Text Chapter 4 Examples (1)

Example 4.1, p.87. Let X represent the uncertainty in the number of delivered source instructions (DSI) of a new software applica Suppose the uncertainty is expressed as the trapezoidal density function seen below. Determine the following A. 𝐸(𝑋) . B. Med(𝑋) . C. 𝑃(𝑋≤𝐸(𝑋)+𝜎_𝑋) . PDF, CDF, Mean, and Variance:

A. a 25,000 parameters of distribution m1 28,000 m2 35,000 b 37,500 n1 3,943,750,000 numerator 1st term n2 2,109,000,000 numerator 2nd term d 58,500 denominator E(X) 31,363.2479 31,363 n1 1.9076563E+14 numerator 1st term n2 7.4677E+13 numerator 2nd term d 117000 denominator [E(X)]^2 983,653,317 Var(X) 8557153.55395 need this for part C Sigma(X) 2925.2612796 2,925 need this for part C B. First we compute the area of the triangle on the LHS. base 3,000 height 0.0001025641 =2/((35000+37500)-(25000+28000)) Area 0.1538 Now we compute the area of the triangle on the RHS. base 2,500 height 0.0001025641 Area 0.1282 P(25,000 <= x <= 28,000) 0.1538 P(35,000 <= x <= 37,500) 0.1282 P(28,000 <= x <= 35,000) 0.7179 P(25,000 <= x <= 35,000) 0.8718 * The text has 34/39 which equals 0.8718. Based on the above, the median includes some portion of the rectangle of the trapezoid. So, we use the CDF with m1 <= x <= m2. a 25,000 m1 28,000 m2 35,000

ation. g: Example 4.2, p.91. If 𝑋 has a uniform distribution, show that 𝑀𝑒𝑑(𝑋)=𝐸(𝑋). The PDF for a uniform distribution is: ?_𝑋 (𝑥)=1/(𝑏−𝑎), 𝑖? 𝑎≤𝑥≤𝑏 Recall that we integrate the PDF to get the CDF (we differentiate the CDF to get the PDF): ?_𝑋 (𝑥)=1/(𝑏−𝑎) ∫1_𝑎^𝑥▒𝑑𝑢=1/(𝑏−𝑎) ├ 𝑢┤|_𝑎^𝑥 ?_𝑋 (𝑥)=1/(𝑏−𝑎) (𝑥−𝑎), 𝑖? 𝑎≤𝑥≤𝑏 The 𝑀𝑒𝑑(𝑋)=0.50. So, we take the CDF and set it equal to 0.50 and solve for 𝑥 . 1/(𝑏−𝑎) (𝑥−𝑎)=0.50 𝑥=1/2 (𝑏−𝑎)+𝑎, 𝑥=1/2 𝑏−1/2 𝑎+𝑎=1/2 𝑏+1/2 𝑎=(𝑏+𝑎)/2 Thus: 𝑀𝑒𝑑(𝑋)=(𝑏+𝑎)/2 Finally, we have that the E(𝑋) is: 𝐸(𝑋)=∫1_(−∞)^∞▒ 𝑥∙? 〖 _𝑋 (𝑥)𝑑𝑥=∫1_𝑎^𝑏▒ 𝑥 〖 /(𝑏−𝑎) 𝑑𝑥=1/(2(𝑏−𝑎))∙├ 𝑥^2 ┤|_𝑎^𝑏= 〗 (𝑏^2−𝑎^2)/(2(𝑏−𝑎))=((𝑏+𝑎)(𝑏−𝑎))/(2(𝑏−𝑎))=(𝑏+𝑎)/2 So, for a uniform distribution Med(𝑋)=E(𝑋)=(𝑏+𝑎)/2.

A. Preliminary information: A random variable 𝑋 is said to have a beta distribution if its PDF is given by: 𝑋~𝐵𝑒𝑡𝑎(𝛼,𝛽,𝑎,𝑏) A random variable ? is said to have a standard beta distribution if its PDF is given by: ?~𝐵𝑒𝑡𝑎(𝛼,𝛽) One may transform from 𝑋~𝐵𝑒𝑡𝑎(𝛼,𝛽,𝑎,𝑏) to ?~𝐵𝑒𝑡𝑎(𝛼,𝛽) by letting 𝑦=(𝑥−𝑎)/(𝑏−𝑎) and we can the opposite direction via 𝑥=𝑦∙(𝑏−𝑎)+𝑎 . The expected values and variances for the beta distribution and standard beta distribution are: Lastly, the mode of ? is given by: 𝑦=(1−𝛼)/(2−𝛼−𝛽) Example 4.4, p. 97. Suppose the activity time 𝑋 (in minutes) to complete assembly of a microcircuit is beta distribute interval 4<𝑥<9 , with shape parameters 𝛼=5, 𝛽=10 . Determine 𝑃(𝑋≤𝑀𝑜𝑑𝑒(𝑋)) . From the above (see preliminary information) we can compute the 𝑀𝑜𝑑𝑒(?) from which we will then determine the 𝑀𝑜𝑑𝑒(𝑋). 𝑀𝑜𝑑𝑒(?)=(1−5)/(2−5−10)=(−4)/(−13)=4/13≈0.308 Now we can transform from the standard beta distribution ? to the beta distribution 𝑋 . We know that to go from the former to the latter we use 𝑦=(𝑥−𝑎)/(𝑏−𝑎) . So, to go from

