Rob owns a company that mows the right-of-way on state highways. He keeps record
ID: 3200563 • Letter: R
Question
Rob owns a company that mows the right-of-way on state highways. He keeps records of number of miles mowed per day versus the temperature and precipitation (Clear or Light Rain, they don’t mow on rainy days). Using prior years’ records yields the following probabilities:
What are the mean, standard deviation and the variance for the below distribution? (Show work)
<60oF
Clear
60- 69oF
Clear
70-79oF
Clear
80-89oF
Clear
ž90oF
Clear
<60oF
Light Rain
60- 69oF
Light Rain
70-79oF
Light Rain
80-89oF
Light Rain
ž90oF
Light Rain
Probability
.10
.15
.20
.20
.10
.10
.08
.04
.02
.01
Miles Mowed
55
65
80
85
75
45
55
75
80
65
<60oF
Clear
60- 69oF
Clear
70-79oF
Clear
80-89oF
Clear
ž90oF
Clear
<60oF
Light Rain
60- 69oF
Light Rain
70-79oF
Light Rain
80-89oF
Light Rain
ž90oF
Light Rain
Probability
.10
.15
.20
.20
.10
.10
.08
.04
.02
.01
Miles Mowed
55
65
80
85
75
45
55
75
80
65
Explanation / Answer
mean = sum (x * P(X))
variance = E(X^2) - mean^2
standard deviation = sqrt(variance)
x 55 65 80 85 75 45 55 75 80 65 P(X) 0.1 0.15 0.2 0.2 0.1 0.1 0.08 0.04 0.02 0.01 x*P(X) 5.5 9.75 16 17 7.5 4.5 4.4 3 1.6 0.65 x^2P(x) 302.5 633.75 1280 1445 562.5 202.5 242 225 128 42.25 mean 69.9 variance 177.49 standard deviation 13.3225373Related Questions
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