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Three example problems for Week 4 Homework.
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Week 4 Three Extra Homework Examples (3x9)/WMA 221 Statistics for Decision Making
Professor Brent HeardNot to be copied or linked to without my permission 3 x 9
WS4DM
(3x9)/W•Disclaimer
▫Please note that I might do things a little differently than your instructor, your tutor, your friends, your mother, your dog, etc.
▫The reason I post these examples is to help you▫There is no requirement for you to even look at
these▫Therefore, if I help you – great. If I don’t – I’m
sorry.▫This is something I simply do as a service to
the students because I like them!
(3x9)/W
•Number 5 Example▫On number 5 in the homework, they are
just trying to get you to use the Binomial to find the probability of certain things happening based on a given n and p.
▫It’s really easy. Let’s look at an example…
(3x9)/W
•Forty-two percent of households say they would feel secure if they had six months of expenses in the bank. You randomly select 9 households and ask them if they would feel secure having six months worth of their expenses in the bank.▫Find the probability exactly 6 would say
yes.▫Find the probability more than 6 would say
yes.▫Find the probability that at most 6 would
say yes.•Solution next page
(3x9)/W
•Based on our question, we know we are asking 9 households, so n=9. Also we know that 42% feel secure having that amount saved, so p=0.42 (the decimal form of 42%)
•The rest is easy using Minitab…▫I use the same approach in Minitab to solve
all three parts of this question▫I recently discovered that it works very
well
(3x9)/W
•Once in Minitab▫Go to Graph >> Probability Distribution
Plot Once There Click the Fourth Option “View
Probability” and then Click “OK”
(3x9)/W• Now in Probability
Distribution Plot▫ Now for this problem,
change the Distribution Dropdown to “Binomial” and input the n (number of trials) and p (Event probability) for this problem (Remember 9 households, so n=9. Forty-two percent, so p = 0.42
(3x9)/W• Now click the Shaded Area Tab and pay attention
(3x9)/W• For this problem we
will be putting in “X values
• For Exact values like P(x=6), we will use “Middle,” for less than problems we will use “Left Tail” and for greater than problems, we will use “Right Tail”
• But pay attention to the wording!
(3x9)/W• Now part “a” was to find P(6) based on n=9 and p=0.42
Therefore, P(6) = 0.08996 or 0.090 rounded to three decimal places
(3x9)/W• Now part “b” was to find “more than 6” based on n=9 and p=0.42. “More
than 6 implies “7 or more.” This is a “Right Tail.”
Therefore, P(x>6) = 0.03338 or 0.033 rounded to three decimal places
(3x9)/W• Now part “c” was to find the probability of “no more than 6” based on n=9
and p=0.42. “No more than 6” implies “6 or less.” This is a “Left Tail.”
Therefore, P(x<=6) is 0.9666 or 0.967 rounded to three decimal places
(3x9)/W
•These are easy, just follow the steps and remember on those “More than 6” type problems, you are dealing with a right tail where “more than 6” means 7 or more…
•Steps again - Graph >> Probability Distribution Plots
•Next page
(3x9)/W
•Order
Pick Tails and values based on problem
See Answer
PracticePracticePractice
(3x9)/W
•Number 13 Example▫On number 13 in the homework, they are
just trying to get you to use the Geometric distribution to find the probability of something happening on the 4th time or 5th or a certain “one shot” time. In other words, when the event happens – game over (You’ve done it for “the first time.”)
▫It’s really easy. Let’s look at an example…
(3x9)/W
•Number 13 Example▫Assume the probability you will make a sale
on any given phone call is 0.11. Find the probability that you A) make your first sale on the third call B) make your first sale on the first, second,
third, fourth, fifth or sixth call. C) Do not make the first sale on one of the
first four calls.
