We now want to quantify the error that we would make in...

We now want to quantify the error that we would make in estimating the annual mean power of the wind turbine by using the Weibull distribution with the parameters directly computed with the power density method at 140m. Compute the annual mean power of the wind turbine using the probability density function of the Weibull distribution $f(V)$ with the parameters $k_{140m} = 2.49$ and $c_{140m} = 7.67 \text{ m/s}$: $P_{mean} = \int_{V=0}^{\infty} P(V)f(V)dV$ Give $P_{mean}$ in kW.

Transcript

00:01 For this problem, to begin, i'll note that the cumulative distribution function for a vable, i believe it's pronounced, but i'm not 100 % sure, i'll be saying it as a vable, the cumulative distribution function for a vable distribution is given by 1 minus e to the power of negative of x over beta to the power of alpha.

00:28 So, to the power of alpha there.

00:32 So in our case, we have that this is going to be 1 minus, and i'll actually write this as exp so i don't have to squish it up into a superscript, so it's 1 minus e to the power of negative x over, well our beta value is 0 .863, and then we put that to the power of 1 .817.

01:01 So, for part a, to find the probability that the wave height is at most 5, pardon me, 0 .5 meters, x greater than or equal to 0 .5, that's just our cumulative distribution function evaluated at 0 .5, which, let's see what we get here.

01:22 1 minus exp, negative bracket 0 .5 over 0 .863 to the power of 1 .817, gives us a result of 0 .3250 when we round to 4 decimal places.

01:41 For part b, the probability that the wave height exceeds its mean by more than one standard deviation, we'll first note that the mean value is going to be beta times gamma 1 plus 1 over alpha.

01:58 So, i'll use my software here to find this, so we know that our beta value is 0 .863, or, pardon me, i just realized i transposed the 3 and the 6 there.

02:10 So actually, in part a, our probability should have been 0 .3099, pardon me.

02:20 So, calculating out our mean, we have 0 .863 times gamma, evaluated at 1 plus 1 over 1 .817.

02:34 So we have that our mean value is equal to 0 .767, roughly.

02:40 The standard deviation is equal to beta times the square root of gamma of 1 plus 2 over alpha minus gamma of 1 plus 1 over alpha squared...

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