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 43.81 ng/mL with a standard deviation of 5.986 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 49.93 ng/mL or more?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 43.81

standard deviation = $\sigma$ = 5.986

P(X $\geq$ 49.93) = 1 - P(x $\leq$ 49.93)

= 1 - P((x - $\mu$ ) / $\sigma$ $\leq$ (49.93 - 43.81) / 5.986)

= 1 - P(z $\leq$ 1.0224)   Using standard normal table,

= 1 - 0.8467

= 0.1533

P(x $\geq$ 49.93) = 0.1533

Likelihood = 0.1533