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

P(X<x)
Mean Stdev x

12.5 2.70 0.4265 12

Week 3 GM 533
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 Area to the Area to the
Distribution Left Right Area between

Mean Stdev P(X<x) x P(X>x) x x1 P(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 Area to the Area
Left Right between

15.32 3.07 0.1792 12.5 0.9584 10

Week 3 GM 533
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 Area to Area to the Area
Distribution the Left Right between

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 for
the Standard Deviation. The
program just rounds. The correct
value 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


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


Mean Stdev P(X<x) x
0.1 0.02 0.2894 0.089

Note: This is not an error. Excel
just rounded the 0.0889 to 0.089.
The correct value is still there.

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.

Week3 lecture gm533

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