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# 09-Error_4_web - Exam#1 Problem#2:ANSWERKEYS The mean...

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The mean lifetime of light-bulbs produced at a certain factory is 14,0 0 0 hours. The standard deviation for the lifetime distribution is 1,00 0 hours “the key number”). Assume that losers are those who purchased light-bulbs lasting less than 10,0 0 0 hours and winners those who purchased light-bulbs lasting longer than 18,0 0 0 hours. Calculate the percentage of the customers who are winners and losers. Tables will be given! Exam # 1  Problem #2: ANSWER KEYS

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The normalized Gaussian curve is obtained when the variable  x  is substituted by a variable  z : σ μ - = x z After the variable transformation, the normalized Gaussian curve  is a Gaussian curve  for  μ  = 0 and  σ   = 1. x _
Table 1. Cumulative distribution function of the normalized Gaussian distribution. | z | P( - ∞ , -z ) P(0, z ) | z | P( - ∞ , -z ) P(0, z ) 0 0.5000 0.0000 0.1 0.4602 0.0398 2.1 0.0179 0.4821 0.2 0.4207 0.0793 2.2 0.0139 0.4861 0.3 0.3821 0.1179 2.3 0.0107 0.4893 0.4 0.3446 0.1554 2.4 0.0082 0.4918 0.5 0.3085 0.1915 2.5 0.0062 0.4938 0.6 0.2742 0.2258 2.6 0.0047 0.4953 0.7 0.2420 0.2580 2.7 0.0035 0.4965 0.8 0.2119 0.2881 2.8 0.0026 0.4974 0.9 0.1841 0.3159 3.9 0.0019 0.4981 1.0 0.1587 0.3413 3.0 0.001350 0.498650 1.1 0.1357 0.3643 3.1 0.000968 0.499032 1.2 0.1151 0.3849 3.2 0.000687 0.499313 1.3 0.0968 0.4032 3.3 0.000483 0.499517 1.4 0.0808 0.4192 3.4 0.000337 0.499663 1.5 0.0668 0.4332 3.5 0.000233 0.499767 1.6 0.0548 0.4452 3.6 0.000159 0.499841 1.7 0.0446 0.4554 3.7 0.000096 0.499904 1.8 0.0359 0.4641 3.8 0.000072 0.499928 1.9 0.0287 0.4713 3.9 0.000048 0.499952 2.0 0.0227 0.4773 4.0 0.000032 0.499968 2.9

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Problem #6 Consider titration a 0.100 M solution of maleic acid (cis- butenedioic acid) with 0.100 M solution of NaOH. K A1 =1.23 × 10 -2 and K A2 =4.66 × 10 -7 . (a) Plot an approximate pH vs. f curve and define the f. (b) What is the majority species derived from maleic acid at f = 2.5? (c) Calculate the value of the overall K a .
0 0.5 1 1.5 2 2.5 3 2 4 6 8 10 12 pH f Titration curve of the 0.100 M solution of a diprotic maleic acid (cis-butenedioic acid) with 0.100 M solution of NaOH The acid has very different values of K a1 and K a2 : K A1 =1.23 × 10 -2 K A2 =4.66 × 10 -7 ; Two end-points are clearly visible!

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Accepted approximations, maleic acid buffers (simple): Before first equivalence C4H2O4H2 = ( C4H2O4H) - + H +, K A1 ; H 2 A = AH - + H + pH = pK A1 + log[C4H2O4H2/[(C4H2O4H)-] Before second equivalence (C4H2O4H)- = (C4H2O4)2- + H+, KA2; HA - = A 2- + H + pH = pK A2 + log[(C4H2O4)2-]/[(C4H2O4H)-] Advanced: The (C4H2O4H) - , amphiprotic, may also: (C4H2O4H) - + H2O = C4H2O4H2 + OH - , K B related to K A2 ; Leads to the equation given by Hadi, (Harris: page 190, eq. 9-10).
Chemistry 222 Fall 2010 STATISTICS, ANALYSIS OF  EXPERIMENTAL ERROR (4)

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Practical advice: 1.We should retain one more  digit for the  average and for the standard deviation than  was present in the original data.   2. The mean and the standard deviation  should both end at the same decimal place.
STATISTICS, ANALYSIS OF EXPERIMENTAL  ERROR Harris, pp. 68 - 116 1. Basic concepts, a sample of data 2. Gaussian distribution (probabilities, statistics) 1. Parameters of the Gaussian distribution the normalized Gaussian curve: a “z” variable,  tabulated,  a histogram ) estimates: populations and samples, s and   σ   2. t-Student’s distribution  1. Q-test 2. Linear regression “If an experiment is repeated a great many times, then the results tend to cluster symmetrically about the average value”. (They will exhibit a bell-shape.)

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DETERMINATION OF PARAMETERS  OF A GAUSSIAN DISTRIBUTION
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09-Error_4_web - Exam#1 Problem#2:ANSWERKEYS The mean...

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