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The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.52 inches and a standard deviation of 0.46 inches


(A) What percentage of the grapefruits in this orchard is larger than 5.45 inches?
(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.45 inches

  • Math -

    Normalize!

    Calculate Z=(X-μ)/σ
    where
    Z = Z score found in standard normal distribution tables
    X = value of random variable, the diameter of 5.45 inchs
    μ = mean diameter of population, 5.52 inches
    σ = standard deviation of population, 0.46 inch.

    Look up a standard normal distribution, which is usually given in percentage of population less than the Z-score, i.e. left tail (but check if this is the case).

    What you need is the right tail, i.e. percentage of population greater than the Z-score. To get this from the table, subtract the left tail from ONE to get the right-tail.

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