ISEN 350 Problem Set M6

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Texas A&M University *

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Feb 20, 2024

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Problem 1: E(cx) = cE(x) : This identity is related to the linearity of expectation. It states that the expected value of a constant times a random variable is equal to the constant times the expected value of the random variable. Mathematically: E(cx) = ∫(cx) * f(x)dx where f(x) is the probability density function of x = c ∫x * f(x) dx = cE(x) So, the identity is verified. E(x + c) = E(x) + c: This identity states that the expected value of the sum of a random variable and a constant is equal to the expected value of the random variable plus the constant. Mathematically: E(x + c) = ∫(x + c) * f(x) dx = ∫x * f(x) dx + ∫c * f(x) dx = E(x) + c ∫f(x) dx = E(x) + c * 1 since ∫f(x) dx = 1 for a valid probability density function = E(x) + c So, the identity is verified. V(x) = E(x^2) - [E(x)]^2: This identity represents the variance of a random variable. It states that the variance of x is equal to the expected value of x squared minus the square of the expected value of x. Mathematically: V(x) = E[(x - E(x))^2] = E[x^2 - 2x*E(x) + (E(x))^2] = E(x^2) - 2E(x*E(x)) + E((E(x))^2) E(xE(x)) = E(x) * E(E(x)) Since E(E(x)) is just E(x) again (the expected value of a constant is the constant itself), we can simplify this further: E(x*E(x)) = E(x) * E(x) = (E(x))^2 Replace E(x*E(x)) with (E(x))^2 in the expression: V(x) = E(x^2) - 2(E(x))^2 + E((E(x))^2) Now, consider the last term, E((E(x))^2). Since E(x) is a constant with respect to the expectation operator E, we have: E((E(x))^2) = (E(x))^2 Replace E((E(x))^2) with (E(x))^2 in the expression: V(x) = E(x^2) - 2(E(x))^2 + (E(x))^2 Simplify the expression: V(x) = E(x^2) - (E(x))^2

V(c + x) = V(x): This identity states that adding a constant to a random variable does not change its variance. Mathematically: V(c + x) = E[(c + x)^2] - [E(c + x)]^2 = E[(c^2 + 2cx + x^2)] - [(c + E(x))^2] = E(x^2) - [E(x)]^2 = V(x) So, the identity is verified. V(cx) = c^2 * V(x) : This identity relates to the variance of a constant times a random variable. Mathematically: V(cx) = E[(cx)^2] - [E(cx)]^2 = E(c^2 * x^2) - [c * E(x)]^2 = c^2 * E(x^2) - c^2 * [E(x)]^2 = c^2 * [E(x^2) - [E(x)]^2] = c^2 * V(x) So, the identity is verified. Problem 2: E[g(X)] = E[2X^2 − 2Xμ] E[g(X)] = 2E[X^2] − 2μE[X] E[X^2] = σ^2 + μ^2 and E[X] = μ: E[g(X)] = 2σ^2 + 2μ^2 − 2μ^2 = 2σ^2 Therefore E[g(X)] is twice the variance of X Problem 3: (a) What is the probability that a sample of 10 covers will contain exactly 2 defectives? 𝑃(? = 2) = (10 ?ℎ???? 2) * (0. 10) 2 * (1 − 0. 10) 8 = 0. 1937 So, the probability that a sample of 10 covers will contain exactly 2 defects is approximately 0.1937. (b) What is the probability that a sample of 10 covers will contain more than 2 defectives? 𝑃(? = 1) = (10 ?ℎ???? 1) * (0. 10) 1 * (1 − 0. 10) 9 = 0. 387420489 𝑃(? = 0) = (10 ?ℎ???? 0) * (0. 10) 0 * (1 − 0. 10) 10 = 0. 3486784401 𝑃(? > 2) = 1 − 0. 1937 − 0. 387420489 − 0. 3486784401 = 0. 07 (c) What is the probability that the first defective cover will be found on the 2nd cover produced, given that all covers are being inspected? 𝑃(? = 2) = (1 − 0. 10) 2−1 * 0. 10 = 0. 09 (d) What is the probability the 4th part cover selected will be defective, given that the 3rd cover selected was defective? 𝑃(4?ℎ ??????𝑖𝑣?|3?? ??????𝑖𝑣?) = 0. 10 (e) What is the probability the 2nd defective cover will be found on the 5th cover sampled? 𝑃(2?? ????𝑙 ??????𝑖𝑣?? ?? 5?ℎ) = (5 ?ℎ???? 1) * (0. 10) * (0. 9) 4 * 0. 10 = 0. 03281

