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 43.24 ng/mL with a standard deviation of 5.616 ng/mL. Assuming the true distribution of blood vitamin D levels follows a normal distribution, what is the 60-th percentile? Recall that the statistic, "percentile", is defined to be the value below which a given percentage of observations are located within the distribution. In other words, for this question find the vitamin D value for which 60% of the observations within the distribution are located to the left of this value.

Answer 1

Solution :

mean = $\mu$ = 43.24

standard deviation = $\sigma$ = 5.616

Using standard normal table,

P(Z < z) = 60%

P(Z < z ) = 0.6

z =0.25

Using z-score formula,

x = z * $\sigma$ + $\mu$

x =0.25 * 5.616 + 43.24

x = 44.64

the distribution are located to the left of this value = 44.64