Question

Suppose that arsenic concentrations in healthy adults are normally distributed with mean 3.2 and standard deviation 1.6. What is the arsenic concentration such that only 14% of the population has a higher arsenic concentration?

A. 3.5
B. 3.0
C. 2.8
D. 2.5

Answer 1

To find the arsenic concentration such that only 14% of the population has a higher concentration, we can use the standard normal distribution table or a calculator to find the corresponding z-score. Converting the z-score back to the original data scale gives us the desired arsenic concentration.

Step-by-step explanation:

To find the arsenic concentration such that only 14% of the population has a higher concentration, we need to find the z-score corresponding to the cumulative probability of 0.14. With a normal distribution, we can use the standard normal distribution table or a calculator to find the z-score. Using the table or a calculator, we can find that the z-score is approximately -1.08.

Now, we can convert the z-score back to the original data scale using the formula: x = µ + (z × σ), where x is the desired arsenic concentration, µ is the mean, z is the z-score, and σ is the standard deviation.

Plugging in the given values: x = 3.2 + (-1.08 × 1.6) = 1.632.

Therefore, the arsenic concentration such that only 14% of the population has a higher concentration is approximately 1.632.