This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Gumbel’s Method – Set 3”.
1. Which of the following is a valid expression for one of the confidence limits of a Gumbel’s variate XT? f(c) is a function of the confidence probability and E is the probable error.
a) \(X_T+\frac{f(c)}{E}\)
b) \(X_T+\frac{E}{(f(c)}\)
c) XT-E.f(c)
d) E.f(c)-XT
View Answer
Explanation: The confidence limits for a Gumbel’s variate are given by two values between which the variate may lie. It depends on the confidence probability and probable error. It is given as,
Confidence limits of XT=XT±E.f(c)
2. What does 80% confidence limits (a and b) of a variate signify?
a) Variate has a 20% probability of lying between a and b
b) Variate has an 80% probability of lying between a and b
c) Variate has less than 80% probability of lying between a and b
d) Variate has more than 80% probability of lying between a and b
View Answer
Explanation: Confidence limits indicate the range between which the calculated variate lies with a given probability that the true variate value also lies within that range. For example, C% confidence limits c1 and c2 indicate that a variate has C% chance of lying between c1 and c2.
3. The 95% confidence limits are approximately three times as large as the 50% confidence limits for a Gumbel variate.
a) True
b) False
View Answer
Explanation: As the confidence probability that a value must lie within a specific interval increases, the size of the interval also naturally increases. For example, in Gumbel confidence limits for a variate for a given sample size, it can be observed that the 80% confidence limits are twice the 50% confidence limits, and 95% confidence limits are almost thrice as large as 50% confidence limits.
4. Which of the following quantities are required to compute the probable error for confidence limit estimation?
a) Frequency factor, mean of sample, sample size
b) Reduced variate, sample size, standard deviation of sample
c) Reduced variate, mean and standard deviation of sample
d) Frequency factor, standard deviation of sample, sample size
View Answer
Explanation: The probable error is given as,
Se=\(b.\frac{σ_(N-1)}{\sqrt{N}}\)
Where, b is a function of the frequency factor, σN-1 is the standard deviation of the sample and N is the sample size.
5. What is the correct expression for the frequency factor term in the probable error formula in estimation of confidence limits for a variate? K is the frequency factor for the variate.
a) \(\sqrt{1+1.1K+1.3K^2}\)
b) \(\sqrt{1+1.3K+1.1K^2}\)
c) \(\sqrt{K+1.1K^2+1.3K^3}\)
d) \(\sqrt{K+1.3K^2+1.1K^3}\)
View Answer
Explanation: The probable error formula required for the estimation of confidence limits involves a term b, which is purely a function of the frequency factor and does not depend on the confidence probability. It is given as,
b=\(\sqrt{1+1.3K+1.1K^2}\).
6. The mean and standard deviation of an annual flood series for a catchment is 220 m3/s and 72 m3/s, respectively. What is the probability of a 400 m3/s flood occurring in the catchment within the next 36 months? Use Gumbel’s method with a reduced mean 0.54 and reduced standard deviation 1.14.
a) 6.7%
b) 8.7%
c) 9.6%
d) 11.4%
View Answer
Explanation: Given \(\overline{x}\)=220 m3/s, σN-1=72 m3/s, XT = 400 m3/s, \(\overline{y}\) =0.54 and Sn = 1.14
Now from Gumbel’s flood frequency equation,
XT=\(\overline{x}\)+K.σN-1 ⇒ 400=220+72K ⇒ K=2.5
⇒\(K=\frac{y_T-\overline{y_n}}{S_n} =2.5 ⇒ \frac{y_T-0.54}{1.14}\)=2.5 ⇒ yT=3.39
⇒\(-ln.ln\frac{T}{T-1}=3.39 ⇒ ln\frac{T}{T-1}=e^{-3.39}=0.03371 ⇒ \frac{ T}{T-1}=e^{0.03371}\)=1.034285
⇒T=\(\frac{1.034285}{1.034285-1}\)=30.17 years
⇒ Probability of occurrence of 400 m3/s flood, p=\(\frac{1}{T}=\frac{1}{30.17}\)=0.03315
⇒ Probability of non-occurrence, q = 1 – p = 1 – 0.03315 = 0.96685
∴Probability of occuring once in next 3 years=1-q3=1-0.966853=0.09619≅9.6%
7. An annual flood series of 100 years for a catchment had a mean of 2100 m3/s and standard deviation of 700 m3/s. What is the maximum ratio of the 68% confidence limits for a flood of return period 100 years? Take reduced mean as 0.56, reduced standard deviation as 1.2065 and confidence probability function as unity.
a) 0.876
b) 0.972
c) 1.008
d) 1.142
View Answer
Explanation: Calculating the terms required for confidence limits,
\(y_T=-ln.ln\frac{T}{T-1}=ln.ln\frac{100}{99}\)=4.60015
\(X_{100}=x ̅+(\frac{y_{100}- \overline{y_n}}{S_n}).σ_{N-1}=2100+(\frac{4.60015- 0.56}{1.2065}).700\)=4444 m3/s
\(K=\frac{y_T-\overline{y_n}}{S_n} = \frac{4.60015-0.56}{1.2065}\)=3.35
b=\(\sqrt{1+1.3K+1.1K^2}=\sqrt{1+(1.3*3.35)+(1.1*3.35^2)}\)=4.207
Se=\(b.\frac{σ_{N-1}}{\sqrt{N}}=4.207*\frac{700}{\sqrt{100}}\)=294.5
⇒ 68% confidence limits =X100±f(c).Se=4444+(1*294.5)=4738.5 and 4149.5
∴Maximum ratio of confidence limits=\(\frac{4738.5}{4149.5}\)=1.142
8. An annual flood series of 40 years for a river had a mean of 4280 m3/s and standard deviation of 1860 m3/s. It was estimated using Gumbel’s method that the flood value for a return period of 250 years was 12390 m3/s. What is the 99% upper confidence limit of this variate? Take function of confidence probability as 2.58 and frequency factor as 4.36.
a) 8405
b) 15445
c) 16375
d) 19124
View Answer
Explanation: Calculating the terms required for confidence limits,
b=\(\sqrt{1+1.3K+1.1K^2}=\sqrt{1+(1.3*4.36)+(1.1*4.36^2)}\)=5.2515
Se=\(b.\frac{σ_{N-1}}{\sqrt{N}}=5.2515*\frac{1860}{\sqrt{40}}\)=1544.42
∴ 99% upper confidence limit =XT+f(c).Se=12390+(2.58*1544.42)=16374.6≅16375 m3/s
Sanfoundry Global Education & Learning Series – Engineering Hydrology.
To practice all areas of Engineering Hydrology, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
