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?
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
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
Use Norminv in excel(%, cup size, stanard dev ) then round to the 2nd decimal
hope this helpful
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
To determine the mean amount of coffee to be dispensed, we need to calculate the cutoff value for the 1 percentile of overfill.
Step 1: Find the z-score corresponding to the percentile value.
Using a standard normal distribution table or a calculator, find the z-score for a cumulative probability of 0.01. The corresponding z-score is approximately -2.33.
Step 2: Calculate the cutoff value.
The formula for calculating the cutoff value is:
cutoff value = mean + (z-score * standard deviation)
Given that the standard deviation is 0.03 oz and the z-score is -2.33, we can now calculate the cutoff value:
cutoff value = mean + (-2.33 * 0.03)
Step 3: Solve for the mean.
Since we want to allow the cup to overfill 1 percent of the time, the cutoff value should be set as the mean amount of coffee to be dispensed. Therefore, we need to solve for the mean in the equation:
mean + (-2.33 * 0.03) = mean
Simplifying the equation:
-0.0699 = 0
The mean value cancels out, leaving a constant equal to zero. This equation implies that there is no solution for the mean that satisfies the given conditions.
Therefore, setting the mean amount of coffee to be dispensed such that the cup overfills 1 percent of the time is not possible in this scenario.