to 𝑋 we use: 𝑥=𝑦∙(𝑏−𝑎)+𝑎 For 4<𝑥<9 we have that 𝑎=4, 𝑏=9 and we have that 𝑦=0.308 . So: 𝑥=0.308(9−4)+4=5.54 Thus the 𝑀𝑜𝑑𝑒(𝑋)=5.54. Now, to compute 𝑃(𝑋≤𝑀𝑜𝑑𝑒(𝑋)) we note that: 𝑃(𝑋≤𝑀𝑜𝑑𝑒(𝑋))=𝑃((𝑋−𝑎)/(𝑏−𝑎)≤(𝑀𝑜𝑑𝑒(𝑋)−𝑎)/(𝑏−𝑎))=𝑃(?≤𝑀𝑜𝑑𝑒(?)) The probability of the minutes to assemble a microcircuit can be determined using the standard beta distribution ?~𝐵𝑒𝑡𝑎(5,10) with the 𝑀𝑜𝑑𝑒(?)=0.308 as this will be the sam for 𝑋~𝐵𝑒𝑡𝑎(5,10,4,9) with the 𝑀𝑜𝑑𝑒(?)=5.54. Using Matlab: betacdf(0.308,5,10) 𝑃(?≤𝑀𝑜𝑑𝑒(0.308))=𝑃(𝑋≤𝑀𝑜𝑑𝑒(5.54))=0.4420 The above tells us that with probability 0.4420 the assembly of the microcircuit will not take longer than 5.54 minutes. One may also use Excel: BETA.DIST(0.308,5,10,TRUE) = 0.4420

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,1,TRUE) Value Mean Sigma Prob P(Cost <= 100) 100 110.42 21.65 0.3152 =NORM.DIST(BL8,BM8,BN8,TRUE RUE) P(Cost <= 140) 140 110.42 21.65 0.9141 =NORM.DIST(BL9,BM9,BN9,TRUE P(110.42 <= Cost <= 140) 0.5989 F(140) - F(100) Example 4.6, p.102. We are given that 𝐶𝑜𝑠𝑡~𝑁(110.42, 〖 21.65 〗 ^2) . We are asked to determine 𝑃(100≤𝐶𝑜𝑠𝑡≤140

Excel Matrix Inverse RHS 1.0000 -1.6450 0.386423 0.613577 30.34 1.0000 1.0360 -0.372995 0.372995 70.55 =MINVERSE(BW51:BX52) Result mean 55.01 =MMULT(BZ51:CA52,CC51:CC52) variance 15.00

A. The P(Cost > 2E(Cost)) = 1 - P(Cost < 2E(Cost) E(Cost) 100 2 E(Cost) 200 <------------ These are for X which is lognormally distributed. Var(Cost) 625 These are given. We need to compute the mean and variance using the equations provided above. X is l distributed and Y is normally distributed. Mean(Y) 4.5749 Var(Y) 0.0606 <-------------These are for Y which is normally distributed. Sigma(Y) 0.2462 These are calculated. Z Value 2.9383 =(LN(CJ26) - CJ32)/CJ34 1 - P(Z <= 2.9383) 0.00165 =1-NORM.S.DIST(CJ36,TRUE) B. P(50 <= Cost <= 150) We need only compute Z values for ln(50) and ln(150) and use th probabilities and make the computation. Z(150) 1.7699 =(LN(150)-$CJ$32)/$CJ$34 Z(50) -2.6920 =(LN(50)-$CJ$32)/$CJ$34 P(Cost <= 150) 0.9616 =NORM.S.DIST(CJ44,TRUE) e is a 5% exceed m's cost. Preliminary information: If 𝑋 is a lognormal random variable with mean 𝐸(𝑋)=𝜇_𝑋 and 𝑉𝑎𝑟(𝑋)=𝜎_𝑋^2 , then: 𝜇_?=𝐸(𝑙𝑛𝑋)=1/2 𝑙𝑛[ 〖 (𝜇_𝑋) 〗 ^4/( 〖 (𝜇_𝑋) 〗 ^2+𝜎_𝑋^2 )] 𝜎_?^2=𝑉𝑎𝑟(𝑙𝑛𝑋)=𝑙𝑛[( 〖 (𝜇_𝑋) 〗 ^2+𝜎_𝑋^2)/ 〖 (𝜇_𝑋) 〗 ^2 ] Example 4.8, p. 112. Suppose the uncertainty in a system's cost is described by a lognormal PDF with 𝐸(𝐶𝑜𝑠𝑡)=100 an 𝑉𝑎𝑅(𝐶𝑜𝑠𝑡)=625 . Determine: A. 𝑃(𝐶𝑜𝑠𝑡>2𝐸(𝐶𝑜𝑠𝑡)) . B. 𝑃(50≤𝐶𝑜𝑠𝑡≤150).

