Respuesta :
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 Â Â Â Â Â Â Â Â 2 Â Â Â Â Â Â Â 3
495 Â Â Â 525 Â Â Â Â Â Â 470
500 Â Â Â Â 515 Â Â Â Â Â 480
505 Â Â Â Â 505 Â Â Â Â Â Â 460
500 Â Â Â Â 515 Â Â Â Â Â Â 470 Â Â Â Â
∑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.
