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# 142 Class Note for M E 345 with Professor Cimbala at PSU

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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Pennsylvania State University taught by a professor in Fall. Since its upload, it has received 42 views.

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Date Created: 02/06/15

M E 345 Fall 2009 Professor John M Cimbala Today we will Do a review example problem PDFs Review the pdf module The Gaussian 0r Normal Probability Density Function Do some example problems Gaussian PDFs and probability estimates Example PDFs Given A sample consists of 1000 length measurements The sample mean is 5365 cm and the sample standard deviation is 125 cm 136 measurements lie between 550 lt x S 570 cm To do a Estimate the probability that a length measurement lies between 550 and 570 cm b Calculate the transformed variables 21 at x1 550 cm and 22 at x2 570 cm c Discuss why it is useful to transform from x to z Solution l 9 W m 6 J ii 013L3e n n low 0 36quot xi 3quot 6 S m to PM A PStltxltm waxen 0436 7 l qlm 2 1 1 x 1 1 a Iquot W 3 9094413 gx cht lttth g ah IF vi PM If mhr 3939 4 90F Lug GilML c 394quot n nk L WW mm 94 Example Con dence level Given Many voltage readings are taken from a power supply The sample mean voltage reading is 171012 V The sample standard deviation of all the readings is 0022 V We assume that the errors in the readings are purely random To do Write the voltage to 95 confidence level Solution 2C 9 13 SIX 6 I t1 LC 5 00 V 6 my a 0mg 1 Example Probability exam scores Given In one of Professor Cimbala s midterm exams the mean was 736 out of 100 possible points and the standard deviation was 92 We assume that the distribution of exam scores is Gaussian The cutoff grade for a D is 60 points To do Predict the percentage of students who failed the exam score lt 60 points Solution a a MA GMMn m r 3 M An PMMIql Wu 1 tu Xq Xri 6 39I3L 3 C c T 1 3 1411 a La u AUF 033 51 3 9m 1 W I 082 A0 05 031359 00011 or Ass 39 7 MAM gym Example Probability Given 100 velocity measurements are taken in a wind tunnel The sample mean velocity is l7 5126 ms The sample standard deviation ofmw readings is 00690 ms We assume that the errors in the readings are purely random a To do Calculate the probability that the velocity of a random measurement is in the range 5126 ms lt Vlt 5200 ms In other words calculate P5l26 lt Vlt 5200 b To do Calculate the probability that the velocity of a random measurement is in the range 5000 ms lt Vlt 5200 ms In other words calculate P5000 lt Vlt 5200 m Solution V V v M I i m m 17 2 F 7 o 1 En L L 5 T 006 F0 Fm Pm A AA Pi 3 vi V f 9V V V Mm Mmm 02m Kemp 3532 003 quot 1 m 1 mm M NM AG y i We mu 0M QM wN Fm ch nLe J39HouLo W My Adme

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