Biostatistics Help Please

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 52.63 ng/mL with a standard deviation of 5.425 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 61.38 ng/mL or more?

Answer 1

o determine the probability that a randomly selected landscaper has a vitamin D level of 61.38 ng/mL or more, given that the blood vitamin D levels are normally distributed with a mean of 52.63 ng/mL and a standard deviation of 5.425 ng/mL, follow these steps:

1. Standardize the value using the z-score formula:
2. z=X−μσz = \frac{X - \mu}{\sigma}z=σX−μ​
3. where:
4. XXX is the value of interest (61.38 ng/mL),
5. μ\muμ is the mean (52.63 ng/mL),
6. σ\sigmaσ is the standard deviation (5.425 ng/mL).
7. Plugging in the values:
8. z=61.38−52.635.425≈8.755.425≈1.61z = \frac{61.38 - 52.63}{5.425} \approx \frac{8.75}{5.425} \approx 1.61z=5.42561.38−52.63​≈5.4258.75​≈1.61
9. Find the probability associated with this z-score. The z-score of 1.61 corresponds to the area under the normal curve to the left of this value. You can find this using the standard normal distribution table or a calculator.
10. For z=1.61z = 1.61z=1.61, the cumulative probability (the area to the left) is approximately 0.9463.
11. Calculate the probability of having a vitamin D level of 61.38 ng/mL or more. This is the complement of the cumulative probability:
12. P(X≥61.38)=1−P(Z≤1.61)P(X \geq 61.38) = 1 - P(Z \leq 1.61)P(X≥61.38)=1−P(Z≤1.61) P(X≥61.38)=1−0.9463≈0.0537P(X \geq 61.38) = 1 - 0.9463 \approx 0.0537P(X≥61.38)=1−0.9463≈0.0537

Therefore, the probability that a randomly selected landscaper has a vitamin D level of 61.38 ng/mL or more is approximately 0.0537, or 5.37%.