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Unformatted text preview: 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. Kyle 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) 0.96040 0.960400 0.00000 0.03960 1.00 0000 1 0.03920 0.999600 0.96040 0.00040 0.03 9600 2 0.00040 1.000000 0.99960 0.00000 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....
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This note was uploaded on 03/25/2011 for the course STA 215 taught by Professor Tba during the Fall '08 term at Grand Valley State University.
 Fall '08
 TBA
 Statistics, Probability

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