Asked by jennifer
a vending machine dispenses coffee into an 8 oz cup. the amount of coffee dispensed into the cup is normally distributed with a standard deviation of 0.03 oz.you can allow the cup to overfill 1 percent of the time. what among should you set as the mean amount of coffee to be dispensed?
Answers
Answered by
drwls
Using a normal distribution function, select the value of x(mean) that makes the probability integral of P(x) from x = 0 to 8.0 equal to 0.99, when sigma = 0.03
Answered by
drwls
You may find the JAVA tool at this website for problems of this type.
http://psych-www.colorado.edu/~mcclella/java/normal/accurateNormal.html
I used it to get an answer of 7.93 oz.
Enter mean of 7.93, sigma = 0.03 and the range of P(x) 0 to 8.0 oz, and you get a probability of 0.990
http://psych-www.colorado.edu/~mcclella/java/normal/accurateNormal.html
I used it to get an answer of 7.93 oz.
Enter mean of 7.93, sigma = 0.03 and the range of P(x) 0 to 8.0 oz, and you get a probability of 0.990
Answered by
Brad
Use Norminv in excel(%, cup size, stanard dev ) then round to the 2nd decimal
hope this helpful
hope this helpful
Answered by
stacy
Find the z-value with a right tail of 1%: z = 2.3263
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Find the x-value using x = zs+u
x = 2.3263*0.01+8 = 8.0233
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Let that be the mean amount of coffee to be dispensed
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Find the x-value using x = zs+u
x = 2.3263*0.01+8 = 8.0233
========
Let that be the mean amount of coffee to be dispensed
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