Question

14.) Although most of us buy milk by the quart or gallon, farmers measure daily production in pounds. Ayrshire cows average 47 pounds of milk a day, with a standard deviation of 6 pounds. For Jersey cows, the mean daily production is 43 pounds, with a standard deviation of 5 pounds. Assume that Normal models describe milk production for these breeds.

Answer 1

69.50%

Explanation:

Given:

Ayrshire cows:

E(X) =μ = 47

SD(X) = σ = 6 Var (X) = 6^2 = 36

Jersey cows:

E(Y) = μ = 43

SD(Y) = σ = 5 Var(Y) = 52 = 25

Properties mean, variance and standard deviation:

E(X +Y) = E(X) E(Y)

V ar(X +Y) = Var(X) + Var(Y)

SD(X +Y) = √Var(X)+Var(Y)

X — Y represents the difference between Ayrshire and Jersey cows.

E(X — Y) = E(X) — E(Y). 47 — 43 = 4

SD(X — Y) = √Var(X)+ Var(Y) =√36+ 25 = √61 = 7.8102

The z-score is the value decreased by the mean, divided by the standard deviation:

z = x-μ /σ = 0-4/ 7.8102 = -0.51

Determine the corresponding probability using table Z in appendix F.

P(X—Y
`\geq` 0) = P(Z > —0.51) = 1—P(Z < —0.51) = 1-0.3050 = 0.6950 = 69.50%