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Please explain your work and reasoning:
Let X be a random variable, normally distributed with mean 100 and variance 64.
Mean = μ = 100,
Variance = σ2 = 64,
Standard Deviation = σ = 8.
1. If there are two random variables, X1 and X2, independent of each other, and each distributed as X as given above, then find the probability that X1+X2 is less than or equal to 220.
X = X1 + X2 is normally distributed with μ = μ1 + μ2 and variance σ^2 = σ1^2 + σ2^2
μ = 100 + 100 = 200, σ^2 = 64 + 64 = 128
σ = √128 = 11.314
z = (X - μ)/σ = (220 - 200)/11.314 = 1.77
P(X ≤ 200) = Area under the Standard Normal Curve to the left of z = 1.77, which is 0.961
2. For the above problem, find L such that X1+X2 is greater than or equal to L is 0.10.
P(X) ≥ 0.10 corresponds to a z score of 1.645
SE = 1.645 * σ = 1.645 * 11.314 = 18.61
L = [μ - SE, μ + SE] = [200 - 18.61, 200 + 18.61] = [181.39, 218.61]
3. If another random variable X3 is also distributed as X as given above and is independent of X1 and X2, what is the expected value of X1+X2+X3? What is the variance of (X1+X2+X3)?
Expected value of X1 + X2 + X3 = μ1 + μ2 + μ3 = 300, Variance of X1 + X2 + X3 = σ1^2 + σ2^2 + σ3^2
= 64 + 64 + 64 = 192.