ISEN 350 Problem Set M6

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Texas A&M University *

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350

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Statistics

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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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