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 40.42 ng/mL with a standard deviation of 5.446 ng/mL. Assuming the true distribution of blood vitamin D levels follows a Gaussian distribution, if you randomly select a landscaper in the US, what is the likelihood that his/her vitamin D level will be between 49.82 and 58.82 ng/mL?

Answer 1

Given,

= 40.42 , = 5.446

We convert this to standard normal as

P( X < x) = P( Z < x - / )

So,

P(49.82 < X < 58.82) = P( X < 58.82) - P( X < 49.82)

= P( Z < 58.82 - 40.42 / 5.446) - P( Z < 49.82 - 40.42 / 5.446)

= P( Z < 3.3786) - P( Z < 1.7260)

= 0.9996 - 0.9578

= 0.0418