Question

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
%	45%	37%	13%	4%	1%

(a) Convert the percentages to probabilities and make a histogram of the probability distribution.

(b) Find the probability that a fisherman selected at random fishing from shore catches one or more fish in a 6-hour period. (Round your answer to two decimal places.)


(c) Find the probability that a fisherman selected at random fishing from shore catches two or more fish in a 6-hour period. (Round your answer to two decimal places.)


(d) Compute μ, the expected value of the number of fish caught per fisherman in a 6-hour period (round 4 or more to 4). (Round your answer to two decimal places.)
μ = fish

(e) Compute σ, the standard deviation of the number of fish caught per fisherman in a 6-hour period (round 4 or more to 4). (Round your answer to three decimal places.)
σ = fish

Answer 1

b)

probability that a fisherman selected at random fishing from shore catches one or more fish in a 6-hour period =P(X>=1) =1-P(X=0) =1-0.45 =0.55

c)

probability that a fisherman selected at random fishing from shore catches two or more fish in a 6-hour period =P(X>=2 ) =0.180

d)

x	P(X=x)	xP(x)	x2P(x)
0	0.450	0.00000	0.00000
1	0.370	0.37000	0.37000
2	0.130	0.26000	0.52000
3	0.040	0.12000	0.36000
4	0.010	0.04000	0.16000
total		0.7900	1.4100
	E(x) =μ=	ΣxP(x) =	0.7900
	E(x2) =	Σx2P(x) =	1.4100
	Var(x)=σ2 =	E(x2)-(E(x))2=	0.785900
	std deviation=	σ= √σ2 =	0.88651

μ =0.79

e)

σ =0.887