Final Exam Solutions - Basic Engineering 3:Probability Statistics | B E 2100, Exams of Engineering

Download Final Exam Solutions - Basic Engineering 3:Probability Statistics | B E 2100 and more Exams Engineering in PDF only on Docsity! BE 2100 : FINAL EXAM 1. A Brinell hardness test involves measuring the diameter of the indentation made when a hardened steel ball is pressed into material under a standard test load. Suppose that the Brinell hardness is determined for each specimen in a sample size of 81. The sample's mean hardness was found to be 64.3 with a sample standard deviation of 6.0. Calculate a 96% confidence interval on the true average Brinell hardness for material specimens of this type. a) [61.93 , 66.67] b) [ 62.93 , 65.67] c) [63.60 , 64.98] d) [63.93 , 64.67] e) [ 64.23, 64.37] 2. Consider the following sample of fat content (in percentage) of n=10 randomly selected hot dogs. 25.2 21.3 22.8 17.0 29.8 21.0 25.5 16.0 20.9 19.5 Assuming that these were selected from a normal population distribution, construct a 90% confidence interval of the population mean fat content. a) [18.95 , 24.85] b) 19.21 , 23.99] c) [19.51 , 24.29] d) [20.21 , 22.99] e) [20.71 , 22.49] 3. Ten percent of the households of a Midwestern suburb for a total of 539 households were surveyed. Of this total of 539 households, it was found that 133 of these households owned at least one firearm. Construct a 98% confidence interval on the proportion of all households in this suburb that own a firearm. a)[.143 , .350] b) [.173 , .320] c) [.203 , .290] d) [.243 , .438] e) [ .292 , .388] 4. The accompanying data present 17 data points on breakdown voltage of electrically stressed flexible printed wiring boards, The data are assumed to be normally distributed. 1470 1510 1690 1740 1900 2000 2030 2100 2190 2200 2290 2380 2390 2480 2500 2580 2700 Construct a 95% confidence interval on the "standard deviation" of the breakdown voltage distribution. a) [251 , 513] b) [261 , 543] c) [271 , 584] d) [301 , 554] e) [321 , 465] 5. We desire to calculate the average stray load loss in watts for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume the stray loss is normally distributed with a standard deviation of 3.0 watts. How large a sample is needed to create a 95% confidence interval to within 0.5 watts of the true mean? a) 67 b) 139 c) 196 d) 246 e) 398 6. Ionizing radiation is being given increasing attention as a method for preserving horticultural products. The process was tested on garlic bulbs. Of 180 that received radiation 153 were still marketable after 240 days. Of 180 untreated bulbs only 119 were marketable after 240 days. The null hypothesis is that radiation was not beneficial. The alternative hypothesis is that radiation does improve the shelf-life (i.e. how long it can be sold) of a garlic bulb. The level of confidence to be used is 0.01. What is the rejection region to be used, what is the value of the test statistic and do you reject or not? a) rejection |z| > 1.88, test stat. z = 4.20 reject b) rejection |z| > 2.33, test stat. z = 2.15 do not reject c) rejection |z| > 1.65, test stat. z = 2.67 reject d) rejection |z| > 1.88, test stat. z = 1.15 do not reject e) rejection |z| > 2.33, test stat. z = 4.20 reject 7. A technology for pipeline rehabilitation uses a flexible liner threaded through existing piper. The average tensile strength of the liner may be influenced by whether or not a specific fusion process is used. However, in this example we are interested only in the variability in the tensile strength as recorded by the “standard deviation.”