Question

Let \(X_{1}\) and \(X_{2}\) be independent with normal distributions \(N(6,1)\) and \(N(7,1)\), respectively. Find \(P\left(X_{1}>X_{2}\right)\). Hint: \(\quad\) Write \(P\left(X_{1}>X_{2}\right)=P\left(X_{1}-X_{2}>0\right)\) and determine the distribution of \(X_{1}-X_{2}\)

Short answer

The probability \(P(X_{1}>X_{2})\) is 0.24.

Step-by-step solution

Calculate Mean and Variance

First, let's calculate the mean and variance of the difference between the two variables, \(X_{1}\) and \(X_{2}\). Let \(Y = X_{1} - X_{2}\). Since the two variables are independent and normally distributed, Y is also normally distributed with mean and variance calculated as follows:\nMean: \( \mu_Y = \mu_{X_{1}} - \mu_{X_{2}} = 6 - 7 = -1\),\nVariance: \( \sigma^2_Y = \sigma^2_{X_{1}} + \sigma^2_{X_{2}} = 1 + 1 = 2 \). Thus, Y has a normal distribution \(N(-1, 2)\).

Calculate Probability

We want to find \( P(Y > 0) = P(X_{1} - X_{2} > 0) \). To standardize Y, we can use the Z-score formula: \( Z = \dfrac{(Y - \mu_Y)}{\sigma_Y} \). To find \( P(Y > 0) \), we will find \( P(Z > \frac{(0 - (-1))}{\sqrt{2}}) = P(Z > \frac{\sqrt{2}}{2}) \) = \( P(Z > 0.707) \).\nUsing Standard Normal Distribution Tables, we find that \( P(Z < 0.707) \approx 0.760 \).\nSince the area under the curve of the standard normal distribution is equal to 1, \( P(Z > 0.7071) = 1 - P(Z < 0.707) = 1 - 0.760 = 0.24 \).\nTherefore, \( P(Y > 0) = 0.24 \).

Conclusion

Therefore, the probability that a random variable from normal distribution \(N(6,1)\) is greater than another random variable from normal distribution \(N(7,1)\) is 0.24. In other words, \(P(X_{1}>X_{2}) = 0.24\).