The Lifetime In Hours X Of A Certain Electrical Componen Q3

The Lifetime In Hours X Of A Certain Electrical Componen Q3 AND Q53 The Lifetime In Hours X Of A Certain Electrical Componen Q3 AND Q53 The Lifetime In Hours X Of A Certain Electrical Componen Q3 AND Q5 3. The lifetime (in hours) X of a certain electrical component follows an exponential distribution with parameter λ = 0.001. Three of these components operate independently in a system. The system fails if at least two of the components fail. Find the probability that the system operates for at least 200 hours without failure.

5. The heights of male students entering the University of Hong Kong, X, are normally distributed with a mean of 170.4 cm. It is known that about 2.5% of the students are taller than 190 cm. (a) What is the standard deviation of X? (b) What is the proportion of male students at the University of Hong Kong with a height of 160 cm or less? (c) What is the probability that the mean height of n = 4 randomly selected male students exceeds 175 cm? (d) Male students are randomly selected one after the other. What is the probability that the fifth selected student is the second selected student with a height below 160 cm? For the profit payoff table below, the decision maker assumes that P(s1) = .15, p(s2) = .50, and p(s3) = .35.

State of Nature Decision s 1 s 2 s 3 d ,000 d ,,000 What alternative would be chosen according to expected value?

Paper For Above instruction This assignment encompasses multiple statistical and probabilistic analyses involving exponential and normal distributions, as well as decision-making under uncertainty. The problems require understanding of survival probabilities, standard deviation determination, cumulative distribution functions, and expected value calculations. This paper aims to systematically address each question with detailed calculations, explanations, and interpretation to demonstrate mastery of the concepts involved.

1. Probability that the system operates at least 200 hours without failure The lifetime X of an electrical component is modeled as an exponential distribution with parameter λ = 0.001. Given three such independent components operating in a system, and the system fails if at least two components fail, we aim to find the probability that the system survives at least 200 hours.

In this context, the probability that a single component operates for at least t hours is given by the exponential survival function:

P(X ≥ t) = e^{-λt} Substituting λ=0.001 and t=200 hours: P(component survives 200 hours) = e^{-0.001 × 200} = e^{-0.2} ≈ 0.8187 Since components operate independently, the number of components that fail by time t follows a binomial distribution: the number of failures among three components, with success probability q = 1 - 0.8187 ≈ 0.1813.

To ensure the system operates for at least 200 hours, at most one component can fail, i.e., the system must have zero or one failed component. Therefore, the probability the system operates at least 200 hours is: P(system survives ≥ 200 hours) = P(0 failures) + P(1 failure) Calculating these: P(0 failures) = (probability all three components survive) = (0.8187)^3 ≈ 0.5493 P(1 failure) = C(3,1) × (0.8187)^2 × 0.1813 ≈ 3 × 0.6703 × 0.1813 ≈ 3 × 0.1214 ≈ 0.3642 Thus, the total probability: ≈ 0.5493 + 0.3642 = 0.9135 Hence, the probability that the system operates for at least 200 hours without failure is approximately 91.35%.

2. Normal distribution of heights and related probabilities Let the height of male students be normally distributed with a mean (µ) of 170.4 cm, and we're told that 2.5% are taller than 190 cm. Using this information, we can find the standard deviation (σ). (a) Standard deviation of heights The 97.5th percentile corresponds to a z-score of approximately 1.96 in the standard normal distribution.

Therefore, we have: 190 = µ + z × σ Plugging in the values: 190 = 170.4 + 1.96 × σ

Solving for σ: σ = (190 - 170.4) / 1.96 ≈ 19.6 / 1.96 ≈ 10 cm (b) Proportion of students with height ≤ 160 cm Now, calculating the probability that a student's height is less than or equal to 160 cm: Standardized z-score: z = (160 - µ) / σ = (160 - 170.4) / 10 = -10.4 / 10 = -1.04 Using standard normal tables or software, P(z ≤ -1.04) ≈ 0.1492 Thus, approximately 14.92% of male students are 160 cm or shorter. (c) Probability mean height of 4 students exceeds 175 cm The sampling distribution of the mean for n=4 students is normally distributed with mean µ=170.4 cm and standard deviation σ/√n = 10 / 2 = 5.

Calculate z-score for the sample mean of 175 cm: z = (175 - 170.4) / 5 = 4.6 / 5 = 0.92 Probability that the sample mean exceeds 175 cm: P( X■ > 175 ) = 1 - P( Z ≤ 0.92 ) ≈ 1 - 0.8212 = 0.1788 (d) Fifth student being the second with height below 160 cm This problem models a negative binomial scenario, where the "success" is drawing a student with height < 160 cm, which occurs with probability p ≈ 0.1492. The probability that the fifth student is the second such student is given by: Number of ways to have exactly 1 success in the first 4 students and the 5th student is a success: P = C(4,1) × p^2 × (1 - p)^3 Calculating: P = 4 × (0.1492)^2 × (1 - 0.1492)^3 ≈ 4 × 0.0223 × 0.8508^3 ≈ 4 × 0.0223 × 0.614 ≈ 4 × 0.0137 ≈ 0.0548

Therefore, the probability is approximately 5.48%.

3. Decision-making based on probability payoffs The payoff table presents three decision options under different states of nature, with assigned probabilities: States of Nature: s1, s2, s3 Probabilities: p(s1)=0.15, p(s2)=0.50, p(s3)=0.35 Decisions and payoffs (in 000s): Decision s1 s2 s3 d1 — — — d2 — — — d3 — — —

Assuming specific payoffs for each decision and state (which should be detailed), the alternative with the highest expected value is selected. The expected value (EV) for each decision is calculated as: EV = Σ [p(s) × payoff(s)] Choosing the decision with the highest EV provides the optimal strategy based on expected payoff.

Conclusion This analysis demonstrates how probabilistic models and statistical calculations inform decision-making and system reliability assessments. Calculations from exponential survival functions, properties of the normal distribution, and probability models provide precise insights into complex engineering and demographic questions. Furthermore, expected value analysis guides optimal decision choices when faced with uncertain future states, exemplifying integral applications of probability theory in practical contexts.

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