CHAPTER 6
5. The information below is the number of daily emergency service calls made by the
volunteer ambulance service of Walterboro, South Carolina, for the last 50 days. To
explain, there were 22 days on which there were 2 emergency calls, and 9 days on which
there were 3 emergency calls.
a.)
Convert this information on the number of calls to a probability distribution.
x
Frequency
P(x)
x * (P)x
(xu)^2 * P(x)
0
8
0.16
0
0.4624
1
10
0.2
0.2
0.098
2
22
0.44
0.88
0.0396
3
9
0.18
0.54
0.3042
4
1
0.02
0.08
0.1058
50
1.7
1.01
b.)
Is this an example of a discrete or continuous probability distribution
? Discrete
distribution
c.)
What is the mean number of emergency calls per day?
1.7
x
P(x)
x * (P)x
0
0.16
0
1
0.2
0.2
2
0.44
0.88
3
0.18
0.54
4
0.02
0.08
1.7
d.)
What is the standard deviation of the number of calls made daily?
1.005
Square root of 1.01 = 1.005
x
P(x)
x u
(xu)^2
(xu)^2 * P(x)
0
0.16
01.7
2.89
0.4624
1
0.2
11.7
0.49
0.098
2
0.44
21.7
0.09
0.0396
3
0.18
31.7
1.69
0.3042
4
0.02
41.7
5.29
0.1058
1.01
15. Industry standards suggest that 10 percent of new vehicles require warranty service
within the first year. Jones Nissan in Sumter, South Carolina, sold 12 Nissans yesterday.
n = 12
Probability = .1
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a.)
What is the probability that none of these vehicles requires warranty service? .
2824
(12 C 0) (.1)^0
(1.1)^(120)
= (1) (1) (.2824)
b.)
What is the probability exactly one of these vehicles requires warranty service?
.
3765
(12 C 1) (.1)^1
(1.1)^(121)
= (12) (.1) (.3138) = .3765
c.)
Determine the probability that exactly two or these vehicles require warranty
service.
.2301
(12 C 2) (.1)^2
(1.1)^(122) = (66) (.01) (.3487) = .2301
d.)
Compute the mean and standard deviation of this probability distribution.
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 Spring '09
 Parker
 Normal Distribution, Standard Deviation, Probability distribution, Probability theory, Discrete probability distribution

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