Question

P: The following wind velocity data were recorded for the selected site:1, 2, 1, 2, 3, 3, 4, 3, 4, 5, 3, 4, 5, 6,7, 5, 4, 3, 2, 2. m/s Bin Nv Frequency Frequency F(v) 1-F(V) L (1. ..... F(v)) 0 0 0 2 0.1 0.1 0.9 0.7 0.45 0.25 0.1 0.2 0.15 0.05 0.05 0.95 0.05 5 2- Find the parameters k and c (shape parameter k and scale parameter c) b-Weibull distribution C-Draw the Weibull distribution d-Analyze the wind data based on your figures.

Answer 1

Solution

part a, b, c

The data for wind velocity is Analysed to be analyzed in this problem

The data is analyzed in Minitab software steps are below

Go to Minitab - Feed the data in the column -Go to Graph section - click on Probability plot - Select the Weibull distribution -Result summary  

a) From the Weibull distribution of the data, we can obtain the Shape factor (K) and scale factor (c) which is found to be 2.366 and 3.900 respectively as we can see the result stats of the data on right side of the plot taking confidence interval of 95%

Probability Plot of Data Weibull - 95% CI 2.366 3.900 Scale & & & R899% AD P.Value 0.364 0.250 Percent MN Data

From the figure/plot drawn above, we can observe that the p-value of the Weibull distribution is >0.25 which is >0.05 ( Assumed level of significance). It means that there is strong evidence not to reject the hypothesis that the data follow the Weibull distribution.

The graphical analysis of the data in a more precise manner is given below

Graphical Summary of Data Summary Report Distribution of Data Examine the center, shape, and variability, Descriptive Statistics Minimum Sth percentie 25th percentile Median 75th percentile 95th percentile Mocium 95% Confidence intervals Mean (2.6988, 4.2012) Median 223524) SEDev (1.2207, 23443) Data in Worksheet Order Investigate any outlers marked in red). Normality Test Decision 0.322 Data 0 15

We can see that the data of wind velocity is normally distributed

I hope that your concept on this is cleared.