Answer:
Sample number 3
Step-by-step explanation:
From the given information:
Sample Service life(hours) Total Mean(X)
1 2 3 4
1 495 500 505 500 2000 500
2 525 515 505 515 2060 515
3 470 480 460 470 1880 470
Total =
Mean =
Thus;
To plot on an X_Bar chart, we have:
Sample Mean (X) UCL LCL
1 500 520 480
5 515 520 480
6 470 520 480
The x-Bar chart is shown in the image attached below. From the image, we realize that the average service life for sample number 3 occurs to be out of the statistical control.
a/b=5/4
a=5k
b=4k
b/c=7/3
b=7x
c=3x
b=4k
b=7x
4k=7x =>
k=7y
x=4y
a/b/c=
=5*7*y/4*7*y/3*4*y=
=35y/28y/12y=
=35/28/12y=
=5/4/12y=
=1,25/12y=
=0.1041666666666667/ y
there may be a mistake in my answer
Check the picture below.
keeping in mind that, log(x) has a domain of x > 0, namely, it will never spit out a number that's 0 or negative, therefore, both graphs go down down and down ever closer and closer to x = 0, but never getting there, and while racing down, f(x) is above g(x), thus is higher.