A. E(X2) 1.6875 <----------- This is given Var(X2) 0.2557 <----------- This is given Sigma(X2) 0.5056 =SQRT(CV17) E(Y) 0.4803 =0.5*LN((CV16^4)/(CV16^2+CV17)) Var(Y) 0.0860 =LN((CV16^2+CV17)/(CV16^2)) Sigma(Y) 0.2932 =SQRT(CV21) Using ln(1.26) we compute a Z value: -0.8497 =(LN(1.26)-CV20)/CV22 P(X2 <= 1.26) 0.1978 =NORM.S.DIST(CY24,TRUE) B. We want the P(X2 <= E(X2)). We first compute the Z value associated with E(X2). lognormally Z value 0.1466 =(LN(1.6875)-CV20)/CV22 P(X2 <= 1.6875) 0.5583 =NORM.S.DIST(CV30,TRUE) hem to get the X is lognormally distributed and we wish to work with Y which is normally distributed. W mean and variance of Y which we determine via X and the formulas provided in Examp nd Example 4.9, p.113. The random variable 𝑋_2 represents the cost of a system's engineering and program manageme Furthermore, the point estimate of 𝑋_2 is denoted 𝑥_( 〖 2𝑃𝐸 〗 _𝑥2 ) and is 1.26. If 𝑋_2 can be approximated by a lognormal distribution, with E(𝑋_2 )=1.6875 and 𝑉𝑎𝑟 〖 (𝑋 〗 _2)=0.255677 , determine: A. 〖 𝑃(𝑋 〗 _2≤𝑥_( 〖 2𝑃𝐸 〗 _𝑥2 )). B. 𝑃 〖 (𝑋 〗 _2≤𝐸( 𝑃 〖 (𝑋 〗 _2)) .

We first use the standard Beta distribution B(5,10) to obtain the values for y at 0.05 and 0.95. y(0.05) 0.152718 =BETA.INV(0.05,5,10) y(0.95) 0.540005 =BETA.INV(0.95,5,10) We substitute the above y values into the equations for a (min) and b (max). x(0.05) 4.76359 given x(0.95) 6.70003 given a 4.0000 min <------------ b 9.0000 max E(Y) 0.3333 <----------- Var(Y) 0.0139 <----------- E(X) 5.6667 <----------- Var(X) 0.3472 <----------- We need the ple 4.8. These are the min and max X values. ent. , Example 4.10, p.117. With 𝛼 and 𝛽 and any two fractiles 𝑥_𝑖 and 𝑥_? , the minimum and maximum possible values of 𝑋 given by the below: 𝑎=(𝑥_𝑖 𝑦_?−𝑥_? 𝑦_𝑖)/(𝑦_?−𝑦_𝑖 ) 𝑏=(𝑥_? (1−𝑦_𝑖)−𝑥_𝑖 (1−𝑦_?))/(𝑦_?−𝑦_𝑖 ) Find the minimum and maximum possible values of 𝑋 if 𝑋~𝐵𝑒𝑡𝑎(5,10,𝑎,𝑏), 𝑥_0.05=4.76359 𝑎𝑛𝑑 𝑥_0.95=6.70003. Find the 𝐸(𝑋)𝑎𝑛𝑑 𝑉𝑎𝑟(𝑋) .

A. In the figure, we can see that it is a Beta Distribution with alpha = 2 and beta = 3.5 given that P(I <= 100,000) = 0.50 and P(I <= 150,000) = 0.95. We have the fractiles and X(0.95) = 150,000. Since alpha = 2 and beta = 3.5, the standard Beta Distribution Y~(2, 3.5). alpha 2 beta 3.5 Y(0.50) 0.3461 =BETA.INV(0.5,$DT$80,$DT$81) Y(0.95) 0.7019 =BETA.INV(0.95,$DT$80,$DT$81) X(0.50) 100,000 X(0.95) 150,000 a 51,366 Min b 191,893 Max B. Mode(Y) 0.2857 Mode(I) 91,516 A. Find the extreme possible values for ? . B. Compute the mode of ? . C. Compute E(?) and 𝜎_?

C. E(Y) 0.3636 Var(Y) 0.0356 E(I) 102,466 Var(I) 703,037,278 Sigma(I) 26,515