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A cola-dispensing machine is set to dispense $9.00$ ounces of cola per cup, with a standard deviation of $1.00$ ounces. The manufacturer of the machine would like to set the control limit in such a way that, for samples of $36$, $5$ percent of the sample means will be greater than the upper control limit, and $5$ percent of the sample means will be less than the lower control point.

a) At what value should the control limit be set?

b) What is the probability that if the population mean shifts to 8.9, this change will not be detected?

c) What is the probability that if the population mean shifts to 9.3, this change will not be detected?

Bombyx mori
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2 Answers2

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The SD of the sample mean is $1/\sqrt{36}=1/6$. The $z$ value corresponding to $0.95$ probability is $\approx 1.645$. So the answer to question a is $9 \pm 1.645/6 \approx 8.725-9.275$.

For question b, we must compute $P(8.725 < x < 9.275)$ taking into account that x is normally distributed with mean $8.9$ and SD $1$. The cumulative probabilities corresponding to $z$ values of $-0.175$ and $0.375$ are $0.431$ and $0.646$. The difference is $0.215$.

For question c, we must compute $P(8.725 < x < 9.275)$ taking into account that x is normally distributed with mean $9.3$ and SD $1$. The cumulative probabilities corresponding to $z$ values of $-0.575$ and $-0.025$ are $0.028$ and $0.490$. The difference is $0.462$.

Anatoly
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In part (c) the answer will be 0 . This is because 9.3 lies outside the control limits i.e. is 8.725 and 9.25 so there will be 0 probability of its occurrence.

Nishant kapoor
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