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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

Respuesta :

shallomisaiah19

Answer:

the answer is  μ=5.28 and σ=1.67

Step-by-step explanation:

[tex]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[/tex]

[tex]E(W^2)=10.34\\\rightarrow Var(W) = E(W^2)-E^2(W)\\=1.4596[/tex]

[tex]E(G) = 2.3\\Var(G)=E(G^2)-E^2(G)\\=6.62-5.29\\=1.33[/tex]

So,

[tex]\mu = E(W+G)=E(W)+E(G)=5.28[/tex]

[tex]\sigma ^2 = Var (W+G) \\=Var(W)+Var(G)=2.79\\\sigma =1.67[/tex]

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

Using the discrete distributions, it is found that the mean and standard deviation are given by:

C) μ=5.28 and σ=1.67

  • The mean of a discrete distribution is given by the sum of all outcomes multiplied by it's respective probability.
  • The variance is given by the sum of the differences squared between each outcome and the mean, multiplied by their respective probabilities.
  • When two variables are added, the mean is the sum of the means, while the standard deviation is the square root of the sum of the variances.

For Worms:

[tex]E_w(X) = 0.05(0) + 0.06(1) + 0.18(2) + 0.35(3) + 0.3(4) + 0.05(5) + 0.01(6) =  2.98[/tex]

[tex]Var_w(X) = 0.05(0 - 2.98)^2 + 0.06(1 - 2.98)^2 + 0.18(2 - 2.98)^2 + 0.35(3 - 2.98)^2 + 0.3(4 - 2.98)^2 + 0.05(5 - 2.98)^2 + 0.01(6 - 2.98)^2 = 1.4596[/tex]

For grubs:

[tex]E_g(x) = 0.05(0) + 0.21(1) + 0.27(2) + 0.38(3) + 0.05(4) + 0.03(5) + 0.01(6) = 2.3[/tex]

[tex]Var_g(x) = 0.05(0 - 2.3)^2 + 0.21(1 - 2.3)^2 + 0.27(2 - 2.3)^2 + 0.38(3 - 2.3)^2 + 0.05(4 - 2.3)^2 + 0.03(5 - 2.3)^2 + 0.01(6 - 2.3)^2 = 1.33[/tex]

For the sum:

[tex]E(X) = E_w(X) + E_g(x) = 2.98 + 2.3 = 5.28[/tex]

[tex]\sigma = \sqrt{Var_w(X) + Var_g(x)} = \sqrt{1.4596 + 1.33} = 1.67[/tex]

Hence option c is correct.

A similar problem is given at https://brainly.com/question/15855314

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