Please see the attached file.

Solution 1
Here it is given that the process standard deviation s = 3, subgroup size n = 5, LCL = 14.15 and UCL = 15.85
Now the Central Line (mean) is CL =(UCL+LCL)/2 =15
Let denote the process mean. Suppose the process mean shift from 15 to 15.2.
Let and denote the sample mean of first and second subsequent sample of 5 observations.
Now the probability that the shift will be detected on first subsequent sample of 5 observations is given by,

, where

= P[Z > 0.48448]
= 1- P[Z< 0.48448] = 1 - 0.686
= 0.314
The probability that the shift will be detected on the second subsequent sample of 5 observations is given by

= P[Z < 0.48448]. P[Z > 0.48448]
= P[Z < 0.48448]. (1-P[Z < 0.48448])
= 0.686*0.314
= 0.2154

Solution 2
Here it is given that the central line of the s chart is .
So an estimate of the process standard deviation is given by

Thus the Natural tolerance = 6 = 6*8.2519 = 49.5115
Here it is given that the lower and upper specification limits of the process as L= 496 and U = 506.
So the Specified tolerance = 506-496 = 10.
Since the natural tolerance is greater than the specified tolerance, the process doesn't meet the specification.

where

= 0.16615 +0.07294 = 0.2391
Thus the 23.91 % of the items are nonconforming.

The average run length (ARL) is defined as the expected number of time periods we have before our chart signals (i.e. sample mean is above UCL or below LCL). To determine the ARL we use the following formula
ARL
where
So if the process mean is shifted to 493, the ARL is given by,
ARL
where

= 0.23358+0.04489 = 0.27847
Thus ARL = 1/0.27847 = 3.591 = 4 (approximately)