Aprocess is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the control chart, so that the probability of a single point plotting outside the control limits when the process is in control is 0.0027. suppose that this chart is being used in phase i and the averages from a set of m samples or subgroups from this process are plotted on this chart. what is the probability that at least one of the averages will plot outside the control limits when m = 5? discuss the results that you have obtained.

First locate the turning point (vertex) of the graph. which is at the coordinates x=5 and y=9 or (5; 9).to determine whether it's a maximum or mininum value, you have to discern whether the vertex's y-value is the highest value or lowest value the graph can reach. since the vertex's y-value, 9 is the highest value the graph can reach, it will be a maximum.another basic way to determine whether the vertex's value is max or min is to see if the graph is smiling or sad. if it's smiling, the turning point/vertex would be a minimum, while a sad parabola would have a maximum y-value at the vertex.