66.8 53.2
Suppose X is a normal random variable with mean 60 and standard deviation 2. A Z score was calculated for a number, and the Z score is -1.3. What is X? A. B. C. D.
39.
1.5 2.5 -1.5 -2.5
Suppose X is a normal random variable with mean 60 and standard deviation 2. A Z score was calculated for a number, and the Z score is 3.4. What is X? A. B. C. D.
38.
2.1 12 1.2 2.4
If X is a normal random variable with mean 60 and standard deviation 2, calculate the Z score if X=57. A. B. C. D.
M Calc
D
If X is a normal random variable with mean 80 and standard deviation 5, calculate the Z score if X=92.
58.7 61.3 62.6 57.4
Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normal tables to find P(Z < 1.3). A. B. C. D.
0.4032 0.9032 0.0968 0.3485
182
Test Bank D
40.
E Calc
C
A. B. C. D. 41.
M Calc
D
42.
M Calc
C
M Calc
0.4821 -0.4821 0.9821 0.0179
Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normal tables to find P(Z > -1.1). A. B. C. D.
44.
0.4918 0.9918 0.0082 0.4793
Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normal tables to find P(Z < -2.1). A. B. C. D.
43.
0.4032 0.9032 0.4893 0.0861
Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normal tables to find P(Z > 2.4). A. B. C. D.
M Calc
B
Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normal tables to find P(1.3 < Z < 2.3).
0.3643 0.8643 0.1357 -0.1357
Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normal tables to find P(-2.25 < Z < -1.1). A. B. C. D.
0.3643 0.8643 0.1235 0.4878
Chapter 6: Continuous Distributions 183 B
45.
M Calc
C
A. B. C. D. 46.
E Calc
A
47.
M Calc
A
M Calc
0.670 -1.254 0.000 1.280
Let Z be a normal random variable with mean 0 and standard deviation 1. The 90th percentile of Z is ____________. A. B. C. D.
49.
0.670 -1.254 0.000 1.280
Let Z be a normal random variable with mean 0 and standard deviation 1. The 75th percentile of Z is ____________. A. B. C. D.
48.
0.3643 0.8521 0.1235 0.4878
Let Z be a normal random variable with mean 0 and standard deviation 1. The 50th percentile of Z is ____________. A. B. C. D.
M Calc
D
Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normal tables to find P(-2.25 < Z < 1.1).
1.645 -1.254 1.960 1.280
Let Z be a normal random variable with mean 0 and standard deviation 1. The 95th percentile of Z is ____________. A. B. C. D.
1.645 -1.254 1.960 1.280
184
Test Bank B
50.
M Calc
C
A. B. C. D. 51.
M Calc
B
52.
M Calc
B
E Calc
0.0987 0.4013 -0.0987 0.5987
Let X be a normal random variable with mean 20 and standard deviation 4. Find P(16 < X < 22). A. B. C. D.
54.
0.2734 0.7734 0.2266 -0.2734
Let X be a normal random variable with mean 20 and standard deviation 4. Find P(X < 19). A. B. C. D.
53.
0.3944 0.8944 0.1056 0.6056
Let X be a normal random variable with mean 20 and standard deviation 4. Find P(X < 17). A. B. C. D.
M Calc
D
Let X be a normal random variable with mean 20 and standard deviation 4. Find P(X < 25).
0.4672 0.0328 0.1498 0.5328
Let X be a normal random variable with mean 20 and standard deviation 4. The 50th percentile of X is ____________. A. B. C. D.
4.000 20.000 22.698 26.579
Chapter 6: Continuous Distributions 185 C
55.
M Calc
A
A. B. C. D. 56.
M Calc
D
57.
M Calc
A
M Calc
25.126 20.000 22.698 26.579
Let X be a normal random variable with mean 40 and standard deviation 8. Find P(32 < X < 44). A. B. C. D.
59.
25.126 20.000 22.698 26.579
Let X be a normal random variable with mean 20 and standard deviation 4. The 95th percentile of X is ____________. A. B. C. D.
58.
25.126 20.000 22.698 26.579
Let X be a normal random variable with mean 20 and standard deviation 4. The 90th percentile of X is ____________. A. B. C. D.
M Calc
D
Let X be a normal random variable with mean 20 and standard deviation 4. The 75th percentile of X is ____________.
0.4672 0.0328 0.1498 0.5328
Let X be a normal random variable with mean 40 and standard deviation 8. Find P(X < 96). A. B. C. D.
1.0000 0.0000 0.0793 0.0575
186
Test Bank B
60.
M Calc
B
Let X be a normal random variable with mean 40 and standard deviation 2. Find P(X < 28). A. B. C. D.
61.
1.0000 0.0000 0.2580 0.0472
A Z score is the number of __________ that a value is from the mean.
E Term
A. B. C. D.
C
Within a range of Z scores from -1 to +1, you can expect to find _______ per cent of the values in a normal distribution.
62.
variances standard deviations units miles
E Term
A. B. C. D.
A
Within a range of Z scores from -2 to +2, you can expect to find _______ per cent of the values in a normal distribution.
63.
95 99 68 34
E Term
A. B. C. D.
C
The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. The life of this bulb is normally distributed. What is the probability that a randomly selected bulb would last longer than 1150 hours?
M Calc
64.
A. B. C. D.
95 99 68 34
0.4987 0.9987 0.0013 0.5013
Chapter 6: Continuous Distributions 187 B
65.
M Calc
C
A. B. C. D. 66.
M Calc
D
67.
H Calc
0.3849 0.8849 0.1151 0.6151
Suppose you are working with a data set that is normally distributed with a mean of 400 and a standard deviation of 20. Determine the value of X such that 60% of the values are greater than X. A. B. C. D.
68.
0.4772 0.9772 0.0228 0.5228
The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. The life of this bulb is normally distributed. What is the probability that a randomly selected bulb would last fewer than 940 hours? A. B. C. D.
H Calc
A
The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. The life of this bulb is normally distributed. What is the probability that a randomly selected bulb would last fewer than 1100 hours?
404.5 395.5 405.0 395.0
Suppose you are working with a data set that is normally distributed with a mean of 400 and a standard deviation of 20. Determine the value of X such that only 1% of the values are greater than X. A. B. C. D.
446.6 353.4 400.039 405
188
Test Bank C
69.
H Calc
C
A. B. C. D. 70.
M Calc
B
71.
H Calc
0.3944 0.8944 0.1056 0.6056
The E.P.A. has reported that the average fuel cost for a particular type of automobile is $800 with a standard deviation of $80. Fuel cost is assumed to be normally distributed. If one of these cars is randomly selected, what is the probability that the fuel cost for this car exceeds $760? A. B. C. D.
72.
432.9 396 367.1 404
The E.P.A. has reported that the average fuel cost for a particular type of automobile is $800 with a standard deviation of $80. Fuel cost is assumed to be normally distributed. If one of these cars is randomly selected, what is the probability that the fuel cost for this car exceeds $900? A. B. C. D.
M Calc
B
Suppose you are working with a data set that is normally distributed with a mean of 400 and a standard deviation of 20. Determine the value of X such that 5% of the values are less than X.
0.1915 0.6915 0.3085 0.8085
The E.P.A. has reported that the average fuel cost for a particular type of automobile is $800 with a standard deviation of $80. Fuel cost is assumed to be normally distributed. We would expect that only 10% of these cars would have an annual fuel cost greater than _______. A. B. C. D.
820.0 902.4 808.0 812.8
Chapter 6: Continuous Distributions 189 A
73.
M Calc
C
The E.P.A. has reported that the average fuel cost for a particular type of automobile is $800 with a standard deviation of $80. Fuel cost is assumed to be normally distributed. If a car is randomly selected, what is the probability that fuel cost would be between $700 and $900? A. B. C. D.
74.
0.7888 0.8944 0.3944 0.1056
The net profit of an investment is normally distributed with a mean of $10,000 and a standard deviation of $5,000. The probability that the investor will not have a net loss is _____________.
M BCalc
A. B. C. D.
B
The net profit of an investment is normally distributed with a mean of $10,000 and a standard deviation of $5,000. The probability that the investor will have a net loss is _____________.
75.
0.4772 0.0228 0.9772 0.9544
M BCalc
A. B. C. D.
A
The net profit of an investment is normally distributed with a mean of $10,000 and a standard deviation of $5,000. The probability that the investor’s net profit will be between $12,000 and $15,000 is _____________.
76.
M BCalc
A. B. C. D.
0.4772 0.0228 0.9772 0.9544
0.1859 0.3413 0.8413 0.4967
190
Test Bank C
77.
