Download Ch. 4 Discrete Random Variables Answer Key 4.1 Two Types of

Ch. 4 Discrete Random Variables
Answer Key
4.1 Two Types of Random Variables
1 Classify Variables as Continuous or Discrete
1) A
2) A
3) A
4) A
5) A
6) A
7) A
8) A
9) A
10) A
11) A
12) possible values of x: {0, 1, 2, 3, 4, 5, 6}; The variable x is discrete since it has a finite number of distinct possible values.
13) natural bounds for x: 0 ounces and 16 ounces; The variable x is continuous since the values of x correspond to the
points in some interval.
4.2 Probability Distributions for Discrete Random Variables
1 Construct Probability Distribution
1) C
2) A
3) This is not a valid probability distribution because the sum of the probabilities is less than 1.
4) This is a valid probability distribution because the probabilities are all nonnegative and their sum is 1.
5) This is not a valid probability distribution because one of the probabilities given is negative.
6) This is not a valid probability distribution because the sum of the probabilities is greater than 1.
7)
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8)
2 Find Probability Given Distribution
1) A
2) A
3) A
4) A
5) A
6) P(x 2 or x 3) p(x 1) p(x 4)
7) P(x 1 or x 2) p(x 1) p(x 2)
p(x
.25
5) .1 03
.20 .45
.2
4.3 Expected Values of Discrete Random Variables
.6
1 Find Expected Value
1) A
2) A
3) A
4) A
5) B
6) Let x cost of fare paid by passenger. The probability distribution for x is:
x
p(x)
$78
1/5
$355
4/5
The expected cost is E(x)
x p(x)
$78
1
5
$355
4
5
$299.60
Since the expected cost is more than the usual one way air fare, the passenger should not opt to fly as a standby.
7) To determine the premium, the insurance agency must first determine the average loss paid on the sports car. Let x
amount paid on the sports car loss. The probability distribution for x is:
x
p(x)
$24,500
.001
$12,000
.01
$5,750
.05
$2,000
.10
$0
.839
Note: These losses paid have already considered the $500 deductible paid by the owner.
The expected loss paid is:
xp(x)
$632
$24,500(.001)
$12,000(.01)
$5,750(.05)
$2,000(.10)
$0(.839)
In order to average $620 profit per policy sold, the insurance company must charge an annual premium of
$632 $620 $1252.00.
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2 Find Mean, Variance, Standard Deviation
1) A
2) A
3) B
4) A
5) A
6) A
7)
1.596;
1.098
8)
9) a.
b.
x p(x)
2(.2)
3(.3)
8(.3)
10(.2)
E(x)
1(.1)
2(.2)
3(.2)
4(.3)
5.7
2.3 2 (.1)
1.3 2 (.2)
0.3 2 (.2)
5(.2)
0.7 2 (.3)
3.3
1.7 2 (.2)
1.27.
c. P(
x
) P(2.03 x 4.57) .2 .3 .5; The Empirical Rule states that about .68 of the data lie within
one standard deviation of the mean for a mound shaped symmetric distribution. For our distribution, this value is
only .5, but it is not a surprise that these numbers aren't closer since our distribution is not symmetric.
4.4 The Binomial Random Variable
1 Understand the Binomial Random Variable
1) A
2) A
3) A
4) A
5) A
6) A
7) A
8) A
9) A
10) D
11) B
12) B
6
6!
13)
(.3)2 (.7)6 2
(.3) 2 (.7)4 15(.09)(.2401)
2
2!(6 2)!
.324
14) Since the probability of success remains the same from trial to trial, the probability of success on the second trial is also
.63.
15) Since the probability of success remains the same from trial to trial, the probability of success on the second trial is .48,
so the probability of failure on the second trial is 1 .48 .52.
2 Find Probability
1) A
2) A
3) A
4) A
5) A
6) A
7) A
8) A
9) A
10) C
11) B
12) C
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13) X is a binomial random variable with n
P(x
5)
14) Let x
p .30.
1
P(x
5)
15 and p
0.4.
1 0.403 (from a binomial probability table)
0.597
the number of the 14 cars with defective gas tanks. Then X is a binomial random variable with n
P(more than half) P(x 7) P(x 8) 1 P(x 7) 1 0.969 0.031
15) Let x the number of the 13 hypertensive patients whose blood pressure drops.
Then X is a binomial random variable with n 13 and p .5.
P(x
11)
P(x
11) P(x 12) P(x 13)
0.011230
16) Let x the number of the 25 ads that resulted in the item being sold within a week.
Then X is a binomial random variable with n 25 and p 0.70.
P(x
16)
0.323 (from a binomial probability table)
A value of x that is less than or equal to 16 will occur in about 32.3% of all such samples.
3 Find Mean/Expected Value, Standard Deviation
1) A
2) A
3) A
4) A
5) A
6) A
7) A
8) C
9) D
4.5 The Poisson Random Variable (Optional)
1 Find Probability
1) A
2) A
3) A
4) A
5) A
6) A
7) A
8) A
9) A
10) A
11) A
12) B
13) D
14) Let x the number of death claims received per day.
Then x is a Poisson random variable with
3.
P(x 7) 1 P(x 6) 0.033509
15) Let x the number of accidents that occur on the stretch of road during a month.
Then x is a Poisson random variable with
7.2.
P(x
2)
P(x
0) P(x
1)
0.006122
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14 and
16) Let x the number of babies born during a one hour period at this hospital.
Then x is a Poisson random variable with
3.
P(x 2) 0.224042
xe
5 7e 5
17)
.1044445
x!
7!
2 Find Mean/Expected Value, Standard Deviation
1) A
2) A
3) A
4) B
5) C
6) a.
b.
3;
3 1.73
c. P(
x
) P(1.27 x 4.73) .22 .22 .17 .61
7) E(x)
42.3; If the daily numbers of customers who sign up for additional services online were averaged for all
days, the result would be 42.3 customers per day.
4.6 The Hypergeometric Random Variable (Optional)
1 Identify the Characteristics of a Hypergeometric Random Variable
1) A
2) B
3) A
4) B
5) 0, 1, 2, 3
6) 0, 1, 2
2 Find Probability
1) A
2) A
3) A
4) A
5) A
6) A
7) A
8) D
9) D
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10) a. P(x
0)
b. P(x
1)
0
6 4
1 4
6(1)
252
10
5
1
42
.024
c. P(x 1) P(x 0) P(x 1) 0
d. P(x 2) 1 P(x 1) 1 .024
11) a.
x
p(x)
0
.071
1
.429
2
.429
3
.071
b. P(x
12) P(x
13) P(x
14) P(x
2)
0)
0)
0)
.071
2 10
0 3
12
3
3 12
0 4
15
4
.429
.429
.024
.976
.024
.929
.545
.363; P(x
20 15
4 3
1)
1
P(x
0)
1
.363
.637
.328
35
7
3 Find Mean, Variance, Standard Deviation
1) A
2) A
3) A
4) B
5) A
6) a.
x
p(x)
0
.083
1
.417
2
.417
3
.083
b.
3(5)
10
c. P( .028
x
1.5;
2
3.028)
5(10
1
5) 3(10
102 (9)
3)
7
;
12
7
12
.764
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