Question

In a certain population of fish, the lengths of the individual fish follow approximately a normal distribu- tion with mean 54.0 mm and standard deviation 4.5 mm. We saw in Example 4.3.1 that in this situation 65.68% of the fish are between 51 and 60 mm long. Suppose a ran- dom sample of four fish is chosen from the population. Find the probability that

(a) allfourfisharebetween51and60mmlong.

(b) the mean length of the four fish is between 51 and 60 mm.

From the Book Statistics for the life Sciences 4th edition Samuels, Witmer, Schaffner

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 54
standard Deviation ( sd )= 4.5
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To find P(a < = Z < = b) = F(b) - F(a)
P(X < 51) = (51-54)/4.5
= -3/4.5 = -0.6667
= P ( Z <-0.6667) From Standard Normal Table
= 0.2525
P(X < 60) = (60-54)/4.5
= 6/4.5 = 1.3333
= P ( Z <1.3333) From Standard Normal Table
= 0.9088
P(51 < X < 60) = 0.9088-0.2525 = 0.6568
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(a)
sample size (n) = 4; standard Deviation ( sd )= 4.5/ Sqrt ( 4 ) =2.25
to find P(a <= Z <=b) = F(b) - F(a)
P(X < 51) = (51-54)/4.5/ Sqrt ( 4 )
= -3/2.25
= -1.3333
= P ( Z <-1.3333) From Standard Normal Table
= 0.0912
P(X < 60) = (60-54)/4.5/ Sqrt ( 4 )
= 6/2.25 = 2.6667
= P ( Z <2.6667) From Standard Normal Table
= 0.9962
P(51 < X < 60) = 0.9962-0.0912 = 0.905
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(b)
the mean length of the four fish is between 51 and 60 mm is 54