A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company produces. He knows from numerous previous samples that this service life is normally distributed with a mean of 500 hours and a standard deviation of 20 hours. On three recent production batches, he tested service life on random samples of four headlamps, with these results SampleService Life (hours) 1495500505500 2525515505515 3470480460470 If he uses upper and lower control limits of 520 and 480 hours, on what sample(s) (if any) does service life appear to be out of control

Sample number 3

Step-by-step explanation:

From the given information:

Sample                   Service life(hours)                Total        Mean(X)

                     1               2            3           4

1                 495         500       505       500       2000        500

2                 525         515       505       515         2060        515

3                 470         480       460       470        1880          470

Total =

Mean =

Thus;

To plot on an X_Bar chart, we have:

Sample        Mean (X)     UCL        LCL

1                      500           520        480

5                     515            520        480

6                     470            520        480

The x-Bar chart is shown in the image attached below. From the image, we realize that the average service life for sample number 3 occurs to be out of the statistical control.

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation: