In: Statistics and Probability
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?
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