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2 answers:
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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
<u>500 515 470 </u>
<u>∑2000 2060 1880</u>
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.
Answer:
<u>Step-by-Step Explanation:</u>
KCFmeans: Keep, Change, Flip
- Keep: the first fraction
- Change: the sign <em>from division to multiplication</em>
- Flip: the second fraction
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= <em>KCF is applied here!</em>
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