Please help me on this question
1 answer:
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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.
we know that
If square GYTD is translated units left and
units up to produce square G'Y'T'D'
that means
the rule of transformation is
<u>Find the coordinates of the image</u>
therefore
<u>the answer is the option</u>
G'(−6, 5), Y'(−2, 5), T'(−2, 1) , and D′(−6, 1)