Kyle Gring
Stats 215
Stats Homework #2
1.
a.
If an airplane has two different alternators with the probability of any one of those
alternators to fail is .02, then I will look at what the probability of both of those to
fail is. The events in this situation are independent because one failing does not
affect the other one from failing. In a visual picture this will be the space inside
the rectangle where the circles overlap. The probability of both of those
alternators to fail is .0004. This answer was found using the binomial probability
calculator where I inputted the sample size which was 2 and entered the
probability of successes (or failures in this case). This can also be found by
multiplying the two independent event probabilities together, which is .
02*.02=.0004.
b.
If an airplane has two different alternators with the probability of one of those
alternators to fail is .02, then I will look at the probability of neither of the
alternators to fail. The events in this situation are independent because one failing
does not affect the other one from failing. In a visual picture this will be the space
outside of both of the circles. The probability that neither of the alternators will
fail is .960400. This answer was found using the binomial probability calculator
where I inputted the sample size which was 2 and entered the probability of
successes (or failure in this case). This can also be found by multiplying the two
independent event probabilities together, which is .98*.98=.960400.
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View Full DocumentKyle Gring
Stats 215
c.
If an airplane has two different alternators with the probability of one of those
alternators to fail is .02, then I will look at the probability that only one will fail.
Then the probability that only one of the alternators to fail is .0392. This answer
was found using the binomial probability calculator where I inputted the sample
size which was 2 and entered the probability of successes (or failures in this case).
This can also be found by taking parts of the circles and cutting out the
intersections which is 2%  0.04% +2%  0.04% = 3.92%.
X
P(X)
P(<=X)
P(<X)
P(>X)
P(>
=X)
0
0.96040
0
0.960400
0.00000
0
0.03960
0
1.00
0000
1
0.03920
0
0.999600
0.96040
0
0.00040
0
0.03
9600
2
0.00040
0
1.000000
0.99960
0
0.00000
0
0.00
0400
Binomial Probabilty
Calculator
Sample Size
2
Probability of Success
0.02
Mean
0.04
Variance
0.0392
Standard Deviation
0.1979
9
2.
A. If my environmental protection group had been selected to do a quality and
safety study on landfills in my area and we has 15 total landfills but could on choose 10 because
of lack of resources I would use a counting rule of combination. If I were to use a combination to
choose the landfills selected I would have 3003 different combinations to choose from. This was
found by using the combination formula where I inputted N which was 15 and R which was 10.
N represents the number we can choose from and R is the total number that we chose. The
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 Fall '08
 TBA
 Statistics, Normal Distribution, Probability, Kyle Gring, Kyle Gring Stats

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