Question

In: Statistics and Probability

Suppose a team of biologists has been studying the Pinedale children’s fishing pond. Let x represent...

Suppose a team of biologists has been studying the Pinedale children’s fishing pond. Let x represent the length of a single trout taken at random from the pond. This group of biologists has determined that x has a normal distribution with mean µ = 10.2 inches and standard deviation σ = 1.4 inches.

a. What is the probability that a single trout taken at random from the pond is between 8 and 12 inches long?

b. What is the probability that the mean length x ̅ of 16 trout taken at random is between 8 and 12 inches?

Solutions

Expert Solution

Solution :

Given that ,

mean = = 10.2

standard deviation = = 1.4   

P(8< x < 12) = P[(8-10.2) /1.4 < (x - ) / < (12-10.2) /1.4 )]

= P(-1.57 < Z <1.29 )

= P(Z <1.29 ) - P(Z< -1.57 )

Using z table   

= 0.9015-0.0582

probability= 0.8433

(B)

n = 16

= 10.2

=  / n= 1.4/ 16=0.35

P(8<     <13 ) = P[(8-10.2) /0.35  < ( - ) /   < (12-10.2) /0.35 )]

= P(-6.29 < Z < -0.57)

= P(Z <-0.57 ) - P(Z <-6.29 )

Using z table

=0.2843-0

=0.2843

probability=0.2843


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