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%	35%	15%	6%	1%

Answer 1

Answer:

To determine the probability that a fisherman selected at random fishing from shore catches one or more fish in a 6-hour period

i.e.,

= P(x > = 1)

=1 - P(x = 0)

=1 - 0.43

= 0.57

To give 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 - 0.35

= 0.22

To give the expected value of the number of fish caught per fisherman in a 6-hour period

i.e.,

μ = 0*0.43+1*0.35+2*0.15+3*0.06+4*0.01

=0.87

Now to compute the standard deviation of the number of fish caught per fisherman in a 6-hour period

i.e.,

σ = sqrt((0-0.87)^2*0.43+(1-0.87)^2*0.35+(2-0.87)^2*0.15+(3-0.87)^2*0.06+(4-0.87)^2*0.01)

Hence standard deviation σ = 0.945