Instructions: Perform all computations in Excel. Submit the completed Excel file for grading. For the normal probability questions, you can use the Z table, or you can use the NORM.DIST() function. .
Scenario 1
You are staffing a security detail for a large manufacturing
plant. Absenteeism is an ongoing problem. (Don’t blame the staff.
You don’t pay very well, and since you started your master’s degree
while still working full time, you have been somewhat irritable. )
The schedule has five employees designated for duty today. Based on
history, the probability that everyone will show up is 65%. The
probability that only 4 will show up is 22%, the probability that 3
will show up is 9%, and the probability that 2 will show up is
4%.
Scenario 2
The armory has four AR15s, three Glock 9s, two shotguns, and three 45 caliber revolvers. Define A as semi-automatic weapons, define B as handguns, and define C as long guns.
Scenario 3:
You are in charge of cybersecurity at your firm. An analyst for your firm determined that the number of daily attacks on your network are normally distributed with a mean of 345 and a standard deviation of 68. Note: I put a nice normal table in the File Cabinet for this exercise. Also, remember that you have to compute the Z score = (X – m)/s. Then, you use the Z score to find the probability. If the probability is for less than, you simply read the probability off the table. If the probability is for greater than, you take the complement of the table probability. Easy!!
Scenario-1)
Probability that 3 or 4 employees will show up is 0.22+0.09=0.31
Probability that there will be only 1 employee on duty is zero
Probability that no more than four employees will show up is
0.04+0.09+0.22=0.35
Scenario-2) Not clear about Armoury
Scenario-3)
Define the standard random variable z as
Where X denotes number of attacks.
1) Using Normal table we have
2) The required probability is 0.50
As 345 is mean and half of data falls below it
3) Required probability is
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