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?

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  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.

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

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  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

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  4. Thanks Kuai that really helped me out.

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