The fill volume of cans filled by a certain machine is normally distributed with

Question

The fill volume of cans filled by a certain machine is normally distributed with mean 12.05oz and standard deviation 0.03oz.
a. What proportion of cans contain less than 12oz?
b. Of a 100 can sample, what is the probability that at least 97 of the cans contain at least 12 oz?
c. The process mean can be adjusted through calibration. To what value should the mean be set so that 99% of the cans will contain 12oz or more?
d. Now that the machine has been calibrated, of a 100 can sample, what is the probability that at least 97 of the cans contain at least 12 oz?

Explanation / Answer

We have a normal distribution with µ = 12.15 and s = 0.10. The z score formula is z = (x - µ)/s Hence, the z score for 12 is z = (12 - 12.15)/0.1 = -0.15/0.1 = -1.5 For a standard normal distribution, p(z < -1.5) = 0.066807201 So roughly 6.7% of the cans have less than 12 ounces.

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