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 % 42% 34% 12% 11% 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 three decimal places.) σ = fish

Answer 1

Answer:

Given,

x	0	1	2	3	4 or more
P(x)	0.42	0.34	0.12	0.11	0.01

b)

P(X >= 1) = P(1) + P(2) + P(3) + P(4 or more)

substitute values

= 0.34 + 0.12 + 0.11 + 0.01

= 0.58

c)

P(X >= 2) = P(2) + P(3) + P(4 or more)

substitute values

= 0.12 + 0.11 + 0.01

= 0.24

d)

Mean = E(X) = x*P(x)

= 0*0.42 + 1*0.34 + 2*0.12 + 3*0.11 + 4*0.01

= 0 + 0.34 + 0.24 + 0.33 + 0.04

= 0.95

e)

consider,

E(X^2) = x^2*P(x)

= 0^2*0.42 + 1^2*0.34 + 2^2*0.12 + 3^2*0.11 + 4^2*0.01

= 0 + 0.34 + 0.48 + 0.99 + 0.16

= 1.97

Variance = E(X^2) - [E(X)]^2

= 1.97 - 0.95^2

= 1.068

Standard deviation = sqrt(variance)

= sqrt(1.068)

= 1.033