Problem 4 (a) A yard of cloth contains 2 or more blemishes. 𝑃(? ≥ 2) = 1 − 𝑃(? = 0) − 𝑃(? = 1) 𝑃(? ≥ 2) = 1 − (? −0.1 0. 1 0 /0!) − (? −0.1 0. 1 1 /1!) = 0. 00467 (b) A bolt contains less than 3 blemishes. Blemishes per bolt = 0.1 * 25 = 2.5 𝑃(? < 3) = (? −2.5 2. 5 0 /0!) + (? −2.5 2. 5 1 /1!) + (? −2.5 2. 5 2 /2!) = 0. 54381 (c) A batch of 10 bolts contains more than 25 blemishes. Blemishes per 10 bolts = 25 𝑃(? > 25) = 1 − 𝑖=0 25 ∑ (? −25 25 𝑖 /𝑖!) = 1 − 0. 55292 = 0. 44708 Problem 5 (a) E(2X - 0.002): E(2X - 0.002) = 2E(X) - E(0.002) E(2X - 0.002) = 2(0.024) - 0.002 = 0.048 - 0.002 = 0.046 So, E(2X - 0.002) = 0.046. (b) V(3X - 0.01): V(3X - 0.01) = (3^2) * V(X) V(X) is the variance of X, which is 0.0045: V(3X - 0.01) = (3^2) * 0.0045 = 9 * 0.0045 = 0.0405 So, V(3X - 0.01) = 0.0405. (c) E(2X^2): E(2X^2) = 2E(X^2) E(2X^2) = 2[V(X) + E(X)^2] E(2X^2) = 2(0.0045 + 0.024^2) = 0.010 E(2X^2) = 0.010 3.23. A random sample of 50 units is drawn from a production process every half hour. The fraction of nonconforming products manufactured is 0.02. What is the probability that p̂ ≤ 0.04 if the fraction nonconforming really is 0.02? 𝑃(?̂ ≤ 0. 04) = 𝑃(𝑥 ≤ 2) = 𝑖=0 2 ∑ ((50 ?ℎ???? 𝑖) * (0. 02) 𝑖 * (0. 98) 50−𝑖 ) = 0. 9216 So, the probability that p̂ ≤ 0.04 is 92.16%

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3.27 A stock brokerage has four computers that are used for making trades on the New York Stock Exchange. The probability that a computer fails on any single day is 0.005. Failures occur independently. Any failed computers are repaired after the exchange closes, so each day can be considered an independent trial. a. What is the probability that all four computers fail in one day? 𝑃(? = 4) = 4 ?ℎ???? 4 * (0. 005) 4 * (1 − 0. 005) (4−4) = 0. 000000000625 So there is approximately 0 probability that all four computers fail in one day. b. What is the probability that at least one computer fails on a day? 𝑃(? >= 1) = 1 − 𝑃(? = 0) 𝑃(? = 0) = 4 ?ℎ???? 0 * (0. 005) 0 * (1 − 0. 005) (4−0) = 1 * 1 * 0. 980149 = 0. 9801 𝑃(? >= 1) = 1 − 𝑃(? = 0) = 1 − 0. 9801 = 0. 0199 So, there is a 0.0199 probability that at least one computer fails on a day. c. What is the mean number of days until a specific computer fails? 𝐸(?) = 1/? = 1/0. 005 = 200 3.33 The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively, what is the probability that a power supply selected at random will conform to the specifications on voltage? 𝑃(4. 95 ≤ ? ≤ 5. 05) = 𝑃(? ≤ 5. 05) − 𝑃(? ≤ 4. 95) = 𝑃(? ≤ 2. 5) − 𝑃(? ≤− 2. 5) 𝑃(? ≤ 2. 5) − 𝑃(? ≤− 2. 5) = 0. 9938 − 0. 0062 = 0. 9876 So, 𝑃(4. 95 ≤ ? ≤ 5. 05) = 0. 9876 3.35 If x is normally distributed with mean μ and standard deviation σ = 4, and given that the probability that x is less than 32 is 0.0228, find the value of μ. Z-value of 0.0228 = -2 z = (x - μ)/σ = (32 - μ)/4 = -2 μ = 32 + 8 = 40 So, μ = 40

3.37 A lightbulb has a normally distributed light output with mean 5000 end foot-candles and standard deviation of 50 end foot-candles. Find a lower specification limit such that only 0.5% of the bulbs will not exceed this limit. The z-score for the 0.5% lower tail is approximately -2.576: − 2.576 = (X − 5000) / 50 X − 5000 = − 2.576 ∗ 50 = −128.8 X = 5000−128.8 X = 4871.2 The required lower specification limit is 4871.2