Question

Bass - Samples: The bass in Clear Lake have weights that are normally distributed with a mean of 2.2 pounds and a standard deviation of 0.5 pounds. Suppose you catch a stringer of 6 bass with a total weight of 16.1 pounds. Here we determine how unusual this is.

(a) What is the mean fish weight of your catch of 6? Round your answer to 1 decimal place.
pounds

(b) If 6 bass are randomly selected from Clear Lake, find the probability that the mean weight is greater than the mean of those you caught. Round your answer to 4 decimal places.


(c) Which statement best describes your situation?

This is not particularly unusual because the mean weight of your fish is only 0.5 pounds above the population average.This is unusual because the probability of randomly selecting 6 fish with a mean weight greater than or equal to the mean of your stringer is less than the benchmark probability of 0.05.     

Answer 1

Answer:

a)

Given,

sample n = 6

Total weight = 16.1

Mean = x / n

= 16.1 / 6

= 2.683

= 2.7

Mean = 2.7

b)

Given,

Mean = 2.2

Standard deviation = 0.5

sample size n = 6

Now consider,

P(xbar > 2.7) = P((xbar-mu)/(s/sqrt(n)) > (2.7 - 2.2)/(0.5/sqrt(6)))

= P(z > 2.45)

= 0.0071428 [since from z table]

= 0.0071

c)

This is strange on the grounds that the likelihood of arbitrarily choosing 6 fish with a mean weight more prominent than or equivalent to the mean of your stringer is not exactly the benchmark likelihood of 0.05.