Question

Eva loves to go fishing. Each time she catches a fish, there is a 70% chance that it is a northern pike and a 30%

chance it is a walleye. Let W be the random variable that represents the number of walleye Eva gets if she

catches 2 fish.

W = # of walleye 0

1

2

P(W)

0. 49

0. 42

0. 09

Calculate the mean of W.

walleye

Uw =

Answer 1

Answer:

Answer: 1.4

Step-by-step explanation:

P(Northern pike) = 0.7

P(walleye) = 0.3

If 2 fishes are caught :

Number of northern pike (x) :

X = 0

P(walleye 1 st) * p(walleye 2nd) = (0.3 * 0.3) = 0.09

X = 1

P(walleye 1st)*P(Northern pike 2nd) OR P(Northern pike 1st)*P(Walleye 2nd)

= (0.3 * 0.7) + (0.7 * 0.3)

= 0.21 + 0.21

= 0.42

P(x = 2)

P(northern pike 1st) * P(northen pike 2nd)

0.7 *0.7 = 0.49

X: _ 0 _ 1 __ 2

P(x): ___ 0.09 ___ 0.42 ____ 0.49

Expected value of northern pikes :

(0* 0.09) + (1 * 0.42) + (2 * 0.49)

0 + 0.42 + 0.98

= 1.4

Answer 2

Expected value of the number of walleye is 0.5.

What is the expected value?

The expected value exists as a long-run average value of random variables. It also demonstrates the probability-weighted average of all possible values. Expected value exists as a generally utilized financial concept.

W = # of walleye 0, 1, 2

P(W) 0. 49, 0. 42, 0. 09

To estimate the expected value of the number of walleye.

E(X) =  number of walleye

E(X) = (0 $\times$ 0.49) + (1 $\times$ 0.42) + (2 $\times$ 0.09)

E(X) = 0 + 0.42 + 0.08

E(X) = 0.5

Therefore, the number of walleye expected value = 0.5.

To learn more about expected value

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