Given the following 25 yrs of peak annual discharge data (cfs), and summary stat

Question

Given the following 25 yrs of peak annual discharge data (cfs), and summary statistics and plots. Given that int defines the endpoints of the histograms, determine the cummulative probability of 15,000 cfs? Slate the equations for the mean, and the standard deviation. Assume that you are not confident in the magnitude of the computed coefficient of skew, determine the population pdf(s), justify your answer, and plot the population pdf(s). x = (5400 9600 10000 6323 8903 5694 3002 15000 6931 8935 2600 19369 4500 8745 2369 5846 5694 2356 1452 4785 13000 4500 2300 7961 2630) min (x)= 1.452 times 10^3 max(x) = 1.937 x 10^4 max(x) - min(x)/6 = 2.986 times 10^3 unit The exponential distn is analytically integrated as follows: integral^b_a lambda middot e^-lambda middot x dx rightarrow e^-a middot lambda - e^-a lambda middot b lambda = 1/x bar, x bar = 0.89, d x bar = 0.89, determine the exceedance probability of 2, illustrate and label the pdf. The mean is 4 and the standard deviation is 1.5 for a random variable x. Assuming a normal distribution, determine the interval probability between x = 1 and x = 3. Given: f(x) = lambda^betax^beya-1e^-lambda x/Gamma (beta) identify the pop pdf, define the pop parameters, and plot pop pdf. Describe the pop pdf skew: is it positive, negative, variable, or zero?

Explanation / Answer

(a) Since the value 15000 falls at 24 the place in 25 data points when sorted in ascending order , the percentile at which this datapoint falls can be taken as its probability i.e. 96percentile=> 0.96%

(b)mean = sum of all data points/(total number of data points)

=167900/25 = 6.176*10^3

standard deviation = sqrt(sum of (x-mean)^2)/n)

=1.275

(c)since the population is almost follwing a right skewed, the distribution is a sort of right normal distribution

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