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 50.46 ng/mL with a standard deviation of 5.151 ng/mL. Assuming the true distribution of blood vitamin D levels follows a normal distribution, what is the 91-st 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 91% of the observations within the distribution are located to the left of this value.
Please show your work

Looking for a Similar Assignment? Order now and Get 10% Discount! Use Coupon Code "Newclient"