Palm_ch7 - Solutions to Problems in Chapter Seven Solutions to Test Your Understanding Problems T7.1-1 The script file for the absolute histogram

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Unformatted text preview: Solutions to Problems in Chapter Seven Solutions to Test Your Understanding Problems T7.1-1 The script file for the absolute histogram is y = [91*ones(1,7), 92*ones(1,8), 93*ones(1,10),... 94*ones(1,6), 95*ones(1,12), 96*ones(1,7)]; x = [91:96]; hist(y,x),ylabel( Absolute Frequency ),xlabel( Thread Strength (newtons) ) The plot is shown in the figure. 90 91 92 93 94 95 96 97 2 4 6 8 10 12 Thread Strength (newtons) Absolute Frequency Figure : for Problem T7.1-1a. The script file for the relative histogram is tests = 50; y = [7, 8, 10, 6, 12]/tests; x = [91:96]; bar(x,y),ylabel( Relative Frequency ),xlabel( Thread Strength (newtons) ) 7-1 90 91 92 93 94 95 96 97 0.05 0.1 0.15 0.2 0.25 Thread Strength (newtons) Relative Frequency Figure : for Problem T7.1-1b. The plot is shown in the figure. T7.1-2 (a) 200(6 / 36) ≈ 33 times. (b) 200(4 + 3 + 2) / 36 ≈ 50 times. (c) 200(1 + 2 + 3 + 4 + 5) / 36 ≈ 106 times. T7.2-1 (a) The script file is y_abs = [1,1,2,0,3,4,5,6,8,11,12,10,11,8,7,5,4,4,4,1,2,0,1]; binwidth = .5; area = binwidth*sum(y_abs); y_scaled = y_abs/area; bins = [64:binwidth:75]; bar(bins,y_scaled),ylabel( Scaled Frequency ),xlabel( Height (inches) ) The plot is shown in the figure. (b) The script file is y_abs = [1,1,2,0,3,4,5,6,8,11,12,10,11,8,7,5,4,4,4,1,2,0,1]; binwidth = .5; bins = [64:binwidth:75]; y_raw = ; 7-2 62 64 66 68 70 72 74 76 0.05 0.1 0.15 0.2 0.25 Height (inches) Scaled Frequency Figure : for Problem T7.2-1a. for i = 1:length(y_abs) if y_abs(i)>0 new = bins(i)*ones(1,y_abs(i)); else new = ; end y_raw = [y_raw,new]; end mu = mean(y_raw),sigma = std(y_raw) Running this file gives a mean of 69.4 and a standard deviation of 2.14. (c) and (d) Running the following script file gives the answers: 43% and 63%. mu = 69.4; sigma = 2.14; b1 = 69; P1 = (1+erf((b1-mu)/(sigma*sqrt(2))))/2 a2 = 68; b2 = 72; P2 = (erf((b2-mu)/(sigma*sqrt(2))) - erf((a2-mu)/(sigma*sqrt(2))))/2 7-3 T7.3-1 Note that y = (15 + 5) x- 5 = 20 x- 5. Thus the session is y = 20*rand(1,1500)-5; mean(y) min(y) max(y) The computed mean, minimum, and maximum should be close to, but not exactly equal to, 5,- 5, and 15, respectively. T7.3-2 In the program given in Example 7.3-1, change the line income = 4000*demand + 1000*(level(k)- demand); to income = 4000*demand + 2000*(level(k)-demand); . The resulting plot is shown in the figure. From this we can see that the optimum production level is 50 units per season and the profit will be $46,000. 25 30 35 40 45 50 2 2.5 3 3.5 4 4.5 5 x 10 4 No. of Units Profit ($) Figure : for Problem T7.3-2. 7-4 T7.3-3 Note that y = 10 x + 7. Thus the session is y = 10*randn(1,1800)+7; mean(y) std(y) The computed mean and standard deviation should be close to, but not exactly equal to, 7 and 10, respectively. The theoretical minimum and maximum of the normal distribution is-∞ and + ∞ respectively, but a computer simulation cannot produce such numbers....
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This note was uploaded on 10/21/2009 for the course ENGR 290 taught by Professor Priritera during the Spring '05 term at S. Alabama.

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Palm_ch7 - Solutions to Problems in Chapter Seven Solutions to Test Your Understanding Problems T7.1-1 The script file for the absolute histogram

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