Probability confusion

Question

A certain automobile manufacturer equips a particular model with either a six-cylinder engine or a four-cylinder engine. Suppose the fuel efficiencies (mpg) for six-cylinder and four-cylinder cars are both normally distributions with means 22 and 28, and standard deviations 2.2 and 1.5, respectively. Assume we randomly choose two cars with six-cylinder engines and denote their mpg by 𝑋1, 𝑋2 , and randomly choose three cars with four-cylinder engines and denote their mpg by 𝑌1,𝑌2,𝑌3.

a. What is the probability that the average mpg from those 2 six-cylinder cars is less than 22 (that is, 𝑃[𝑋(x bar)< 22])?

b. Find 𝑃[𝑌(y bar) < 𝑋(x bar)].

Answer 1

This problem is easier than it may appear due to the fact that the sum and difference of normals is normal

f6 is N with mean 22 and sd 2.2

f4 is N with mean 28 and sd 1.5

Then, the average of 2 6 cylinders X is N with mean 22 and sd 2.2/√2

The average of 3 four cylinders Y is N with mean 28 and sd 1.5/√3

a) As the mean is 22 and the median of the normal equals the mean, P(X < 22) = .5

b) As X has variance (2.2²/2) and Y has variance (1.5²/3)

P(Y < X) = P(Y - X < 0)

Y - X has mean 28 - 22 = 6 and variance (2.2²/2) + (1.5²/3) = 3.17

Then, P(Y - X < 0) = P(z < -6/sqrt(3.17)) = P(z < -3.33694) = .000376