If f(x) = 6x – 9, evaluate f(x)/3 for x = -5 a. –7 b. –13 c. 7 d. 13
1 answer:
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Answer:
Given the data in the question;
Samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.
a)
For a p chart ( control chart for fraction nonconforming), the center line and upper and lower control limits are;
UCL = p" + 3√[ (p"(1-P")) / n ]
CL = p"
LCL = p" - 3√[ (p"(1-P")) / n ]
here, p" is the average fraction defective
Now, with the 30 samples of size 100
p" = [∑(6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.)] / [ 30 × 100 ]
p" = 234 / 3000
p" = 0.078
so the trial control limits for the fraction-defective control chart are;
UCL = p" + 3√[ (p"(1-P")) / n ]
UCL = 0.078 + 3√[ (0.078(1-0.078)) / 100 ]
UCL = 0.078 + ( 3 × 0.026817 )
UCL = 0.078 + 0.080451
UCL = 0.1585
LCL = p" - 3√[ (p"(1-P")) / n ]
LCL = 0.078 - 3√[ (0.078(1-0.078)) / 100 ]
LCL = 0.078 - ( 3 × 0.026817 )
LCL = 0.078 - 0.080451
LCL = 0 ( SET TO ZERO )
Diagram of the Chart uploaded below
b)
from the p chart for a) below, sample 28 violated the first western electric rule,
summary report from Minitab;
TEST 1. One point more than 3.00 standard deviations from the center line.
Test failed at points: 28
Hence, we conclude that the process is out of statistical control
Lets Assume that assignable causes can be found to eliminate out of control points.
Since 28 is out of control, we should eliminate this sample and recalculate the trial control limits for the P chart.
so
p" = 0.0745
UCL = p" + 3√[ (p"(1-P")) / n ]
UCL = 0.0745 + 3√[ (0.0745(1-0.0745)) / 100 ]
UCL = 0.0745 + ( 3 × 0.026258 )
UCL = 0.0745 + 0.078774
UCL = 0.1532
LCL = p" - 3√[ (p"(1-P")) / n ]
LCL = 0.0745 - 3√[ (0.0745(1-0.0745)) / 100 ]
LCL = 0.0745 - ( 3 × 0.026258 )
LCL = 0.0745 - 0.078774
UCL = 0 ( SET TO ZERO )
The second p chart diagram is upload below;
NOTE; the red circle symbol on 28 denotes that the point is not used in computing the control limits
