Biologists are analyzing soil to check for the number of worms and grubs in a wildlife preserve. Let the random variable W represent the num

Question

Biologists are analyzing soil to check for the number of worms and grubs in a wildlife preserve. Let the random variable W represent the number of worms found in 1 square foot of soil, and let the random variable G represent the number of grubs found in 1 square foot of soil. The following tables show the probability distributions developed by the biologists for W and G.

W 0 1 2 3 4 5 6
Probability 0.05 0.06 0.18 0.35 0.30 0.05 0.01
G 0 1 2 3 4 5 6
Probability 0.05 0.21 0.27 0.38 0.05 0.03 0.01

Assume that the distributions of worms and grubs are independent. What are the mean, μ, and standard deviation, σ, for the total number of worms and grubs in 1 square foot of soil?

A) μ=5 and σ=1.67

B) μ=5 and σ=2.36

C) μ=5.28 and σ=1.67

D) μ=5.28 and σ=2.36

E) μ=5.28 and σ=2.79

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Katherine 2 weeks 2021-09-08T10:47:13+00:00 1 Answer 0

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    2021-09-08T10:48:58+00:00

    Answer:

    the answer is  μ=5.28 and σ=1.67

    Step-by-step explanation:

    E(W) =0\times0.05+1\times0.06+2\times0.18+3\times0.35+4\times0.3+5\times0.05+6\times0.01\\=0.06+0.36+1.05+1.2+0.25+0.06\\=2.98

    E(W^2)=10.34\\\rightarrow Var(W) = E(W^2)-E^2(W)\\=1.4596

    E(G) = 2.3\\Var(G)=E(G^2)-E^2(G)\\=6.62-5.29\\=1.33

    So,

    \mu = E(W+G)=E(W)+E(G)=5.28

    \sigma ^2 = Var (W+G) \\=Var(W)+Var(G)=2.79\\\sigma =1.67

    Therefore, the answer is  μ=5.28 and σ=1.67

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45:7+7-4:2-5:5*4+35:2 =? ( )