Green Version
Business 201-301
Summer 2009
Exam II
Green Version
Name:____________________
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Green Version
1. The Dog Days of Summer
For each part below, calculate the following: i. Mean, ii. Standard Deviation, iii. Pr(X=4), iv. Pr(X>4).
MARK YOUR ANSWERS TO EACH PART CLEARLY ELSE I WILL SIMPLY SUBTRACT CREDIT.
a. Suppose there is a 75% probability that any given day in June will reach or exceed 95 degrees.
Take a sample of 6 days from a binomial distribution defining a day over 95 degrees as a success.
(16 pts.)
-Mean = n*p = 6*0.75 = 4.5
-Standard deviation = √(n*p*q) = √(6*0.75*0.25) = 1.06
6!
Pr(x=4) = 4!(6−4)! 0.754 0.256−4 = 0.296 𝑜𝑟 29.6%
6!
6!
Pr(x>4) = 5!(6−5)! 0.755 0.256−5 + 6!(6−6)! 0.756 0.256−6 = 0.355 + 0.177 = 0.533 𝑜𝑟 53.3%
b. Suppose that the average June in Missouri contains 27 days that exceed 95 degrees. Divide up
this Poisson segment and answer parts i-iv with respect to one week1. (16 pts.)
-Mean = 27/4 = 6.75
-Standard deviation = √6.75 = 2.59
-Pr(x=4) =
6.754 𝑒 −6.75
4!
= 0.101 𝑜𝑟 10.1%
- Pr(x>4) = Pr(x=5) + Pr(x=6) + Pr(x=7) +…+Pr(x=∞)=0.802 or 80.2%
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You may assume there are four weeks in June.
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Green Version
c. Suppose the number of days in June that exceed 95 degrees is distributed uniformly between 0
and 30. (16 pts.)
-Mean =
𝑏+𝑎
2
=
30+0
2
= 15
(30−0)2
- Standard Deviation = √
12
= 8.66
-Pr(x=4) = 0, this is a continuous distribution
-Pr(x>4) = height*width = 1/30*26 = 26/30 = 0.867 or 86.7%
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Green Version
2. The Future of Energy
Suppose that a sample of 25 windmills has produced a mean of 3.5 megawatts of energy.
a. Assuming that the population standard deviation of windmills is 1.2 megawatts, construct a 90%
confidence interval of the mean energy production of windmills. Also sketch this confidence
interval. (6 pts.)
Since we know the population’s standard deviation, we are using the Z.
1.2
90% 𝐶𝐼 = 3.5 ± 1.645
= 3.10 𝑡𝑜 3.89
√25
b. Assuming that you only know the sample standard deviation of windmills is 1.4 megawatts,
construct a 90% confidence interval of the mean energy production of windmills. Also sketch
this confidence interval. (6 pts.)
Now that we are using the sample standard deviation, we use a t with n-1 degrees of freedom.
1.4
90% 𝐶𝐼 = 3.5 ± 1.7109
= 3.08 𝑡𝑜 3.91
√25
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Green Version
c. Conduct a hypothesis test (α=0.10) that the average windmill can produce 5 megawatts of
energy using either the population standard deviation or the sample standard deviation quoted
to you in previous parts. (24 pts.)
Indicate your choice: I am using the _________________standard deviation.
population/sample
Population
Sample
H0: μ = 5
HA: μ ≠ 5
H0: μ = 5
HA: μ ≠ 5
Zcrit = ±1.645
tcrit = ±1.7109
If Zstat > 1.645 or < -1.645
Reject H0. Else, fail to reject.
If tstat > 1.7109 or < -1.7109
Reject H0. Else, fail to reject.
3.5−5
Zstat = 1.2
⁄
√25
3.5−5
= −6.25
tstat = 1.4
⁄
√25
Reject H0.
= −5.35
Reject H0.
Both: It would appear that the average windmill in not capable of producing 5 megawatts of energy.
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