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 47.84 ng/mL with a standard deviation of 4.465 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 50.08 ng/mL or more?

Answer 1

Solution :

Given that ,

mean = = 47.84

standard deviation = = 4.465

P(X 50.08) = 1 - P(x 50.08)

= 1 - P((x - ) / (50.08 - 47.84) / 4.465)

= 1 - P(z < 0.5017)

= 1 - 0.6921

= 0.3079

P(X 50.08) = 0.3079

Likelihood = 0.3079