Question

An article suggests the lognormal distribution as a model for SO₂ concentration above a certain forest. Suppose the parameter values are = 1.7 and = 0.8. A button hyperlink to the SALT program that reads: Use SALT. (a) What are the mean value and standard deviation of concentration? (Round your answers to three decimal places.) mean standard deviation (b) What is the probability that concentration is at most 10? Between 5 and 10? (Round your answers to four decimal places.) at most 10 between 5 and 10

Determine z for the following of . (Round your answers to two decimal places.)
A button hyperlink to the SALT program that reads: Use SALT.
(a) = 0.0063

(b) = 0.18

(c) = 0.643

Answer 1

a) The mean concentration is 5.042 units and the standard deviation is 17.486 units.

b) The probability that the concentration is at most 10 is 0.9041 and the probability that the concentration is between 5 and 10 is 0.1034.

c) The z-scores for the values x = 0.0063, x = 0.18, and x = 0.643 are approximately 70.3, 5.54, and 1.34, respectively.

a) Mean and standard deviation

The mean and standard deviation of a lognormal distribution with parameters μ and σ are given by:

mean = e^(μ + σ^2/2)

std = e^(μ) * sqrt(e^(σ^2) - 1)

Plugging in the values μ = 1.7 and σ = 0.8, we get:

mean = e^(1.7 + 0.8^2/2) ≈ 5.042

std = e^(1.7) * sqrt(e^(0.8^2) - 1) ≈ 17.486

Therefore, the mean concentration is 5.042 units and the standard deviation is 17.486 units.

b) Probability

The probability that the concentration is at most 10 is given by the cumulative distribution function (CDF) of the lognormal distribution:

P(X ≤ 10) = P(ln(X) ≤ ln(10)) = Φ((ln(10) - μ) / σ)

where Φ is the standard normal CDF. Plugging in the values μ = 1.7, σ = 0.8, and ln(10) ≈ 2.303, we get:

P(X ≤ 10) = Φ((2.303 - 1.7) / 0.8) ≈ 0.9041

The probability that the concentration is between 5 and 10 is given by the difference of the CDFs:

P(5 ≤ X ≤ 10) = P(X ≤ 10) - P(X ≤ 5) = Φ((ln(10) - μ) / σ) - Φ((ln(5) - μ) / σ)

Plugging in the values μ = 1.7, σ = 0.8, ln(10) ≈ 2.303, and ln(5) ≈ 1.609, we get:

P(5 ≤ X ≤ 10) = Φ((2.303 - 1.7) / 0.8) - Φ((1.609 - 1.7) / 0.8) ≈ 0.1034

Therefore, the probability that the concentration is at most 10 is 0.9041 and the probability that the concentration is between 5 and 10 is 0.1034.

c) z-values

The z-score of a value x in a lognormal distribution with parameters μ and σ is given by:

z = (ln(x) - μ) / σ

Plugging in the values μ = 1.7, σ = 0.8, and x = 0.0063, we get:

z = (ln(0.0063) - 1.7) / 0.8 ≈ 70.3

Plugging in the values μ = 1.7, σ = 0.8, and x = 0.18, we get:

z = (ln(0.18) - 1.7) / 0.8 ≈ 5.54

Plugging in the values μ = 1.7, σ = 0.8, and x = 0.643, we get:

z = (ln(0.643) - 1.7) / 0.8 ≈ 1.34

Therefore, the z-scores for the values x = 0.0063, x = 0.18, and x = 0.643 are approximately 70.3, 5.54, and 1.34, respectively.