I need help in find Q100. The mean is 8855.824 And the standard deviations is 57

Question

I need help in find Q100. The mean is 8855.824 And the standard deviations is 5709.88 odav/pid-4789659-dt-content-rid-25582817 1/courses/16139.1181c/2017 AP Step 3: Estimation of the 100-year flood To estimate the 100-year flood (Q1a) on the Chattooga River near Clayton, GA, we will assume the annual maximum flows are well described by a log-Pearson III distribution. The log-Pearson II distribution is the recommended distribution for fitting annual maximum streamflows in the United States, since it has been repeatedly shown to produce a good fit to annual maximum streamflow data in many different regions. The log-Pearson III distribution is an extremely flexible distribution for strictly positive random variables. If the coefficient of skewness in logarithmic space of the flows is zero, then the log-Pearson III turns into a lognormal distribution, another distribution that is commonly used to describe many environmental variables. Qim is the annual maximum streamflow that is exceeded on average once every 100 years, and is defined as: 100 where O is a random variable representing annual maximum streamflow. Employing a log- Pearson III distribution requires a logarithmic transformation of the data: Y, = In(Q,) We will first estimate Ys and then determine Qo as: Q100 = exp(Yo) These parameters are the mean of the log annual maxinum flows, ', the standard dntatio" of the log annual maximum of The log-Pearson IIl distribution is a 3-parameter distribution. nlows, ay, and a frequency factor, Kp which is a function of the coefficient of skewness the log annual maximum flows and the non-exceedence prohabilay of interest (in this case 0.99). The method of mornents estimators ofpv and ' are: o,-1) R readily provides these sanaple statisties t5 and S,) for the data. While tables have ho developed fot estimating K an accepted methodology is to instcad use the W transformaton to estimate Kp . ,26.36)

Explanation / Answer

Q100 is defined as P(q<Q100)=0.99. Given the mean as u and standard deviation as d. the z variable is (q-u)/d.

Thus probability becomes P((Q100-u)/d<k)=0.99, k can be found from z table as k=2.33, thus Q100=u+2.33*d.

putting the values Q100=8855.824+2.33*5709.88=22159.8444.

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