Template:Example: Standard Actuarial Example - Revision history

Lisa Hacker: Redirected page to Non-Parametric Life Data Analysis

2012-08-10T08:10:37Z

Redirected page to Non-Parametric Life Data Analysis

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	[[~~Category: For Deletion~~]]		#REDIRECT [[Non-Parametric_Life_Data_Analysis]]

Kate Racaza: Replaced content with 'Category: For Deletion'

2012-07-24T07:36:08Z

Replaced content with 'Category: For Deletion'

← Older revision		Revision as of 07:36, 24 July 2012
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	~~'''Standard Actuarial Example'''~~		[[Category: For Deletion]]

	~~Find reliability estimates for the data in the~~ [[~~Simple-Actuarial Example Data\|Actuarial-Simple Example]~~] ~~using the actuarial-standard method.~~

	~~<br>'''Solution'''~~

	~~The solution to this example is similar to that in the [[Simple-Actuarial Example Data\|Actuarial-Simple Example~~]], with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in the equation for the actuarial-standard method. Applying this equation to the data, we can generate the following table:
	~~<center><math>\begin{matrix}~~
	~~Start & End & Number of & Number of & Adjusted & {} & {} \\~~
	~~Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \prod\mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\~~
	~~0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\~~
	~~50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\~~
	~~100 & 150 & 2 & 2 & 43 & 0.953 & 0.918 \\~~
	~~150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844 \\~~
	~~200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791 \\~~
	~~250 & 300 & 1 & 2 & 28 & 0.964 & 0.762 \\~~
	~~300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702 \\~~
	~~350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604 \\~~
	~~400 & 450 & 3 & 4 & 15 & 0.800 & 0.484 \\~~
	~~450 & 500 & 1 & 2 & 9 & 0.889 & 0.430 \\~~
	~~500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298 \\~~
	~~550 & 600 & 1 & 0 & 4 & 0.750 & 0.223 \\~~
	~~600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\~~
	~~\end{matrix}</math></center>~~
	~~<br>As can be determined from the preceding table, the reliability estimates for the failure times are:~~
	~~<center><math>\begin{matrix}~~
	~~Failure Period & Reliability \\~~
	~~End Time & Estimate \\~~
	~~50 & 96.2% \\~~
	~~150 & 91.8% \\~~
	~~200 & 84.4% \\~~
	~~250 & 79.1% \\~~
	~~300 & 76.2% \\~~
	~~350 & 70.2% \\~~
	~~400 & 60.4% \\~~
	~~450 & 48.4% \\~~
	~~500 & 43.0% \\~~
	~~550 & 29.8% \\~~
	~~600 & 22.3% \\~~
	~~650 & 4.5% \\~~
	~~\end{matrix}</math></center>~~

Kate Racaza at 22:18, 9 March 2012

2012-03-09T22:18:34Z

← Older revision		Revision as of 22:18, 9 March 2012
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	'''Standard Actuarial Example'''		'''Standard Actuarial Example'''

	Find reliability estimates for the data in the [[Simple-Actuarial Example Data\|Simple~~-Actuarial~~ Example]] using the standard ~~actuarial~~ method.		Find reliability estimates for the data in the [[Simple-Actuarial Example Data\|Actuarial-Simple Example]] using the actuarial-standard method.

			<br>'''Solution'''

	~~'''Solution'''~~		The solution to this example is similar to that in the [[Simple-Actuarial Example Data\|Actuarial-Simple Example]], with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in the equation for the actuarial-standard method. Applying this equation to the data, we can generate the following table:

	The solution to this example is similar to that of [[Simple-Actuarial Example Data\|Simple~~-Actuarial~~ Example]], with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in equation for the standard ~~actuarial~~ method. Applying this equation to the data, we can generate the following table:

	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	Start & End & Number of & Number of & Adjusted & {} & {} \\		Start & End & Number of & Number of & Adjusted & {} & {} \\
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	600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\		600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\
	\end{matrix}</math></center>		\end{matrix}</math></center>
			<br>As can be determined from the preceding table, the reliability estimates for the failure times are:

	As can be determined from the preceding table, the reliability estimates for the failure times are:

	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	Failure Period & Reliability \\		Failure Period & Reliability \\

