Question 1171023: Let X be the life span of one light bulb of a particular brand and Y be the lifespan of a bulb of another brand. Suppose that X is N(965, 19^2) and Y is N(995, 31^2). If one of each brand of bulb is randomly selected, Find P(X > Y).
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**Given Information:**
* X ~ N(965, 19²) (X is the lifespan of the first brand, normally distributed with mean 965 and variance 19²)
* Y ~ N(995, 31²) (Y is the lifespan of the second brand, normally distributed with mean 995 and variance 31²)
We want to find P(X > Y).
**1. Define a New Variable**
Let W = X - Y. Then, we need to find P(W > 0).
**2. Find the Mean and Variance of W**
* Mean of W (μW) = E(W) = E(X - Y) = E(X) - E(Y) = 965 - 995 = -30
* Variance of W (σW²) = Var(W) = Var(X - Y) = Var(X) + Var(Y) (since X and Y are independent)
* σW² = 19² + 31² = 361 + 961 = 1322
* Standard deviation of W (σW) = √1322 ≈ 36.359
Therefore, W ~ N(-30, 1322).
**3. Standardize W**
We need to find P(W > 0). To do this, we standardize W using the z-score formula:
* Z = (W - μW) / σW
* Z = (0 - (-30)) / √1322 = 30 / √1322 ≈ 30 / 36.359 ≈ 0.825
**4. Find the Probability**
We want to find P(W > 0), which is equivalent to P(Z > 0.825).
* P(Z > 0.825) = 1 - P(Z < 0.825)
Using a standard normal distribution table or a calculator, we find:
* P(Z < 0.825) ≈ 0.7953
Therefore:
* P(Z > 0.825) = 1 - 0.7953 = 0.2047
**Final Answer:**
P(X > Y) ≈ 0.2047
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