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.)

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

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

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!

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?

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

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

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

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?

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

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

Week 3 GM 533

• Pretty Cool Eh?

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.

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

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?

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

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.

Week 3 GM 533

• You can also use the Excel Normal Distribution Calculator File to work with Proportions. Just read the questions carefully.

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.

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

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.

Week 3 GM 533

• Here is a similar problem where you have to do a little more math.

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?

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

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.

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

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%

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.

Week 3 GM 533

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