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 30.8 ng/mL with a standard deviation of 4.371 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 between 36.84 and 39.73 ng/mL

Answer 1

0.0631 = 6.31% probability that his/her vitamin D level will be between 36.84 and 39.73 ng/mL

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The population average level of vitamin D in US landscapers was found to be 30.8 ng/mL with a standard deviation of 4.371 ng/mL

This means that
`\mu = 30.8, \sigma = 4.371`

What is the likelihood that his/her vitamin D level will be between 36.84 and 39.73 ng/mL?

This is the pvalue of Z when X = 39.73 subtracted by the pvalue of Z when X = 36.84.

X = 39.73

`Z = (X - \mu)/(\sigma)`

`Z = (39.73 - 30.8)/(4.371)`

`Z = 2.04`

`Z = 2.04` has a pvalue of 0.9793

X = 36.84

`Z = (X - \mu)/(\sigma)`

`Z = (36.84 - 30.8)/(4.371)`

`Z = 1.38`

`Z = 1.38` has a pvalue of 0.9162

0.9793 - 0.9162 = 0.0631

0.0631 = 6.31% probability that his/her vitamin D level will be between 36.84 and 39.73 ng/mL