Can someone plz help me with this
1 answer:
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Answer:
69.50%
Step-by-step 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 0) = P(Z > —0.51) = 1—P(Z < —0.51) = 1-0.3050 = 0.6950 = 69.50%