With the expected value (the mean =0.89), provide a conclusion about what this numerical information is telling you, using the data below.

If you were a company putting out recommendations to the consumer as to the longevity of the parts, how would you explain this information to them? Would you suggest that they replace this part before or after this time?

Keep in mind that the value obtained for the expected value is really a time value. picture below for reference

G43 Compute the expected value & variance. X fx f (t) = BtB-1e(-t) B A B C D E F G H 1 Trail run Time of failure Sort X In(x) F(x) In(-In(1-F(x))) 1 18 1 0 0.013889 4.269681149 B = 1.73 2 19 5 1.609438 0.041667 3.156849494 B 1.73 f(t) = 1.73the(-t) 1.73 3 10 10 2.302585 0.069444 -2.631457288 VOU AWN 4 22 12 2.484907 0.097222 -2.280052381 Now, take your now fully defined function and compute: 5 20 15 2.70805 0.125 2.013418678 (i) The expected value E = So t . f (t) dt 6 23 15 2.70805 0.152778 -1.797019751 8 7 21 18 2.890372 0.180556 -1.613804055 8 9 8 24 19 2.944439 0.208333 -1.454081455 E = t . 1.73the (-t)1.73 at 10 9 1 20 2.995732 0.236111 -1.311806993 11 10 24 20 2.995732 0.263889 1.18294803 E=1.175 12 11 22 20 2.995732 0.291667 1.064673327 13 12 24 21 3.044522 0.319444 -0.95491249 (ii) The variance VAR = S(t - u)2 . f (t) at 14 13 25 21 3.044522 0.347222 -0.852099496 15 14 20 21 3.044522 0.375 -0.755014863 u= t1/1.73 ent = 0.891 16 15 25 22 3.091042 0.402778 -0.662683927 17 16 24 3.091042 0.430556 -0.574308609 18 17 23 22 3.091042 0.458333 0.489219929 19 18 24 22 3.091042 0.486111 0.406843748 VAR = +2 . f (t) at - 12 20 19 5 22 3.091042 0.513889 -0.326675105 21 20 25 23 3.135494 0.541667 -0.248258101 22 21 22 23 3.135494 0.569444 0.171169278 VAR = t1/1.73 . e-t dt - 12 23 22 21 3.135494 0.597222 0.095002909 24 23 24 23 3.135494 0.625 -0.019356889 0 25 24 23 23 3.135494 0.652778 0.056182104 VAR = 1.07637 - u2 26 25 22 24 3.178054 0.680556 0.13205572 VAR = 1.07637 - 0.8912 IN 26 15 24 3.178054 0.708333 0.208755483 VAR = 1.07637 - 0.793881 28 27 25 24 3.178054 0.736111 0.286852083 VAR = 0.2825 29 28 23 24 3.178054 0.763889 0.367038004 30 29 21 24 3.178054 0.791667 0.45019365 31 30 25 24 3.178054 0.819444 0.537496821 Then record the values for: 32 31 23 25 3.218876 0.847222 0.630617758 32 15 25 3.218876 0.875 0.732099368 i. the mean / = 0.891 33 34 33 20 25 3.218876 0.902778 0.846192666 ii. the standard deviation o = 0.5315 35 34 26 25 3.218876 0.930556 0.981039808 36 35 22 25 3.218876 0.958333 1.156269006 37 36 12 26 3.258097 0.986111 1.453173762 38