When a machine for dispensing a cola drink into bottles is in statistical control, the amount dispensed has a mean of 500 ml (milliliters) and a standard deviation of 4 ml.
a. In constructing a control chart to monitor this process with periodic samples of size 4, how would you select the target line and the upper(3sd above mean) and lower control limits(3sd below mean)?
I have no problem with this question. By simple calculation, mean=500, UL=506 and LL=494 (with sd=2). Its the next question that bugs me...
b. If the process actually deteriorates and operates with a mean of 491 ml and a standard deviation of 6 ml, what is the probability that the next value plotted on the control chart indicates a problem with the process, falling more than 3 standard deviations from the target?
My thought is that since its 3 sd, I have to use Z-score of 3. From z-score table the probability of the value within 3 sd is about 0.99, thus probability of falling more than 3 sd from mean is about 0.01. However the answer given is 0.83. How do I get this answer instead?
The deteriorated process, sampling four times, has a mean of 491 and a standard deviation of $\frac6{\sqrt4}=3$. We need the probability that the average from this process falls outside the good interval of 494 to 506 that was worked out in the first part.
The z-scores of 494 and 506 are $$z_1=\frac{494-491}3=+1$$ $$z_2=\frac{506-491}3=+5$$ and the probability under the standard normal curve between these two z-values is 0.158655. The probability outside this interval, which is also the probability that a problem is indicated, is thus $1-0.158655=0.841345$.
This is the correct answer. The answer you were given may have used different rounding rules.