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 41.37 ng/mL with a standard deviation of 4.43 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 31.47 ng/mL or more? Answer:

Answer 1

Solution :

Given that ,

mean = = 41.37

standard deviation = = 4.43

P(X 31.47) = 1 - P(x 31.47)

= 1 - P((x - ) / (31.47 - 41.37) / 4.43)

= 1 - P(z < -2.2348)   Using standard normal table,

= 1 - 0.0127   

= 0.9873

P(X 31.47) = 0.9873

Likelihood = 0.9873