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I have a statistics question?

The weight of steers in a herd are distributed normally the variance is 40,000 and the mean steer weight is 1500 LBS. find the probability that a randomly selected steer weighs more then 1,900 LBS. Explanation would be great but answer works too lol i lost my notes

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  • Anonymous
    9 years ago
    Favorite Answer

    P(X > 1900)

    = P(Z > (1900 - 1500) / 200)

    = P(Z > 2)

    = 1- Φ(2)

    = 1 - 0.97725

    = 0.02275

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  • Anonymous
    9 years ago

    Ans:

    mean=1500

    sd=sqrt(40,000)

    test statistic, z= (x- mean )/sd

    Need to find P(x>1900)=?

    if x=1900 , then z=(1900-1500)/sqrt(40,000)=2

    P(x>1900)=P(z>2)=0.5-P(0<z<2)

    =0.5-0.4772

    =0.0228(Ans.)

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