Question 1204858: A vending machine is designed to dispense a mean of 7.6 oz of coffee into an 8-oz cup. If the standard deviation of the amount of coffee dispensed is 0.3 oz and the amount is normally distributed, find the percent of times the machine will dispense less than 7.62 oz.
Answer by ikleyn(52778)   (Show Source):
You can put this solution on YOUR website! .
A vending machine is designed to dispense a mean of 7.6 oz of coffee into an 8-oz cup.
If the standard deviation of the amount of coffee dispensed is 0.3 oz and the amount
is normally distributed, find the percent of times the machine will dispense less than 7.62 oz.
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This "percent of times", in mathematical language, is called "the probability".
It is the area under this specified normal curve on the left of the raw mark z= 7.62.
To find it, you may use your regular calculator TI-83/84, namely, its standard function
"normalcdf". The suffix "cdf" stands for "cumulative distribution function".
The format is
area = normalcdf(z1, z2, mean, SD).
In this problem z1 = -9999 (emulates "-oo", or negative infinity); z2 = 7.62, mean = 7.6, SD = 0.3.
You get then
area = normalcdf(-9999, 7.62, 7.6, 0.3) = 0.5266.
It is your ANSWER. The probability is 0.5266.
It is the same as to say "the percent of times" is 52.66%.
Alternatively, you may use online free of charge calculator at this web-site
https://onlinestatbook.com/2/calculators/normal_dist.html
which is intended for the same purposes.
This online calculator has perfect and intuitively clear interface, so even a beginner student
can use it without any difficulties. Surely, this online calculator will get you the same answer.
The calculator shows the area of the interest, so after using it, you will understand the meaning
of the calculation procedure in full. This online calculator is a perfect teacher, at the same time.
To be honest, I personally never use calculators TI-83/84 for such problems, and use exclusively
this referred online calculator. But in school, they probably require using regular calculators.
Solved.
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