Posted by **Michael** on Sunday, October 28, 2012 at 9:48pm.

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 -
**MathMate**, Sunday, October 28, 2012 at 10:14pm
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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