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3 November, 03:37

The average age of doctors in a certain hospital is 48.0 years old. suppose the distribution of ages is normal and has a standard deviation of 6.0 years. if 9 doctors are chosen at random for a committee, find the probability that the average age of those doctors is less than 48.8 years. assume that the variable is normally distributed.

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  1. 3 November, 06:41
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    True mean = mean (or average) + / - Z*SD/sqrt (sample population)

    Now,

    True mean has to be less than 48.8 years,

    Mean (average) = 48.0 years,

    SD = 6.0 years, and

    Sample size (n) = 9 doctors

    Making Z the subject of the formula

    Z = (True mean - mean) / (SD/sqrt (n))

    Substituting,

    Z = (48.8-48.0) / (6.0/sqrt (9)) = 0.4

    From Normal distribution probabilities table,

    At Z = 0.4, P (x<0.4) = 0.6554 0r 65.54%
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