mont4e_sm_ch03_mind

# mont4e_sm_ch03_mind - ⎛125 ⎞⎛ 375 ⎞ ⎜ ⎜ 5...

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Unformatted text preview: ⎛125 ⎞⎛ 375 ⎞ ⎜ ⎜ 5 ⎟⎜ 0 ⎟ 2.3453E8 ⎟⎜ ⎟ ⎠⎝ ⎠= = 0.00092 f X (5) = ⎝ 2.5524 E11 ⎛ 500 ⎞ ⎜ ⎜5⎟ ⎟ ⎝ ⎠ b) x f(x) 3-150. 0 1 2 3 4 5 6 7 8 9 10 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000 Let X denote the number of totes in the sample that exceed the moisture content. Then X is a binomial random variable with n = 30. We are to determine p. If P(X ≥ 1) = 0.9, then P(X = 0) = 0.1. Then which results in p = 0.0739. ⎛ 30 ⎞ 0 ⎜ ⎟( p) (1 − p )30 = 0.1 , giving 30ln(1−p)=ln(0.1), ⎜0⎟ ⎝⎠ 3-151. Let t denote an interval of time in hours and let X denote the number of messages that arrive in time t. Then, X is a Poisson random variable with λ = 10t. Then, P(X=0) = 0.9 and e-10t = 0.9, resulting in t = 0.0105 hours = 37.8 seconds a) Let X denote the number of flaws in 50 panels. Then, X is a Poisson random variable with λ = 50(0.02) = 1. P(X = 0) = e-1 = 0.3679. b) Let Y denote the number of flaws in one panel, then P(Y ≥ 1) = 1 − P(Y=0) = 1 − e-0.02 = 0.0198. Let W denote the number of panels that need to be inspected before a flaw is found. Then W is a geometric random variable with p = 0.0198 and E(W) = 1/0.0198 = 50.51 panels. 3-152. c) P (Y ≥ 1) = 1 − P (Y = 0) = 1 − e = 0.0198 Let V denote the number of panels with 1 or more flaws. Then V is a binomial random variable with n=50 and p=0.0198 −0.02 ⎛ 50 ⎞ ⎛ 50 ⎞ P(V ≤ 2) = ⎜ ⎟0.0198 0 (.9802) 50 + ⎜ ⎟0.01981 (0.9802) 49 ⎜0⎟ ⎜1⎟ ⎝⎠ ⎝⎠ ⎛ 50 ⎞ + ⎜ ⎟0.0198 2 (0.9802) 48 = 0.9234 ⎜2⎟ ⎝⎠ Mind Expanding Exercises 3-153. The binomial distribution P(X=x) = n! px(1-p)n-x x!(n − x )! n! [λ/n]x[1 – (λ/n)]n-x x!(n − x )! If n is large, then the probability of the event could be expressed as λ/n, that is λ=np. We could re-write the probability mass function as: P(X=x) = Where p = λ/n. P(X=x) Now we can re-express: n × (n − 1) × (n − 2) × (n − 3)...... × (n − x + 1) λ x (1 – (λ/n))n-x x x! n 3-29 [1 – (λ/n)]n-x = [1 – (λ/n)]n[1 – (λ/n)]-x And the first part of this can be re-expressed further as [1 – (λ/n)]n = So: P(X=x)= ((1 - λ n ) ) − n/λ −λ Now: In the limit as n → ∞ n × (n − 1) × (n − 2) × (n − 3)...... × (n − x + 1) λ x (1 - λ n )−n/λ x x! n ( ) −λ [1 – (λ/n)]-x n × (n − 1) × (n − 2) × (n − 3)...... × (n − x + 1) n x ≅1 In the limit as n → ∞ [1 – (λ/n)]-x ≅ 1 Thus: x P(X=x) = λ (1 - λ n )− n/λ x! z ( ) −λ We know from exponential series that: ⎛ 1⎞ Limit z → ∞ ⎜1 + ⎟ = e ≅ 2.7183 ⎝ z⎠ − n/λ = e. Thus, In our case above –n/λ = z, so (1 - λ n ) P(X=x) = λ −λ e x! n! e −λ λx px(1 – p)n-x = x!(n − x )! x! x It is the case, therefore, that Limit n → ∞ The distribution of the probability associated with this process is known as the Poisson distribution. The pmf can be expressed as: f(x) = e −λ λx x! 3-154. Sow that To begin, as ∞ ∑ (1 − p)i −1 p = p∑ (1 − p)i −1 , by definition of an infinite sum this can be rewritten i =1 i =1 i =1 ∞ ∑ (1 − p)i −1 p = 1 using an infinite sum. ∞ ∞ p ∑ (1 − p)i −1 = i =1 p p = =1 1 − (1 − p ) p 3-30 3-155. E ( X ) = [(a + (a + 1) + ... + b](b − a + 1) a −1 ⎡b ⎤ i − ∑i⎥ ⎢∑ i =1 ⎦ = ⎣ i =1 2 2 (b − a + 1) ⎡ b(b + 1) (a − 1)a ⎤ − ⎢ ⎥ 2⎦ ⎣2 = (b − a + 1) ⎡ (b − a + b + a ) ⎤ ⎢ ⎥ 2 ⎦ =⎣ = (b + a) 2 b b+ a 2 2 (b − a + 1) ⎡ (b + a)(b − a + 1) ⎤ ⎢ ⎥ 2 ⎦ =⎣ (b − a + 1) b ⎡b 2 (b − a + 1)(b + a ) 2 ⎤ i − (b + a )∑ i + ⎥ ∑ [i − ] ⎢ ∑ 4 i =a ⎣ i =a ⎦ i =a = V (X ) = b + a −1 b + a −1 2 b(b + 1)(2b + 1) (a − 1)a(2a − 1) ⎡ b(b + 1) − (a − 1)a ⎤ (b − a + 1)(b + a ) − − (b + a ) ⎢ ⎥+ 6 6 2 4 ⎣ ⎦ = b − a +1 2 (b − a + 1) − 1 = 12 3-156. Let X denote the number of nonconforming products in the sample. Then, X is approximately binomial with p = 0.01 and n is to be determined. If P ( X ≥ 1) ≥ 0.90 , then P ( X Now, P(X = 0) = ( )p (1 − p) n 0 0 = 0) ≤ 0.10 . (1 − p)n ≤ 0.10 , and n = (1 − p ) n . Consequently, n≤ ln 0.10 = 229.11 . Therefore, n = 230 is required ln(1 − p ) 3-157. If the lot size is small, 10% of the lot might be insufficient to detect nonconforming product. For example, if the lot size is 10, then a sample of size one has a probability of only 0.2 of detecting a nonconforming product in a lot that is 20% nonconforming. If the lot size is large, 10% of the lot might be a larger sample size than is practical or necessary. For example, if the lot size is 5000, then a sample of 500 is required. Furthermore, the binomial approximation to the hypergeometric distribution can be used to show the following. If 5% of the lot of size 5000 is nonconforming, then the probability of zero nonconforming product in the sample is approximately 7 × 10 −12 . Using a sample of 100, the same probability is still only 0.0059. The sample of size 500 might be much larger than is needed. 3-31 3-158. Let X denote the number of panels with flaws. Then, X is a binomial random variable with n =100 and p is P( X < 5) = P( X ≤ 4) = P( X = 0) + P( X = 1) + P( X = 2) + P( X = 3) + P( X = 4) = (100 ) p 0 (1 − p)100 + (100 ) p1 (1 − p )99 + (100 ) p 2 (1 − p )98 0 1 2 + (100 ) p 3 (1 − p)97 + (100 ) p 4 (1 − p)96 3 4 = 0.034 3-159. Let X denote the number of rolls produced. Revenue at each demand 1000 2000 3000 0.3x 0.3x 0.3x 0 ≤ x ≤ 1000 mean profit = 0.05x(0.3) + 0.3x(0.7) - 0.1x 0.05x 0.3(1000) + 0.3x 0.3x 1000 ≤ x ≤ 2000 0.05(x-1000) mean profit = 0.05x(0.3) + [0.3(1000) + 0.05(x-1000)](0.2) + 0.3x(0.5) - 0.1x 0.05x 0.3(1000) + 0.3(2000) + 0.3x 2000 ≤ x ≤ 3000 0.05(x-1000) 0.05(x-2000) mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000) + 0.05(x-2000)](0.3) + 0.3x(0.2) - 0.1x 0.05x 0.3(1000) + 0.3(2000) + 0.3(3000)+ 3000 ≤ x 0.05(x-1000) 0.05(x-2000) 0.05(x-3000) mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000)+0.05(x-2000)]0.3 + [0.3(3000)+0.05(x3000)]0.2 - 0.1x 0 0.05x Profit 0.125 x 0.075 x + 50 200 -0.05 x + 350 Max. profit \$ 125 at x = 1000 \$ 200 at x = 2000 \$200 at x = 3000 \$200 at x = 3000 the probability of one or more flaws in a panel. That is, p = 1 − e −0.1 = 0.095. 0 ≤ x ≤ 1000 1000 ≤ x ≤ 2000 2000 ≤ x ≤ 3000 3000 ≤ x The bakery can make anywhere from 2000 to 3000 and earn the same profit. 3-160.Let X denote the number of acceptable components. Then, X has a binomial distribution with p = 0.98 and n is to be determined such that P( X ≥ 100 ) ≥ 0.95 . P( X ≥ 100 ) 102 0.666 103 0.848 104 0.942 105 0.981 Therefore, 105 components are needed. n 3-32 ...
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