What is the probability that exactly 4/7 parts have greater lengths than 26cm given normal distribution,mean?

Suppose that the lengths of automotive parts on a production line can be regarded as independent observations from a normal distribution with mean 25cm and standard deviation 4cm. Seven parts are randomly selected from the line. What is the probability that exactly four of these parts have lengths greater than 26cm.

Not sure what i am doing wrong. using sd as 4/(7^.5).

Any help would be good. Some steps would be good. Thanks

1 Answer

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  • Faz
    Lv 7
    1 decade ago
    Favorite Answer

    First, calculate the probability that one of these parts have lengths greater than 26cm.

    z = (26-25)/4 = 0.25

    P(Z>0.25) = 1 - P(Z<0.25)

    From tables or calculator, P(Z<0.25) = 0.598706, so P(Z>0.25) = 0.401294

    Since the probabilities of the lengths of the parts are independent observations, use the binomial distribution:

    = (7C4) . ( 0.401294)^4 . ( 1 - 0.401294)^3

    = 0.194787

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