Date: Na Statistics Chapter 6 Part II Test 1.) Suppose a team of biologists has

Question

Date: Na Statistics Chapter 6 Part II Test 1.) Suppose a team of biologists has been studying off the coast of Cuttyhunk. Let x represent the length of a single sea bass taken at random from the pond. This group of biologists has determined that x has a normal distribution with mean, = 8.4 inches and standard deviation, .76 inches. (a) Newly passed regulations by allow sea bass longer than 6 inches to be kept. What is th probability that you catch a fish that does not need to be returned to the sea? (b) What is the probability that the mean length R of five sea bass taken at random is larger than 6 inches? (c) What is the probability that the mean length of five sea bass taken at random is between 10.2 inches?

Explanation / Answer

Here Mean = 8.4 inches

Standard deviation = 0.76 inches

(a) Here,,

Pr(x > 6 inches) = 1 - Pr(x < 6 inches) = 1 - NORM(x < 6 inches ; 8.4 inches ; 0.76 inces)

Z = (6 - 8.4)/0.76 = -3.16

Pr(x > 6 inches) = 1 - Pr(x < 6 inches) = 1 - NORM(x < 6 inches ; 8.4 inches ; 0.76 inces)

=1 - Pr(Z < -3.16) = 1 - 0.0008 = 0.9992

(b) Here sample size n = 5

standard error of sample mean = /sqrt(n) = 0.76/sqr(5) = 0.34

Pr(x > 6 inches) = 1 - Pr(x < 6 inches) = 1 - NORMSDIST (x < 6 inches ; 8.4 ; 0.34 inches)

Z = (8.4 - 6)/ 0.34 = 7.06

Pr(x > 6 inches) = 1 - Pr(x < 6 inches) = 1 - NORMSDIST (x < 6 inches ; 8.4 ; 0.34 inches) = 1 - Pr(Z < 7.06)

= 1.000

(c) Pr(6 inches < x < 10.2 inches) = Pr(x < 10.2 inches) - Pr(x < 8 inches)

Z = (10.2 - 8.4)/0.34 = 5.29 ; Z1 = 7.06

Pr(6 inches < x < 10.2 inches) = Pr(x < 10.2 inches) - Pr(x < 8 inches) = 1 - 0 = 1

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