A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod
1 answer:
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.
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Answer:
The voltage in exponential form would be .
The original voltage in the circuit is of a volt.
Step-by-step explanation:
Let x represent the original voltage.
We have been given that the voltage in an electrical circuit is multiplied by itself each time it is reduced. The voltage is 27/125 of a volt and it has been reduced 3 times.
The original voltage multiplied by itself for 3 times would be .
Now, we will equate with 27/125 as:
We know that 27 is cube of 3 and 125 is cube of 5, so we can represent our equation as:
Therefore, the voltage in exponential form would be
Now, we will take cube root of both sides to find original voltage in the circuit as:
Using property , we will get:
Therefore, the original voltage in the circuit is of a volt.