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Suppose X indicates one hydrologic variable following a lognormal distribution with a mean of 350 m³/s and a standard deviation of 70 m³/s. If Y denotes the log-transformed series of X, i.e., Y = ln(X), then approximately the mean and standard deviation of Y will be:

a) 4.15 and 1.80, respectively.
b) 5.84 and 0.20, respectively.
c) 8.25 and 3.25, respectively.
d) 11.50 and 3.80, respectively.

Consider a 50-year data set of annual maximum 24-hour rainfall at a particular place that follows a normal distribution with a mean of 101.3 mm and a standard deviation of 60 mm. Determine the magnitude of annual maximum rainfall with a return period of 20 years:

a) 162.6 mm
b) 150.0 mm
c) 178.5 mm
d) 200.0 mm

Answer :

To solve this problem, we need to understand the transformation between a lognormal distribution and its associated normal distribution.

Part 1: Mean and Standard Deviation of Y = ln(X)

Given:

  • X follows a lognormal distribution with a mean [tex]\mu_X = 350[/tex] m³/s and standard deviation [tex]\sigma_X = 70[/tex] m³/s.
  • Y is the natural log transformation of X, i.e., [tex]Y = \ln(X)[/tex].

Steps:

  1. Convert Parameters of Lognormal to Normal: The variables [tex]\mu[/tex] and [tex]\sigma[/tex] for a lognormal distribution relate to its normal mean ([tex]\mu_Y[/tex]) and standard deviation ([tex]\sigma_Y[/tex]) using:

    \mu_Y = \ln\left(\frac{\mu_X^2}{\sqrt{\mu_X^2 + \sigma_X^2}}\right)
    \sigma_Y = \sqrt{\ln\left(1 + \frac{\sigma_X^2}{\mu_X^2}\right)}

  2. Calculate Mean and Standard Deviation for Y:

    • [tex]\mu_Y = \ln\left(\frac{350^2}{\sqrt{350^2 + 70^2}}\right) \approx \ln(338.2) \approx 5.84[/tex]
    • [tex]\sigma_Y = \sqrt{\ln\left(1 + \frac{70^2}{350^2}\right)} \approx \sqrt{0.04} \approx 0.20[/tex]

Thus, the correct choice for the mean and standard deviation of Y [tex]= \ln(X)[/tex] is b) 5.84 and 0.20, respectively.

Part 2: Magnitude of Annual Maximum Rainfall with Return Period of 20 Years

Given:

  • The rainfall data follows a normal distribution with mean [tex]\mu = 101.3[/tex] mm and standard deviation [tex]\sigma = 60[/tex] mm.
  • Return period T is 20 years.

Steps:

  1. Convert the Return Period to Probability of Exceedance (P):

    • [tex]P = \frac{1}{T} = \frac{1}{20} = 0.05[/tex] (probability of exceedance in any given year)

  2. Determine the Z-score for a 20-year return period:

    • We use the inverse of the standard normal distribution to find the Z-score corresponding to a cumulative probability of [tex]1 - P = 0.95[/tex].
    • Consulting a standard normal (Z) table, [tex]Z \approx 1.645[/tex].

  3. Calculate the Magnitude of Rainfall for the 20-year Return Period:

    • Using the formula: X_T = \mu + Z \cdot \sigma
    • [tex]X_{20} = 101.3 + 1.645 \times 60 \approx 199.9 \approx 200 \text{ mm}[/tex]

Therefore, the magnitude of the annual maximum rainfall with a 20-year return period is d) 200.0 mm.

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