How to use normal distribution to calculate probability that one player will score more points than-CodePudding

Please help me set this problem up in R:

Alex is expected to score a mean of 2 points, normally distributed with a SD of 2 points. Bob is expected to score a mean of 1 point, normally distributed with a SD of 3 points.

What is the probability that Alex scores more points than Bob?

CodePudding user response：

If you want simulation, using rnorm function that generates normal distribtion.

n <- 1000000
Alex <- rnorm(n, 2, 2)
Bob <- rnorm(n, 1, 3)
sum(Alex>Bob)/n

[1] 0.610427

CodePudding user response：

Written in R code, the text book solution is

mu_Alex <- 2
sd_Alex <- 2
mu_Bob <- 1
sd_Bob <- 3

The question asks for P(A > B) = P(A - B > 0).
Let D = A - B and compute the mean and variance of the difference. Don't forget that the variance is a quadratic operator, so the variances add up. Then take the square root.

mu_Diff <- mu_Alex - mu_Bob
var_Diff <- sd_Alex^2   sd_Bob^2
sd_Diff <- sqrt(var_Diff)

Transform to standard gaussian. (This is not quite right, what I'm transforming to standard gaussian is the zero in A - B > 0)

z_Diff <- (0 - mu_Diff)/sd_Diff

And get the upper tail, since we want P(D > 0).

pnorm(z_Diff, mean = mu_Diff, sd = sd_Diff, lower.tail = FALSE)
#[1] 0.6384329

Park's simulation is not very far from the exact value.