gm533 week 3 lecture march 2012
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Week 3 GM 533 Lecture ChartsTRANSCRIPT
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Week 3 Lecture
GM 533
Applied Managerial Statistics
B. Heard(These may not be copied, reproduced, or posted in an online classroom without my permission. Students may download one copy for personal use.)
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Week 3 GM 533
• Your Online Course explains how to do Normal Distribution calculations in Minitab
• I am going to share with you an approach I used with a visually impaired student who could not “see” Minitab to use it.
• You can use Minitab or Excel with these calculations, I simply like to expose you to different methods of solving problems
• Being able to use Excel is important, because almost every person has access to Excel in the workplace
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Week 3 GM 533
• At this time, I am going to request that you download an Excel File from a link to Google Docs at the “Statcave”
▫ The file is titled Normal_Distribution_Made_EZ
▫ It is a template to do Normal Distribution calculations
▫ You DON’T have to use this in your work, but you may find it very helpful
▫ I use it at work, I know it works correctly
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Week 3 GM 533
• Most of your problems on your Checkpoint quiz this week deal with the Normal Distribution
• The following examples should help you a lot!
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Week 3 GM 533
• The population of fish lengths in “Lazy Lake” is normally distributed with a mean of 12.5 inches and a standard deviation of 2.7 inches.What is the probability that a fish caught in Lazy Lake at random will be less than 12 inches long?
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Week 3 GM 533• Open the Excel File
Normal_Distribution_Made_EZ
• Input 12.5 for the Mean and 2.7 for the Standard Deviation
Normal Distribution
Mean Stdev
12.5 2.70
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Week 3 GM 533
• What is the probability that a fish caught in Lazy Lake at random will be less than 12 inches long?
• We want to know the area to the left because we want to know the probability a fish will be less than 12 inches.
• Enter 12 in the green box under the x in the boxes under “Area to the Left”
• Never enter any values in areas that are not green – they are there to do calculations
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Week 3 GM 533• We can see the answer is 0.4265, that is the probability a
fish would be less than 12 inches given that the distribution is normal with the given mean and standard deviation!
Normal Distribution Area to the Left
Mean StdevP(X<x)
x
12.5 2.70 0.4265 12
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Week 3 GM 533
• The population of fish lengths in Lazy Lake is normally distributed with a mean of 12.5 inches and a standard deviation of 2.7 inches.What is the probability that a fish caught in Lazy Lake at random will be between 11.3 and 13.7 inches long?
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Week 3 GM 533
• What is the probability that a fish caught in Lazy Lake at random will be between 11.3 than 13.7 inches long?
• We want to know the area between because we want to know the probability a fish will be between 11.3 and 13.7 inches long.
• Enter 11.3 on the left and 13.7 on the right in the green boxes under the x1 and x2 in the boxes under “Area between”
• Never enter any values in areas that are not green –they are there to do calculations
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Week 3 GM 533• We can see that the probability would be 0.3433
or 34.33% that a fish is between 11.3 and 13.7 inches
Normal
Distribution
Area to the
Left
Area to the
Right Area between
Mean Stdev P(X<x) x P(X>x) x x1P(x1<X<x2) x2
12.5 2.70 11.30 0.3433 13.70
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Week 3 GM 533
• Pretty Cool Eh?
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Week 3 GM 533
• Over the last year, 87% of Americans bought something that was “not on their list” at the grocery store. Assume these purchases were normally distributed. The mean amount spent on these items was $15.32 with a standard deviation of 3.07
• Find the probability someone spent less than $12.50 on “non-list” items.
• Find the probability someone spent more than $10.00 on “non-list” items.
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Week 3 GM 533• You can see the answers below
Area to the
Left
Area to the
Right
Area
between
Mean Stdev P(X<x) x P(X>x) x x1P(x1<X<x2) x2
15.32 3.07 0.1792 12.5 0.9584 10
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Week 3 GM 533
• The population of fish lengths in Lazy Lake is normally distributed with a mean of 12.5 inches and a standard deviation of 2.7 inches. A sample of 9 fish were caught randomly from the lake.What is the probability that of those fish caught in Lazy Lake at random will average between 12.2 than 13.1 inches long?
