A survey of 250 lobster fisherman found that they catch an average of 32 pounds of lobster per day with a standard deviation of four pounds. If a random sample of 30 lobster fisherman is selected, what is the probability that their average catch is less than 31.5 pounds?

Answer 1

Z = (score-mean)/SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

Answer 2

I am not understanding how to set this up. I have n=30 SD=4 xbar=250 after that I am lost

Answer 3

n = 30
x = 31.5
μ = 32
SEm = SD/√n
SEm = 4/√30
SEm = 0.73
z = ( x - μ ) / SEm
z = (31.5-32)/0.73
z = -0.68
P(z < -0.68) = 0.2483

Answer 4

Thanks Kuai that really helped me out.