Tan Version
Business 201-301
Summer 2009
Exam II
Tan Version
Name:____________________
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Tan Version
1. The Dog Days of Summer
For each part below, calculate the following: i. Mean, ii. Standard Deviation, iii. Pr(X=3), iv. Pr(X>3).
MARK YOUR ANSWERS TO EACH PART CLEARLY ELSE I WILL SIMPLY SUBTRACT CREDIT.
a. Suppose there is a 70% probability that any given day in June will reach or exceed 95 degrees.
Take a sample of 5 days from a binomial distribution defining a day over 95 degrees as a success.
(16 pts.)
-Mean = n*p = 5*0.7 = 3.5
-Standard Deviation = √(n*p*q) = √(5*0.7*0.3) = 1.02
5!
-Pr(x=3) = 3!(5−3)! 0.73 0.35−3 = 0.308 or 30.8%
5!
5!
-Pr(x>3) = Pr(x=4) + Pr(x=5) =4!(5−4)! 0.74 0.35−4 + 5!(5−5)! 0.75 0.35−5 =0.360 + 0.168 = 0.528 or
52.8%
b. Suppose that the average June in Missouri contains 24 days that exceed 95 degrees. Divide up
this Poisson segment and answer parts i-iv with respect to one week1. (16 pts.)
-Mean = 24/4 = 6 days per week on average.
-Standard Deviation = √6 = 2.44
-Pr(x=3) =
63 𝑒 −6
=
3!
0.08 or 8%
-Pr(x>3) = Pr(x=4) + Pr(x=5) + Pr(x=6) +…+Pr(x=∞)=0.848 or 84.8%
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You may assume there are four weeks in June.
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Tan 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=3) = 0,this is a continuous distribution.
-Pr(x>3) = height*width = 1/30*27 = 27/30 or 0.9 or 90%
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Tan 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 95%
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
95% 𝐶𝐼 = 3.5 ± 1.96
= 3.02 𝑡𝑜 3.97
√25
b. Assuming that you only know the sample standard deviation of windmills is 1.4 megawatts,
construct a 95% 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
95% 𝐶𝐼 = 3.5 ± 2.0639
= 2.92 𝑡𝑜 4.07
√25
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Tan Version
c. Conduct a hypothesis test (α=0.05) 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 = ±1.96
tcrit = ±2.0639
If Zstat > 1.96 or < -1.96
Reject H0. Else, fail to reject.
If tstat > 2.0639 or < -2.0639
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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