The net profit of an investment is normally distributed with a mean of $10,000 and a standard deviation of $5,000. The probability that the investor’s net gain will be at least $5,000 is _____________.
M BCalc
A. B. C. D.
A
Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. The probability that the project will be completed within 185 workdays is ______.
78.
0.1859 0.3413 0.8413 0.4967
M BCalc
A. B. C. D.
D
Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. The probability that the project will be completed within 215 workdays is _____.
79.
0.0668 0.4332 0.5000 0.9332
M BCalc
A. B. C. D.
A
Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. The probability that the project will not be completed within 215 work-days is _____.
80.
M BCalc
A. B. C. D.
0.0668 0.4332 0.5000 0.9332
0.0668 0.4332 0.5000 0.9332
Chapter 6: Continuous Distributions 191 C
81.
Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. The probability that the project will be completed within ____ workdays is 0.99.
M BCalc
A. B. C. D.
B
The length of steel rods produced by a shearing process are normally distributed with = 120 inches and = 0.05 inch. Industry standards require the rods to be between 119.90 and 120.15 inches, inclusively. The probability that a rod produced by this process will conform to industry standards is ______________.
82.
211 187 223 200
M BCalc
A. B. C. D.
C
The length of steel rods produced by a shearing process are normally distributed with = 120 inches and = 0.05 inch. Industry standards require the rods to be between 119.90 and 120.15 inches, inclusively. Any rod longer than 120.15 inches is re-sheared. The probability that a rod produced by this process will require re-shearing is ___________.
83.
0.9542 0.9759 0.9974 0.6826
M BCalc
A. B. C. D.
B
The length of steel rods produced by a shearing process are normally distributed with = 120 inches and = 0.05 inch. Industry standards require the rods to be between 119.90 and 120.15 inches, inclusively. Any rod shorter than 119.90 inches is scrapped (used in the next melt). The probability that a rod produced by this process will be scrapped is ___________.
84.
M BCalc
A. B. C. D.
0.0458 0.0228 0.0013 0.0241
0.0458 0.0228 0.0013 0.0241
192
Test Bank A
85.
The weights of aluminum castings produced by a process are normally distributed with = 2 pounds and = 0.10 pound. Design specifications require the castings to weigh between 1.836 and 2.164 pounds, inclusively. The probability that a casting produced by this process will conform to design specifications is _________.
M BCalc
A. B. C. D.
C
The weights of aluminum castings produced by a process are normally distributed with = 2 pounds and = 0.10 pound. Design specifications require the castings to weigh between 1.836 and 2.164 pounds, inclusively. Any casting weighing less than 1.836 pounds is scrapped. The probability that a casting produced by this process will be scrapped, due to under-weight, is _________.
86.
0.8990 0.4495 0.9974 0.9500
M BCalc
A. B. C. D.
C
The weights of aluminum castings produced by a process are normally distributed with = 2 pounds and = 0.10 pound. Design specifications require the castings to weigh between 1.836 and 2.164 pounds, inclusively. Any casting weighing more than 2.164 pounds is re-worked. The probability that a casting produced by this process will be re-worked, due to over-weight, is _________.
87.
0.1010 0.4495 0.0505 0.0010
M BCalc
A. B. C. D.
B
Let X be a binomial random variable with n=20 and p=.8. If we use the normal distribution to approximate probabilities for this, we would use a mean of _______.
E Calc
88.
A. B. C. D.
0.0010 0.1010 0.0101 0.0505
20 16 3.2 8
Chapter 6: Continuous Distributions 193 C
89.
M Calc
B
Let X be a binomial random variable with n=20 and p=.8. If we use the normal distribution to approximate probabilities for this, we would use a standard deviation of _______. A. B. C. D.
90.
16 3.2 1.79 0.16
Let X be a binomial random variable with n=20 and p=.8. If we use the normal distribution to approximate probabilities for this, a correction for continuity should be made. To find the probability of more than 12 successes, we should find _______.
M Term
A. B. C. D.
B
Let X be a binomial random variable with n=20 and p=.8. If we use the normal distribution to approximate probabilities for this, a correction for continuity should be made. To find the probability of 12 successes or more, we should find _______.
91.
P(X>12) P(X>12.5) P(X>11.5) P(X12) P(X>11.5) P(X>12.5) P(X
View more...
Comments