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
%	43%	36%	15%	5%	1%

Answer 1

Ans:

The 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)

=1-0.43

=0.57

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

=1-P(x=0)-P(x=1)

=1-0.43-0.36
=0.21

(d) The expected value of the number of fish caught per fisherman in a 6-hour period
μ = 0*0.43+1*0.36+2*0.15+3*0.05+4*0.01

=0.8500

(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 = sqrt((0-0.85)^2*0.43+(1-0.85)^2*0.36+(2-0.85)^2*0.15+(3-0.85)^2*0.05+(4-0.85)^2*0.01)

=0.9206