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: Sample Service Life (hours)
1 495 500 505 500
2 525 515 505 515
3 470 480 460 470
1) What is the sample mean service life for sample 2?
A. 495 Hours
B. 460 Hours
C. 500 Hours
D. 515 Hours
E. 525 Hours
2) What is the mean of the sampling distribution of sample means when the service life is in control?
A. 250 Hours
B. 470 Hours
C. 515 Hours
D. 495 Hours
E. 500 Hours
3) What is the standard deviation of the sampling distribution of sample means for whenever service life is in control?
A. 20 Hours
B. 10 Hours
C. 11.55 Hours
D. 5 Hours
E. 6.67 Hours
4) If he uses upper and lower control limits of 520 and 480 hours, what is his risk (alpha) of concluding that service life is out of control when it is actually under control (Type I error)?
A. 0.0026
B. 0.3174
C. 0.6826
D. 0.0456
E. 0.9544
5) 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?
A. sample 1
B. sample 2
C. sample 3
D. both samples 2 and 3
E. all samples are in control

1) 515 hours ( D )

2) 495 hours ( D )

3) 10 hours ( B )

4) 0.0456 ( D )

5) sample 3 ( C )

Explanation:

Standard deviation = 20 hours

Mean = 500 hours

1) Determine the sample mean service life for sample 2

Sample mean sample service life of sample 2 = Average of service life of sample 2

= ( 525 + 515 + 505 + 515 ) / 4

= 2060 / 4 =  515

2) Calculate the mean of sampling distribution of sample means

∑ mean service life samples / 3 ( 1 )

sample 1 = (495 + 500 + 505 + 500 ) / 4 =  500

sample 2 = ( 525 + 515 + 505 + 515 ) / 4 = 515

sample 3 = (470 + 480 + 460 + 470) = 470

back to equation 1

= ( 500 + 515 + 470 ) / 3 = 495

3) Determine STD of sampling distribution of sample means

Std = 20

n = 4

∴ std of sampling distribution = 20 / √4 =  10 hours

4) calculating  the risk alpha

upper control  ( UCL ) = 520 , lower control ( LCL ) = 480

mean = 500

Z value for upper control = ( 520 - 500) / 10 = 2

Z value for lower control = ( 480 - 500 ) / 10 = -2

confidence level of -2 and 2 ( using z-table ) =  0.9544

∴ risk alpha = 1 - 0.9544 = 0.0456

5) Service life appears to be out of control in ; sample 3

UCL = 520 , LCL = 480

Because mean value of sample 3 = 470 lies outside the lower control limit

x=3

step-by-step explanation:

this is really simple i saw that there was only one term (5x) and thought if that was 15 what would i have to multiply everything else to get it there? well if 5x became 15 then x would have to be 3 so plug the 3 in the x for the other equations and you get 15 for the other ones. so there you go. : )