Question

Movie stars and U.S. presidents have fished Pyramid Lake. It is one of the best places in the lower 48 states to catch trophy cutthroat trout. 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
%	44%	35%	14%	6%	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 =

(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 two decimal places.)
σ =

Answer 1

B)

Probabality that a fisherman selected at random fishing from shore catches one or more fish in a 6 hour period

= P(x = 1) + P(x = 2) + p(x = 3) + p( x >= 4)

= 0.35 + 0.14 + 0.06 + 0.01

= 0.56

C)

Probabality that a fisherman selected at random fishing from shore catches two or more fish in a 6 hour period

= P(x = 2). + P(x = 3) + p(x >= 4)

= 0.14 + 0.06 + 0.01

= 0.21

D)

Expected value = μ

= x p(x)

= 0 x 0.44 + 1x 0.35 + 2 x 0.14 + 3 x 0.06 + 4 x 0.01

= 0.85

E)

Variance =Σ[x^2 * p(x)] - μ^2

= [0^2 x 0.44 +1^2 x 0.35 +2^2 x 0.14 +3^2 x 0.06 +4^2 x 0.01]- 0.85^2

= 1.61 - 0.7225

= 0.8875

Standard deviation = √(variance)

= √0.8875

= 0.94