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A particular lake is known to be one of the best places to catch a certain type of fish. In this table, x = number of fish caught in a 6-hour period. The percentage data are the percentages of fishermen who caught x fish in a 6-hour period while fishing from shore.
x 0 1 2 3 4 or more
% 43% 35% 15% 6% 1%
(a) Find the probability that a fisherman selected at random fishing from shore catches one or more fish in a 6-hour period. (Enter a number. Round your answer to two decimal places.
(b) Find the probability that a fisherman selected at random fishing from shore catches two or more fish in a 6-hour period. (Enter a number. Round your answer to two decimal places.)
(c) Compute μ, the expected value of the number of fish caught per fisherman in a 6-hour period (round 4 or more to 4). (Enter a number. Round your answer to two decimal places.)
μ = fish
(d) Compute σ, the standard deviation of the number of fish caught per fisherman in a 6-hour period (round 4 or more to 4). (Enter a number. Round your answer to three decimal places.)
σ = fish

Sagot :

ogorwyne

Answer:

a. 0.57

b. 0.22

c. 0.87

d. 0.9450

Step-by-step explanation:

A.

P(X>=1)

= 0.35 + 0.15 + 0.06 + 0.01

= 0.57

This is the probability of catching one or more fishes.

b.

P(X>=2)

0.15 + 0.06 + 0.01

= 0.22

This is the probability of catching 2 or more fishes.

C.

Compute μ

μ = E(X)

= (0x0.43) + (1x0.35) + (2x0.15) + (3x0.06) + (4x0.01)

= 0+0.35+0.3+0.18+0.04

= 0.87

D.

The standard deviation

Sd = √V(X)

V(X) = E(X²) - (E(X))²

E(X²) = (0²x0.43)+(1²x0.35)+(2²*0.15)+(3²*0.06)+(4²*0.01)

E(X²) = 0.35+0.6+0.54+0.16

= 1.65

V(X) = 1.65 - (0.87)²

= 1.65 - 0.7569

= 0.8931

Sd = √0.8931

= 0.9450

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