. (Assume normality.) No fusion s1 = 277.3 and n= 11 Fused s2 = 205.9 and n =11 The null hypothesis is that there is no difference in variability and the alternative hypothesis is there is a difference. The level of confidence to be used is 0.05. What is the rejection region to be used, what is the value of the test statistic and do you reject or not? a) rejection F > 1.02, test stat. F = 1.81 reject b) rejection F > 3.72, test stat. F = 1.81 do not reject c) rejection F > 4.43, test stat. F = 1.81 reject d) rejection F > 3.72, test stat. F =3.81 do not reject e) rejection F > 4.43, test stat. F = 3.81 do not reject 8. The ultimate load (kN) for two different types of beams is given below Sample Sample Sample Type Size mean Standard Dev. Fiberglass 8 33.4 2.2 Carbon Grid 6 42.8 4.3 Assume normality and that the variances of the two populations are equal. (In other words, calculate a pooled variance.) The null hypothesis is that there is no difference in means and the alternative is that the two means are unequal. The level of confidence to be used is 0.05. What is the rejection region to be used, what is the value of the test statistic and do you reject or not? a) rejection |t| > 4.88, test stat. t = 3.36 do not reject b) rejection |t| > 2.18, test stat. t = 4.36 reject c) rejection |t| > 2.18, test stat. t = 5.36 reject d) rejection |t| > 3.06, test stat. t = 4.36 reject e) rejection |t| > 3.06, test stat. t = 5.36 reject 9. Let the time to complete an online purchase at sahara.com be a random variable (in minutes) with the following probability density function. f(t) = 2t/55 3 < t < 8 SOLUTIONS: 1. Large sample size, normal distn x ± Z 2 α *S/ n = 64.3 2.05*(6.0/9) = (62.93,65.67) ± 2. Small sample size, t x ± t 2 α *S/ n = 21.9 (4.13/10 )*t± 0.5 0.05= (19.51,24.29) 3. Large sample size: CI for a Population Proportion (Page 388 of text) p = 133/539 = 0.246 p̂ ± Z 2 α * n pq = 133/539 ± 2.33* 539 754.0*246.0 = (0.203,0.290) 4. (n-1)*S2/ (n-1)*S2025.0χ ≤ 2σ ≤ 2/ 2975.0χ (17-1)*370.62/28.85 ≤ (17-1)*370.62σ ≤ 2/6.91 = (276.0, 563.9) 5. n = (Z 2 α *σ /H) = (1.96*3/0.5) = 139 2 2 6. One tail test: Reject Region : Z > Zα = Z0.01= 2.33; p= (153+119)/(180+180) = 0.756 test statistic Z = )180/1180/1(*244.0*756.0 180/119180/153 + − = 4.20 Reject Null Hypothesis since Z > Zα 7. Two tail test of two population variances (Page 476 of text) Degrees of freedom for s1 = 11-1 = 10 Degrees of freedom for s2 = 11-1 = 10 Rejection region F>F 2 α = F10,10,0.025= 3.72 Test statistic F = S12/ S22 = 277.32/205.92 = 1.81 Do not reject Null Hypothesis since F is not greater than Fα 8. Two tailed test for sample means, small sample test (Page 452 of text) Sp2 = 286 2.2*)18(3.4*)16( 22 −+ −+− = 10.52 Test Statistic t = )6/18/1(*52.10 4.338.42 + − = 5.36 Rejection region t > t0.025 (2.179, From tables) Reject since test statistic t > t0.025 9. P(t>5) = = t∫ 8 5 55/2 dtt 2/55]58=0.71 10. P(t<5+2 | t>5) = P(t<7 t>5)/P(t>5) = P(5<t<7)/P(t>5) = ∩ P(5<t<7) = = t∫ 7 5 55/2 dtt 2/55]57 = 0.436 Thus P(5<t<7)/P(t>5) = 0.436 / 0.71 = 0.61 11. E(t) = = 2*t∫ 8 3 55/2* dttt 3/(55*3) ]38 = 5.88 12. P(x >0.28) = P(Z > (0.28-0.30)/0.06) = p(Z > -0.3333) = 0.5 + 0.1293 = 0.6293 13. 1µ = 2; 1σ = 0.2; 2µ = 2.1; 2σ = 0.3; E(X2-X1) = 2.1-2 = 0.1 12 XX −σ = 22 3.02.0 + = 0.36 P(0.05<X2-X1<0.15) = P( 36.0 1.015.0 36.0 1.005.0 − << − Z ) = P(-0.139 <Z<0.139) = 0.11 14. Binomial , p=0.2, q = 1-0.2 = 0.8, n=20; P(Y≥ 6) = 1-P(Y 5) = 1-0.80 = 0.20 (From the Binomial Distribution Tables) ≤ 15. Geometric distribution P(4) = p*q4-1= 0.2*0.83= 0.10 16. Poisson Distribution λ = 4*1.5 = 6; P(Y<= 8) = 0.85 17. Source of Variation Sum of Squares Degrees of freedom Mean Square Fcritical- F(v1,v2,α ) Treatments 0.0608 2 0.0608/2 = 0.0304 0.0304/0.308 =0.987 F(2,12,0.05) = 3.89 Error 0.3701 12 0.3701/12 = 0.0308 Total .4309 14 As Fcritical is less than Fα mixtures do not have significant effect in the degree of soiling 18. 4 Measurements Hour 1 2 3 4 Avg Range 1. 36 35 34 33 34.5 3 2. 31 31 34 28 31 6 3. 30 30 32 31 30.75 2 4. 32 33 33 42 35 10 5. 32 34 37 36 34.75 5 6. 36 32 31 35 33.5 5 7. 33 33 36 34 34 3 8. 23 33 36 29 30.25 13 9. 43 36 35 38 38 8 10. 36 35 38 37 36.5 3 11. 34 38 41 45 39.5 11 12. 36 40 39 38 38.25 4 Center Line = 34.67; R = 6.0833; A2 = 0.729 UCL = 34.67 + 6.0833*0.729 = 39.105 LCL = 34.67 - 6.0833*0.729 = 30.235 Average of 11 observation is out of bounds and thus the process is not under control