CHEE 231 Assignment 3 Solutions.pdf

CHEE 231 Assignment 3 Solutions.pdf - Data Analysis and...

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Data Analysis and Experimental Data, CHEE 231 Assignment 3 Solutions September 29, 2015 Solution to Problem 1 Mean = 150g Standard deviation = 0.1g 5 . 1 1 . 0 15 . 0 1 . 0 150 15 . 150 X Z or 5 . 1 1 . 0 15 . 0 1 . 0 150 85 . 149 Z From table 1 in Montgomery on page 487 93313 . 0 5 . 1 Z P However, this represents: Z = 1.5 P= 0.93313 And we want Z = 1.5 P= ? Z = -1.5 First, work with only half the graph. Thus, P = 0.93313 - 0.5 = 0.43313
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Then, work with the whole graph Thus P = 0.43313*2 = 0.86626 Therefore, 86.63% of the cue balls are acceptable. Solution to Problem 2 𝑧 = 𝑥 − 𝜇 𝜎 (a) 𝑧 = 0.62 − 0.5 0.05 = 2.4 From the table of Z values, P(X>0.62)= 1-0.991802 =0.00819 (b) 𝑧 = 0.47 − 0.5 0.05 = −0.6 From the table of Z values, P(X<0.47)= 0.274253 = 0.63 − 0.5 0.05 = 2.6 From the table of Z values, P(X<0.63)= 0.995339 0.995339 - 0.274253 = 0.721086 (c) z=1.209 corresponds to a P of 0.901475
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1.209 = 𝑥 − 0.5 0.05 𝑥 = 0.560 Solution to Problem 3 a) Mean = 350 b) Variance = 100 c) ?~𝑁(𝜇 = 350, 𝜎 2 = 100); 𝑃(270 < ? < 360) = 𝑃 ( 270 − 350 10 < ? − µ 𝜎 < 360 − 350 10 ) = 𝑃(−8 < ? < 1) = 𝑃(−8 < ? < 0) + 𝑃(0 ≤ ? < 1) = 0.5 + 0.3413 = 0.8413 𝑃(? > 400) = 𝑃 ( ? − µ 𝜎 < 400 − 350 10 ) = 𝑃(? > 5) = 0 Solution to Problem 4 𝑃(? > 𝑥) = 0.95 0.95 = 1 − 𝑃(? < 𝑥)
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