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19 August, 21:25

Consider a normal distribution with mean 30 and standard deviation 2. What is the probability that a value selected at random from this distribution is greater than 30

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  1. 19 August, 22:43
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    The probability that the value selected at random is greater than 30 will be found as follows:

    z-score is given by:

    z = (x-mu) / sig

    z = (30-30) / 2=0

    thus

    P (x>30) = 1-P (X<30) = 1-P (z<0) = 1-0.5=0.5

    Answer P (x>30) = 0.5
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