Assignment 2 solutions

Assignment 2 solutions - CE93-Engineering Data Analysis...

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Unformatted text preview: CE93-Engineering Data Analysis Joan Walker, Spring 2011 Assignment 2 solutions 1. The following data represent the lifetimes (in hours) of a sample of 40 transistors: 112 121 126 108 141 104 136 134 121 118 143 116 108 122 127 140 113 117 126 130 134 120 131 133 118 125 151 147 137 140 132 119 110 124 132 152 135 130 136 128 a. Compute the sample mean and sample median mean = 127.425 median = 127.5 b. Are the data approximately normal? According to the histogram, the data looks approximately normal c. Compute mean absolute deviation and sample standard deviation. mean absolute deviation = 9.675 sample standard deviation = 11.873 Note: if the student divides by n instead of n-1 (biased estimate of standard deviation), the answer they will get is 11.724 d. What percentage of the data fall within given by Chebyshev’s inequality. ? Compare your answer to the bound CE93-Engineering Data Analysis Joan Walker, Spring 2011 From this interval, 34 of 40 data points, or 85% of data points, fall within 1.5 standard deviations from the mean. According to Chebyshev’s inequality, at least 0.5556 of the data (55.56%) falls within 1.5 standard deviations from the mean: This shows an example where data tends to be much less dispersed than the lower bound described by Chebyshev’s inequality 2. Problem 3.3 in Ross. a. EF = {7} b. E U FG = {1,3,4,5,7} c. EGc = {3,5,7} d. EFc U G = {1,3,4,5} e. Ec (F U G) = {4,6} f. EG U FG = {1,4} 3. Problem 3.14 in Ross. For two events that are not independent, P(E or F) = P(E) + P(F) – P(EF). This is shown below in the Venn diagram: E F EF The intersection of E and F is subtracted from the above equation because it is counted twice when P(E) and P(F) are added together. To find the probability that exactly one event E or F occurs, the intersection must be subtracted again. Therefore, the probability of exactly E or F = P(E) + P(F) – P(EF) – P(EF) = P(E) + P(F) – 2P(EF). 4. The designs for a plane specify the wingspan, b, to be between 21 and 30 ft. To maintain lift during slow flight, the aspect ratio, (AR = b2/S, where b is the wingspan and S is the wing area), must be at least 9. Due to construction limitations, the wing area must be at least 49 ft2. a. Draw the sample space for the plane’s wingspan and wing area (hint: draw the sample space using b2 instead of b) CE93-Engineering Data Analysis Joan Walker, Spring 2011 b2 (ft2) 900 441 S (ft2) 49 100 b. Identify the following events within the sample space in (a). i. b2 (ft2) 900 576 441 S (ft2) 64 49 ii. ; 100 CE93-Engineering Data Analysis Joan Walker, Spring 2011 b2 (ft2) 900 588 441 S (ft2) 49 75 100 iii. b2 (ft2) 900 441 S (ft2) 70 49 iv. 100 , and b2 (ft2) 900 576 441 S (ft2) 49 64 100 CE93-Engineering Data Analysis v. Joan Walker, Spring 2011 . b2 (ft2) 900 441 S (ft2) 70 49 100 c. Make new drawings to show the following: i. ; b2 (ft2) 900 441 S (ft2) 49 ii. 100 => darkest intersection of the below graph. CE93-Engineering Data Analysis Joan Walker, Spring 2011 b2 (ft2) 900 441 S (ft2) 49 iii. Among and 70 100 - null. , , and , which pairs of events are mutually exclusive? are mutually exclusive because their spaces do not overlap. 5. A cantilever beam is designed to hold loads at its midpoint and end. Loads Wa, or Wb, or both can be applied at the midpoint or at the end. The bending moment at the fixed end, M1, depends on the loading pattern. Two loads can be placed at one point at the same time. At least one load must be placed each time. Wa=200 kg and Wb=350 kg. a. Determine the sample space of M1. Possible situations: Situation Load @2 Load @3 M1 1 Wa - 2000 kg∙m 2 - Wa 4000 kg∙m 3 Wb - 3500 kg∙m 4 - Wb 7000 kg∙m CE93-Engineering Data Analysis Joan Walker, Spring 2011 5 Wa Wb 9000 kg∙m 6 Wb Wa 7500 kg∙m 7 Wa Wb - 5500 kg∙m 8 - Wa Wb 11000 kg∙m Sample Space for : {2000 kg∙m, 4000 kg∙m, 3500 kg∙m, 7000 kg∙m, 9000 kg∙m, 7500 kg∙m, 5500 kg∙m, 11000 kg∙m} b. Define the following events: i. ii. iii. Are these events mutually exclusive? Why or why not? & and & are mutually exclusive but include M1=9,000 kg∙m. & are not because they both c. The probabilities of the locations where loads Wa and Wb are placed are as follows: i. ii. iii. iv. The locations where loads Wa and Wb are placed are independent of one another. Knowing this, what are the probabilities associated with the events E1, E2, and E3? Use the notation Wx @ 0 to represent load x not being placed at all. =1-0.35-0.5=0.15 =1-0.2-0.45=0.35 First, let’s write the probability of Wb not being placed, using total probability theorem (note , because one load must be placed somewhere). CE93-Engineering Data Analysis Joan Walker, Spring 2011 The other solutions similar to this are used several times in the solution that follows. Each situation’s probability is detailed below: 1.) 2.) 3.) 4.) 5.) CE93-Engineering Data Analysis Joan Walker, Spring 2011 6.) 7.) 8.) In summary, the probabilities for each situation are sh own in the following table: Solution 2, nonindependence assumption Situation Load @2 1 Wa 2 - Load @3 Wa M1 3 4 5 6 7 8 Wb Wb Wa Wa Wb 3500 kg∙m 7000 kg∙m 9000 kg∙m 7500 kg∙m 5500 kg∙m 11000 kg∙m Total= 1 Wb Wa Wb Wa Wb - P(situation) 2000 kg∙m 4000 kg∙m 0.144 0.206 0.0461 0.104 0.142 0.0905 0.0634 0.203 = 0.144 + 0.206 + 0.0461 = 0.396 = 0.142 + 0.0905 = 0.233 = 0.142 + 0.203 = 0.345 d. Find , , , and . = 0.396 + 0.233-0 = 0.629 CE93-Engineering Data Analysis Joan Walker, Spring 2011 = 0.396 + 0.233 + 0.345 – 0 – 0 – (P{5} = 0.142) = 0.832 = 0.396 = 0.345 – 0.142 = 0.203 6. Travel between cities A and B can be conducted through two alternate routes, as shown in the figure below. One route consists of Highways 1 and 2 and a tunnel, the other route consists of Highway 3. Let highway is open, tunnel is open. Also, let travel between cities A and B is possible. a. Describe the sample space of all possible travel conditions. This should list all possible combinations of events , , , and their complements. Show event in the sample space. b. Write an expression for event E in terms of operations on events , , , and . c. Use de Morgan’s rule to derive an expression for event (the complement of ) in terms of the complementary events , , , and . Describe this result in words. Travel from city A to city B is not available when at least one among Highway 1, Tunnel and Highway 2 is closed and Highway 3 is closed. ...
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