An Xത chart with three-sigma limits has parameters as follows: UCL = 104; Center Line = 100; LCL = 96; n = 5 Suppose the process quality characteristic being controlled is normally distributed with a true mean of 98 and a standard deviation of 8. What is the probability that the control chart would exhibit lack of control by at least the third point plotted?

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

Step-by-step explanation:

We consider "lack of control by at least the third point plotted" if at least one of the three first points is over the UCL or under the LCL.

The probability of one point of being over UCL=104 is:

The probability of one point of being under LCL=96 is:

Then, the probability of exhibit lack of control is:

The probability of having at least one point out of control in the first three points is:

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

28

step-by-step explanation:

8+8+6+6

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.