Question

PLEASE DO NOT ROUND YOUR ANSWER. THANKS

1. 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 554 ng/mL with a standard deviation of 4.191 ng/mL. Assuming the true distribution of blood vitamin D levels follows a normal distribution, if you randomly select a landscaper in the US, what is the probability that his/her vitamin D level will be 59.88 ng/mL or more?

Answer 1

µ =    55.4                  
σ =    4.191                  
                      
P ( X ≥   59.88   ) = P( (X-µ)/σ ≥ (59.88-55.4) / 4.191)              
= P(Z ≥   1.069   ) = P( Z <   -1.069   ) =   

0.142544457648
(answer)

Thanks in advance!

revert back for doubt

Please upvote