week3 lecture gm533

Week 3 Review for GM 533

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 attempt to transfer an Excel File here in Webex

– 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 Lake Big Catch 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 Lake Big Catch 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 Lake Big Catch 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 Lake Big Catch 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 Lake Big Catch at random will be between 11.3 than 13.7 inches long?

Week 3 GM 533

• What is the probability that a fish caught in Lake Big Catch 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 Lake Big Catch 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 Lake Big Catch 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 – 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 0.53

Normal

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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