The UCL and LCL for an x chart are 25 and 15 The central line is 20, and the process variability is considered to be in statistical control. The results of the next six sample means are 18, 23, 17, 21, 24, and 16. What should you do

Complete Question

The UCL and LCL for an X-bar chart are 25 and 15 respectively. The central line is 20, and the process variability is considered to be in statistical control. The results of the next six sample means are 18, 23, 17, 21, 24, and 16. What should you do?

a. Explore the assignable causes because the second, fourth, and fifth samples are above the mean.

b. Explore the assignable causes because there is a trend.

c Nothing: the process is in control.

d Explore the assignable causes because there is a run

The correct option is c

Step-by-step explanation:

From the question we are told that

   The data for the next six means is 18, 23, 17, 21, 24, and 16.

    The central  line is   k =  20  

Generally the mean of the next six mean is mathematically represented as

=>    

Generally given that the mean of the next six means is very close to the central line , it means that the distribution of the process is more centered (i.e it well within the limit of the  central line ), and what this tell us is that the process is in control so we do nothing

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

you answer is a

step-by-step explanation: