Question

for certain rainfall data, the standard deviation Sx = 3.4 mm. If the Normal distribution was used and the 10 year rain x10 = 34.1 mm, then the 2 year rain x2 is:

6.82 mm

29.74 mm

31.72 mm

38.18 mm

Answer 1

Given, normal distribution is used in rainfall data.

Normal distribution is characterized by the mean and standard deviation.

The depth of rainfall with specific probability of exceedance can be estimated using the following equation:

……………………………….equation (A)

Where,

k = frequency factor (depends on the selected probability of exceedance)

s = standard deviation of the sample data

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Frequency factor for 10 year rain and 2 year rain

Frequency factor is given by the following equation:

k = frequency factor

Where, p = (1/T),

Where, T = return period

As per the question, we need frequency factor for return period 10 years and 2 years Inorder to get the required answers.

Frequency factor for 10 year rain:

T = return period = 10 years

p = (1/T)

p = 1/10

p = 0.1

w = 2.145967

k = 1.281729 for return period = 10 years

Frequency factor for 2 year rain:

T = return period = 2 years

p = (1/T)

p = 1/2

p = 0.5

w = 1.17741

k = -0.0000001 for return period = 2 years

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As per the question,

For 10 year rain or (given)

s = 3.4 mm (given)

Return period = 10 years and probability of exceedance = 10% (given)

As per the normal distribution, frequency factor corresponding to probability of exceedance = 10% is: k = 1.281729 (calculated above)

Substituting the above values in the equation (A),

Thus, the mean of the sample data = 29.742

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s = 3.4 mm (given)

(Calculated above)

Return period = 2 years and thus probability of exceedance = 50%.

As per the normal distribution, frequency factor corresponding to probability of exceedance = 50% is: k = -0.0000001 (calculated above)

Substituting the above values in the equation (A),

The 2 year rain = 29.742 or 29.74 mm

Answer: 29.74 mm