Harry Guo at 23:28, 5 March 2012

2012-03-05T23:28:19Z

← Older revision		Revision as of 23:28, 5 March 2012
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	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	Start & End & Number of & Number of & Adjusted & {} & {} \\		Start & End & Number of & Number of & Adjusted & {} & {} \\
	Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\		Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \prod\mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\
	0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\		0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\
	50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\		50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\

Harry Guo at 23:53, 20 February 2012

2012-02-20T23:53:32Z

← Older revision		Revision as of 23:53, 20 February 2012
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	Find reliability estimates for the data in the [[Simple-Actuarial Example Data\|Simple-Actuarial Example]] using the standard actuarial method.		Find reliability estimates for the data in the [[Simple-Actuarial Example Data\|Simple-Actuarial Example]] using the standard actuarial method.


	'''Solution'''		'''Solution'''

Harry Guo at 23:53, 20 February 2012

2012-02-20T23:53:23Z

← Older revision		Revision as of 23:53, 20 February 2012
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	'''Standard Actuarial Example'''		'''Standard Actuarial Example'''

	Find reliability estimates for the data in Example 10 using the standard actuarial method.		Find reliability estimates for the data in the [[Simple-Actuarial Example Data\|Simple-Actuarial Example]] using the standard actuarial method.

	'''Solution'''		'''Solution'''

	The solution to this example is similar to that of Example 10, with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in equation for the standard actuarial method. Applying this equation to the data, we can generate the following table:		The solution to this example is similar to that of [[Simple-Actuarial Example Data\|Simple-Actuarial Example]], with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in equation for the standard actuarial method. Applying this equation to the data, we can generate the following table:

	<center><math>\begin{matrix}		<center><math>\begin{matrix}

Harry Guo at 23:45, 20 February 2012

2012-02-20T23:45:21Z

← Older revision		Revision as of 23:45, 20 February 2012
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	'''Solution'''		'''Solution'''

	The solution to this example is similar to that of Example 10, with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in ~~Eqn. (standact)~~. Applying this equation to the data, we can generate the following table:		The solution to this example is similar to that of Example 10, with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in equation for the standard actuarial method. Applying this equation to the data, we can generate the following table:

	<center><math>\begin{matrix}		<center><math>\begin{matrix}

Harry Guo at 19:06, 20 February 2012

2012-02-20T19:06:29Z

← Older revision		Revision as of 19:06, 20 February 2012
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	As can be determined from the preceding table, the reliability estimates for the failure times are:		As can be determined from the preceding table, the reliability estimates for the failure times are:

			<center><math>\begin{matrix}
			Failure Period & Reliability \\
			End Time & Estimate \\
			50 & 96.2% \\
			150 & 91.8% \\
			200 & 84.4% \\
			250 & 79.1% \\
			300 & 76.2% \\
			350 & 70.2% \\
			400 & 60.4% \\
			450 & 48.4% \\
			500 & 43.0% \\
			550 & 29.8% \\
			600 & 22.3% \\
			650 & 4.5% \\
			\end{matrix}</math></center>

Harry Guo: Created page with ''''Standard Actuarial Example''' Find reliability estimates for the data in Example 10 using the standard actuarial method. '''Solution''' The solution to this example is simi…'

2012-02-20T19:03:50Z

Created page with ''''Standard Actuarial Example''' Find reliability estimates for the data in Example 10 using the standard actuarial method. '''Solution''' The solution to this example is simi…'

New page

'''Standard Actuarial Example'''

Find reliability estimates for the data in Example 10 using the standard actuarial method.

'''Solution'''

The solution to this example is similar to that of Example 10, with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:

<center><math>\begin{matrix}
Start & End & Number of & Number of & Adjusted & {} & {} \\
Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\
0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\
50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\
100 & 150 & 2 & 2 & 43 & 0.953 & 0.918 \\
150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844 \\
200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791 \\
250 & 300 & 1 & 2 & 28 & 0.964 & 0.762 \\
300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702 \\
350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604 \\
400 & 450 & 3 & 4 & 15 & 0.800 & 0.484 \\
450 & 500 & 1 & 2 & 9 & 0.889 & 0.430 \\
500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298 \\
550 & 600 & 1 & 0 & 4 & 0.750 & 0.223 \\
600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\
\end{matrix}</math></center>

As can be determined from the preceding table, the reliability estimates for the failure times are: