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%

(a)

Convert the percentages to probabilities and make a histogram of the probability distribution. (Select the correct graph.)

(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

a)

x	0	1	2	3	4 or more
P[X]	0.43	0.35	0.15	0.06	0.01

b)

Pr [ X is one or more ] = 1 - Pr[ X is None]

= 1 - Pr [ X = 0 ]

= 1 - 0.43

= 0.57

c)

Pr[ X is 2 or more ] = 1 - Pr [ X is less than 2 ]

= 1 - Pr [ X < 2 ]

= 1 - [ Pr [ X = 0 ] + Pr [ X = 1 ] ]

= 1 - [ 0.43 + 0.35 ]

= 1 - 0.78

= 0.22

d]

The following table shows the provided outputs of the discrete random variables, along with the corresponding probabilities:

X	p(X)
0	0.43
1	0.35
2	0.15
3	0.06
4	0.01

Now, we need to multiply the corresponding X outcomes with the corresponding probabilities, in order to compute the population mean μ:

X	p(X)	X⋅p(X)
0	0.43	0⋅0.43=0
1	0.35	1⋅0.35=0.35
2	0.15	2⋅0.15=0.3
3	0.06	3⋅0.06=0.18
4	0.01	4⋅0.01=0.04

Therefore, the population mean is calculated as follows:

e)

Now, we need to multiply the corresponding squares X^2 of the outcomes with the corresponding probabilities, in order to compute the population variance σ2:

X	p(X)	X⋅p(X)	X2⋅p(X)
0	0.43	0⋅0.43=0	0^2⋅0.43=0
1	0.35	1⋅0.35=0.35	1^2 ⋅0.35=0.35
2	0.15	2⋅0.15=0.3	2^2 ⋅0.15=0.6
3	0.06	3⋅0.06=0.18	3^2 ⋅0.06=0.54
4	0.01	4⋅0.01=0.04	4^2 ⋅0.01=0.16

Therefore, we first compute the expected value of X^2:

Therefore, the population variance is computed as follows:

And finally, taking square root to the variance we get that the population standard deviation is

Therefore, based on the data provided for the discrete distribution, the population mean is μ=0.87 and the population standard deviation is σ=0.945

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