Question

A research study examined the blood vitamin D levels of the entire US population of landscape gardeners. The population average level of vitamin D in US landscapers was found to be 34.15 ng/mL with a standard deviation of 5.672 ng/mL. Assuming the true distribution of blood vitamin D levels follows a normal distribution, if you randomly select a landscaper in the US, what is the likelihood that his/her vitamin D level will be 49.72 ng/mL or less?

Answer 1

Given: = 34.15 ng/ml, = 5.672 ng/ml

To find the probability, we need to find the Z scores first.

Z = (X - )/ [/n]. Since n = 1, Z = (X - )/

To find the required probabilities in excel we use the function NORMSDIST, which returns the left tailed probability.

If we need to find a right tailed probability, first find the left tailed probability and then do 1 - the the left tailed value.

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(b) For P( X 49.72) = P(X < 49.72)

Z = (49.72 – 34.15/5.672 = 2.75

The required probability for P(X < 49.72) is = 0.9970

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