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Question: a research study examined the blood vitamin d levels of...

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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.

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