Question

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 4701. What is the sample mean service life for sample 2? A) 460 hours.B) 495 hours.C) 515 hours.D) 525 hours. 2. What is the mean of the sampling distribution of sample means for whenever service life is in control? A) 250 hours.B) 470 hours.C) 495 hours.D) 500 hours.E) 515 hours. 3. 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.

Answer 1

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1                2              3

495      525            470

500         515           480

505        505            460

500         515             470        

∑2000     2060         1880

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4=  470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using  μ ± σ , μ±2 σ  or μ ± 3 σ.

In this question the limits are determined by using  μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4=  470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other  two means are within the given UCL and LCL.

This can be shown by the diagram.