Downloaded from https://www.everyspec.com AMCP 706.196 Example No. 11 Use the relay data in Example No. 10 in par. 2-4. 1. See if they might reasonably have come from an exponential distribution, i. e., FP,(x) - 1 - exp(- x/#). For aim- plicity of notation, define F= Fb*(Xi). Procedure ram-We 1. Calculate the sample 1. Sample mean =. 16, 994 w 0. mean, 2. Find the critical value 2. For N = 50, and s-confidence '. 90% we have of d from Table 2-12; d - 0.17. 90% s -confidence seems reasomnble, 3. Since we estimated the 3. Newd 0.17 x 0.8 -0.14. exponential parameter from the data, multiply d Oy 0. 4. Begin with lowest value, 4. FP - 1 - exp(- first failure/sanmle mean) t -- 1.
(F 4 denotei the Fj - 1 - exp(- 1283/16994) - 0.073. actual Cdf at sample- point L.) 0/50 - 0.00; 1/50 - 0.02. di - 0.073 0.14; OK. Ba. Calculate iA and ij from i.. 0 1 Eqs. 2-46 and 2-47. (J/50) - 0.14 - = 0.073 Use the i to denote the actual intersections in i . 10%, A - 11 Fig. 2-12(B). They will generally be frac- F 4 5 (1/50) + 0.14 - 0.10 tional values. F(X 1 ) 0 0. 16 0. 16 1 - exp(- xg /1,994) x, - 2962, a - 5. b. Find the next io b. io w min{b, 111, 5, x5 - 3437. =mlninA, dal; evaluate i F 4 . Compare with l 0aN Fs - 0. 183 (by same procedure as in stop 4) and with (YO - 1)/N. i 0 /N - 6/50 - 0. 10; (UO - 1)/N 0 0. 08.
The maximum difference is dg -0.103 which wo is less than 0.14; OK.