Penny has an existing fish tank. Penny adds fish and takes away fish from the ta

Question

Penny has an existing fish tank. Penny adds fish and takes away fish from the tank every Thursday. There are three types of fish that can go in the tank, tetra 1, tetra 2, and tetra 3 fish. A random tetral has a mean weight of 6 and a standard deviation of 2.5. A random tetra 2 has a mean weight of 8 and a standard deviation of 3. A random tetra 3 has a mean weight of 11 and a standard deviation of 3.5. Fish weights are known to be normally distributed Summary Tetra 1/A Tetra 2/B Tetra 3/C 1e1) Fish Type Fish Added(Subtracted) Mean Weight Standard Deviation of weight 2.5 3.5 Suppose on Thursday, 4 random Tetra 1 fish are added to the tank, 5 random Tetra 3 fish are added to the tank, and 10 random Tetra 2 fish are taken out of the tank. Let A be the total change in weight of Tetra 1 fish on Thursday. Let B be the total change in weight of Tetra 2 fish on Thursday. Let C be the total change in weight of Tetra 3 fish on Thursday. Let W be the total change in weight of fish in the tank on Thursday a)What is the expected value of A? b)What is the standard deviation of A? c) What is the expected value of W? d) What is the variance of W? e) What is the standard deviation of W? f) If all variables are normally distributed then what is the probability that W is -5? g) If all variables are normally distributed then what is the probability that W is > 20?

Explanation / Answer

a) A is change in weight of Tetra 1 fish. E(A) = 4*E(weight of 1 T-1 fish) = 24

b) SD of A = sqrt[ (4^2)*(Var of 1 T-1 fish) ] = 10

c) E(W) = E(A) + E(B) + E(C) = 4*(6) - 10*(8) + 5*(11) = -1

d) Var(W) = Var(A) + Var(B) + Var(C) = (4^2)*(2.5)^2 + (10^2)*(3)^2 + (5^2)*(3.5)^2 = 1306.25

e) SD(W) = sqrt[ Var(W) ] = 36.1421

f) Z-score for (W = -5) = [ { (-5) - (-1) } / (36.1421) ] = -0.11067 => P(W < -5) = 0.4562 [Answer]

g) Z-score for (W = 20) = [ { 20 - (-1) } / (36.1421) ] = 0.58104 => P(W > 20) = 0.2810 [Answer]

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