Question

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%

(b) 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.)

(c) 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.)

(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). (Enter a number. 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). (Enter a number. Round your answer to three decimal places.) σ = fish

Answer 1

Solution :

x	P(x)	x * P(x)	x2 * P(x)
0	0.43	0	0
1	0.35	0.35	0.35
2	0.15	0.3	0.6
3	0.06	0.18	0.54
4	0.01	0.04	0.16
Sum	1	0.87	1.65

(b)

The probability that a fisherman selected at random fishing from shore catches one or more

fish in a 6-hour period is,

= P(x 1)

= 0.35 + 0.15 + 0.06 + 0.01

= 0.57

(c)

The probability that a fisherman selected at random fishing from shore catches two or

more fish in a 6-hour period is ,

= P(x 2)

= 0.15 + 0.06 + 0.01

= 0.22

(d)

Mean = = X * P(X) = 0.87 fish

(e)

Standard deviation =

=X ² * P(X) - ²

=  1.65 - 0.87²

= 0.945 fish