SOLUTION: 3. The Edwards's Theater chain has studied its morie customers to determine how much money they spend on concessions. The study revealed that the spending distribution is approxima

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Question 1207223: 3. The Edwards's Theater chain has studied its morie customers to determine how much money they spend on concessions. The study revealed that the spending distribution is approximately normally distributed with a man of 84.11 and a standard deviation of 81.37.
a) What percentage of customers will spend less than $2.50 on concessions"
(b) What percentage of the customers will spend between 83.00 to $4.20 cm concession?
(c) You got free tea if you spend more than 86.00. How many people are likely to get tea if 100 people attended the movie.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the calculator at can help you find the answer to these questions.

here are the results.







this calculator can work directly off the raw scores or off the z-scores.

to show you how it works off the z-scores, we'll take the second problem that calculates the probability of between 4.2 and 83.

to work off z-scores, you set the mean to 0 and the standard deviation to 1.

you find the z-score for 4.2 and the z-score for 83.

z = (x-m)/s is the z-score formula.
z is the z-score.
x is the raw score.
m is the raw mean.
s is the standard deviation.

for raw score of 4.2, formula becomes z = (4.2 - 84.11) / 81.37 = -.982057 rounded to 6 decimal places.

for raw score of 83, formula becomes z = (83 - 84.11) / 81.37 = -.013641 rounded to 6 decimal places.

set the calculator mean to 0 and the calculator standard deviation to 1 and find the probability of getting a z-score between -.982057 and -.013641.

the results are shown below.



the calculator tells you that the probability is .3325.
this is the same as the probability we got earlier when working directly with raw scores.
prior to the user of calculators, you has to get the z-score first and then work off the z-score.
the calculators do that work for you, but it's still useful to know how to work with the z-scores,