Assignment On Supplementary Work
Added on 2022-10-04
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Running Head: Supplementary Work
EG-285: Supplementary Work
Student’s Name
Institution Affiliation
EG-285: Supplementary Work
Student’s Name
Institution Affiliation
![Assignment On Supplementary Work_1](/_next/image/?url=https%3A%2F%2Fdesklib.com%2Fmedia%2Fimages%2Fwc%2Fab4df57b8e7e4099b04286e7ea37e5ce.jpg&w=3840&q=10)
Running Head: Supplementary Work
Questions on Data Set 1.
Question one
Actual Times to failures need to have a Poisson distribution.
Question Two
a. less than 1.07 years,
P ( x<1.07 ) =1−e−0.935(1.07)=0.632
b. more than 1.34 years
P ( x>1.34 ) =e−0.935 ( 1.07 )
¿ 0.8 6
c. Between 0.5 and 0.75 years.
P ( 0.5< x <0.75 )= ( 1−e−0.935 (0.75 ) )− ( 1−e−0.935 ( 0.5 ) )
¿ 0.13 1
Question Three
Weibull probability plot
-0.600 -0.400 -0.200 0.000 0.200 0.400 0.600 0.800
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Weibull Probability Plot
X
Y(pdf)
Questions on Data Set 1.
Question one
Actual Times to failures need to have a Poisson distribution.
Question Two
a. less than 1.07 years,
P ( x<1.07 ) =1−e−0.935(1.07)=0.632
b. more than 1.34 years
P ( x>1.34 ) =e−0.935 ( 1.07 )
¿ 0.8 6
c. Between 0.5 and 0.75 years.
P ( 0.5< x <0.75 )= ( 1−e−0.935 (0.75 ) )− ( 1−e−0.935 ( 0.5 ) )
¿ 0.13 1
Question Three
Weibull probability plot
-0.600 -0.400 -0.200 0.000 0.200 0.400 0.600 0.800
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Weibull Probability Plot
X
Y(pdf)
![Assignment On Supplementary Work_2](/_next/image/?url=https%3A%2F%2Fdesklib.com%2Fmedia%2Fimages%2Fjm%2Fde8ebdef5f6844098d1f5b338c9faf6e.jpg&w=3840&q=10)
Running Head: Supplementary Work
Exponential probability plot
0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0.000
0.100
0.200
0.300
0.400
0.500
0.600 Exponetial probability plot
X
Y (pdf)
From the two charts, it’s evident that the exponential probability plot forms a linear
pattern as opposed to Weibull that forms a non-linear pattern. This suggests that the
times between failures are exponentially distributed as opposed to Weibull.
Question Four
i. The 63rd quantile of the times between failure s
The 63rd quantile of the times between failures is 1.19
ii. ii. the value for .
Question Five
The confidence interval is given by
CI =Q63 ± z0.95∗S
√n , where s=standard deviation , n=smaple ¿ ¿
Q=quantile , z0.95=zscore at 95 % , n=smaple ¿ ¿
Q63=1.19 years
Exponential probability plot
0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0.000
0.100
0.200
0.300
0.400
0.500
0.600 Exponetial probability plot
X
Y (pdf)
From the two charts, it’s evident that the exponential probability plot forms a linear
pattern as opposed to Weibull that forms a non-linear pattern. This suggests that the
times between failures are exponentially distributed as opposed to Weibull.
Question Four
i. The 63rd quantile of the times between failure s
The 63rd quantile of the times between failures is 1.19
ii. ii. the value for .
Question Five
The confidence interval is given by
CI =Q63 ± z0.95∗S
√n , where s=standard deviation , n=smaple ¿ ¿
Q=quantile , z0.95=zscore at 95 % , n=smaple ¿ ¿
Q63=1.19 years
![Assignment On Supplementary Work_3](/_next/image/?url=https%3A%2F%2Fdesklib.com%2Fmedia%2Fimages%2Ffh%2F1401d0d16b6246c2a9fceec168a2f333.jpg&w=3840&q=10)
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