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Week 3 GM 533
• This one is a little different because we are dealing with a sample of 9
▫ We need to adjust our standard deviation by dividing it by the square root of the sample size
▫ We will use 2.7/√9 = 2.7/3 = 0.9
▫ Use 0.9 as your standard deviation
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Week 3 GM 533
• So the probability would be 0.3781 or 37.81%
Normal
Distribution
Area to
the Left
Area to the
Right
Area
between
Mean Stdev P(X<x) x P(X>x) x x1P(x1<X<x2) x2
12.5 0.90 12.20 0.3781 13.10
If you are taking a multiple choice test, sometimes your answer may be slightly different due to rounding. – YOU WILL BE ABLE TO TELL THE CORRECT ANSWER.
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Week 3 GM 533
• You can also use the Excel Normal Distribution Calculator File to work with Proportions. Just read the questions carefully.
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Week 3 GM 533
• In a recent telephone survey among Happy County residents, 1000 residents participated. Based on the survey, it was predicted that 53% of residents approve of a new city park. For argument’s sake, assume that 55% of the residents in the county support the new park (p = 0.55). Calculate the probability of observing a sample proportion of residents 0.53 or higher supporting the new park. We are assuming a normal distribution.
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Week 3 GM 533
• We must first calculate the standard deviation
▫ It is calculated using the following formula
▫ Square Root ((p)(q)/sample size)
▫ q is just 1 minus p
▫ So we have
Square Root ((0.55)(1 – 0.55)/1000)
Which is Square Root ((0.55)(0.45)/1000)
Which is Square Root (0.0002475)
Which is 0.0157 USE THIS VALUE FOR YOUR STANDARD DEVIATION AND p FOR YOUR MEAN
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Week 3 GM 533• As you see, we get 0.8986 for the probability
being over 53% or 0.53Normal
Distribution Area to the Left Area to the Right
Mean Stdev P(X<x) x P(X>x) x
0.55 0.02 0.8986 0.53
Note: I did input 0.0157 forthe Standard Deviation. Theprogram just rounds. The correctvalue is still there.
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Week 3 GM 533
• Here is a similar problem where you have to do a little more math.
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Week 3 GM 533
• The local Animal Recovery Center notes that over the past 12 years, studies have shown that 10 % of adopted pets are returned. The local university’s polling group just conducted a study of 225 adoptions from the Animal Recovery Center.What is the probability that less than 30 adoptions resulted in the pet being returned?
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Week 3 GM 533
• We must first calculate the standard deviation▫ It is calculated using the following formula▫ Square Root ((p)(q)/sample size)▫ q is just 1 minus p▫ Our “p” is 10% or 0.10 based on the study▫ So we have Square Root ((0.10)(1 – 0.10)/225)
Which is Square Root ((0.10)(0.90)/225)
Which is Square Root (0.0004)
Which is 0.02 USE THIS VALUE FOR YOUR STANDARD DEVIATION AND p FOR YOUR MEAN
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Week 3 GM 533
• We must also calculate our critical value or value that we are interested in.
▫ We wanted to know the probability of less than 30 of the 225 returning their pets.
30/225 is 0.1333 (13.33%)
We will use this value to find the probability of less that 13.33% of the pets being returned.
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Week 3 GM 533• From the spreadsheet, we can see the probability
is 0.9520
Normal Distribution Area to the Left
Mean Stdev P(X<x) x
0.1 0.02 0.9520 0.133
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Week 3 GM 533• What about less than 20 of the 225 are returned?
• The only thing that changes is the value or proportion we are interested in.
• It is now 20/225 is 0.0889 or 8.89%
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Week 3 GM 533• The probability would be 0.2894
Normal Distribution Area to the Left
Mean Stdev P(X<x) x
0.1 0.02 0.2894 0.089
Note: This is not an error. Exceljust rounded the 0.0889 to 0.089.The correct value is still there.
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Week 3 GM 533
• Come see me at the “Statcave”
• www.facebook.com/statcave
• You DO NOT have to be a Facebook person to see these.
• If you are, become a fan.
• IT IS NOT REQUIRED TO BE ON FACEBOOK. IT’S SOMETHING I DO FOR FUN.
• I post charts there because it is easy for me to do.