let X~normal(0,16) and Y~normal(1,9), find P(X>Y)
To find P(X > Y), we need to compare the distributions of X and Y.
First, let's calculate the mean and variance of X and Y:
For X ~ Normal(0, 16), the mean (μ) is 0 and the variance (σ^2) is 16.
For Y ~ Normal(1, 9), the mean (μ) is 1 and the variance (σ^2) is 9.
Next, let's standardize the distributions by calculating the z-scores for X and Y:
The z-score formula is:
z = (x - μ) / σ
For X:
z_x = (X - 0) / √16
= X / 4
For Y:
z_y = (Y - 1) / √9
= (Y - 1) / 3
Now, we want to find P(X > Y), which is the same as finding P(X - Y > 0).
Let's define Z = X - Y.
To find P(Z > 0), we need to calculate the z-score for Z at 0:
z_z = (Z - μ_Z) / σ_Z
The mean and variance of Z can be calculated as follows:
Mean of Z:
μ_Z = μ_X - μ_Y = 0 - 1 = -1
Variance of Z:
σ_Z^2 = σ_X^2 + σ_Y^2 = 16 + 9 = 25
Now we can calculate the z-score for Z at 0:
z_z = (0 - (-1)) / √25
= 1 / 5
= 0.2
To find P(Z > 0), we can look up the z-score 0.2 in the standard normal distribution table or use a calculator.
Using a standard normal distribution table, we find that the area to the left of 0.2 is 0.5793. Therefore, the area to the right of 0.2 is 1 - 0.5793 = 0.4207.
P(Z > 0) = 0.4207
Therefore, P(X > Y) is also 0.4207.