question archive 1) A particular lake is known to be one of the best places to catch a certain type of fish
1) A particular lake is known to be one of the best places to catch a certain type of fish
Subject:StatisticsPrice:2.87 Bought7
1) 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
2.
Compute C5,3. (Enter an exact number.)
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ANSWER:
b). =0.57
c). =0.22
d). =0.87
e). =0.945
2). =10
Step-by-step explanation
x 0 1 2 3 4 or more
0.43 0.35 0.15 0.06 0.01
a). select appropriate histogram by seeing above table
b). catches 1 or more fish
=P(X?≥? 1)=0.35+0.15+0.06+0.01=0.57
c). 2 or more fish
=P(X?≥? 2)=0.15+0.06+0.01=0.22
d). expected value u=xP(X)
u=0*0.43+1*0.35+2*0.15+3*0.06+4*0.01=0.87
e). standard deviation σ
Var(X)=E(X2)-u2
E(X2)=0*0.43+12*0.35+22*0.15+32*0.06+42*0.01=1.65
=1.65-0.872=0.8931
so σ=?0.8931?? =0.945
2). C5,3=5!/3!(5-3)!=5*4/2=10