(3x9)/W
•Easy in Minitab▫Go to Graph >> Probability Distribution
Plot Once There Click the Fourth Option “View
Probability” and then Click “OK”
(3x9)/W• Now in Probability
Distribution Plot▫Now for this
problem, change the Distribution Dropdown to “Geometric” and input Event probability for this problem (Remember the probability is 0.11)
(3x9)/W• Now click the Shaded Area Tab and pay attention
(3x9)/W• Now in Shaded Area Tab
after choosing Geometric distribution and inputting correct Event Probability▫ Part A) - Make your first
sale on the third call
(3x9)/W
•Answer•0.08713 or 0.087
rounded to three decimals
(3x9)/W• Now in Shaded Area Tab
after choosing Geometric distribution and inputting correct Event Probability
Part B) - make your sale on the first, second, third, fourth, fifth or sixth call. (THIS MEANS MAKE YOUR SALE IN ONE OF THE FIRST SIX CALLS – SO IT IS “LEFT TAIL” WITH 6 AS YOUR STOPPING POINT)
(3x9)/W
•Answer•0.5030 or 0.0503
rounded to three decimals
(3x9)/W• Now in Shaded Area Tab
after choosing Geometric distribution and inputting correct Event Probability
Part C) - Do not make a sale on the first four calls. (THIS MEANS MAKE YOUR FIRST SALE ON THE FIFTH CALL OR LATER – SO THIS IS A “RIGHT TAIL” STARTING WITH 5.)
(3x9)/W
•Answer•0.6274 or 0.627
rounded to three decimals
(3x9)/W
•Last Question - Are any of these “Unusual?”
•“NO” – To be considered “Unusual” the probability would have to be less than 0.05
•In other words, 0.0671, 0.124, 0.053 are not unusual. However values like 0.047, 0.003, 0.019 are because they are less than 0.05 or “5%.”
(3x9)/W
•Number 15 Example▫On number 15 in the homework, they are
just trying to get you to use the Poisson to find the probability of something happening. Read about the differences in Geometric and Poisson problems, you should be able to spot them pretty easy. With a Poisson, they usually give you an average over a given timeframe.
▫Let’s look at an example…
(3x9)/W• Number 15 Example
▫A major hurricane has really strong winds and if I wanted to confuse you I would tell you how strong the winds have to be, but I don’t want to confuse you. During the last century the mean number of hurricanes to hit Gilligan’s Island per year was about 0.52 (SCREAMS POISSON- “Over the last century the average number of hurricanes per year”) Find the following
Probability exactly one hurricane will hit Gilligan’s Island Probability at most one hurricane will strike Gilligan’s Island Probability more than one hurricane will hit Gilligan’s Island
(3x9)/W
•Again back to Minitab (Should sound familiar)▫Go to Graph >> Probability Distribution
Plot Once There Click the Fourth Option “View
Probability” and then Click “OK”
(3x9)/W• Now in Probability
Distribution Plot▫Now for this
problem, change the Distribution Dropdown to “Poisson” and input the mean for this problem (Remember the number of hurricanes was 0.52)
(3x9)/W• Now click the Shaded Area Tab and pay attention
(3x9)/W• Now in Shaded Area Tab
after choosing Geometric distribution and inputting correct Event Probability▫ Part A) – Exactly one
hurricane
(3x9)/W•Answer•0.3092 or 0.309
rounded to three decimals
•This is “NOT UNUSUAL” because it is greater than 0.05
(3x9)/W• Now in Shaded Area
Tab after choosing Geometric distribution and inputting correct Event Probability
Part B) – at most one hurricane
NOTE AT MOST ONE IS ZERO OR ONE, SO IT IS LEFT TAILED WITH AN ENDPOINT AT 1
(3x9)/W• Answer• 0.9037 or 0.904 rounded
to three decimals▫ This is the probability of
at most one hurricane hitting… which means it combines the probabilities of seeing none or zero and one
▫ Again, this is “NOT UNUSUAL” – as a matter of fact there is about a 90% chance of getting either zero or one hurricanes
(3x9)/W• Now in Shaded Area Tab
after choosing Geometric distribution and inputting correct Event Probability
Part C) – More than one hurricane – THIS MEANS “TWO OR MORE” Therefore we have a Right Tail with 2 on the left endpoint.
(3x9)/W• Answer• 0.09633 or 0.096 rounded to
three decimals – again not unusual
• However note that if I add my results from part b and part c together I should get one because I have “covered” all possibilities (In that the probability of seeing 1 or less plus the probability of seeing 2 or more would be everything▫ (THIS IS JUST AN ATTEMPT
TO HELP YOU BETTER UNDERSTAND PROBABILITY DISTRIBUTIONS)
(3x9)/W
•Last Question - Are any of these “Unusual?”
•“NO” – To be considered “Unusual” the probability would have to be less than 0.05
(3x9)/W
•Hope you enjoyed this… Let me know if these help!
•More examples next week….•Visit me at www.facebook.com/statcave
for Stats•Or www.facebook.com/cranksmytractor
for my column that runs in newspapers in the Southern US