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S2 Continuous Random Variables Question

ImageUploadedByStudent Room1444238387.456590.jpg How would I answer this question?


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Reply 1
Original post by Abby5001
ImageUploadedByStudent Room1444238387.456590.jpg How would I answer this question?


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Anyone?


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(a) For a function to be a probability density function, its integral over its domain has to equal a certain constant.

(b) Once you've got k, do you remember the formula for the mean of a continuous distribution?
Reply 3
Can you explain a in a simpler way please I'm at the stage where I have 3^k+1=k+1 ?


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Original post by Abby5001
Can you explain a in a simpler way please I'm at the stage where I have 3^k+1=k+1 ?
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The condition you need to satisfy is 013xkdx=1\int_{0}^{1} 3 x^{k} dx = 1

If you do this integral, you can then solve for k. Does that help?
Reply 5
Original post by Gregorius
The condition you need to satisfy is 013xkdx=1\int_{0}^{1} 3 x^{k} dx = 1

If you do this integral, you can then solve for k. Does that help?


I get that part I just dont understand how you would solve 3^k+1=K+1


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Original post by Abby5001
I get that part I just dont understand how you would solve 3^k+1=K+1
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You need to take a closer look at that integral as the equation you're trying to solve isn't the right one.

013xkdx=301xkdx=3[xk+1k+1]01=3k+1\int_{0}^{1}3 x^{k} dx = 3 \int_{0}^{1} x^{k} dx = 3 \left[\frac{x^{k+1}}{k+1}\right]_{0}^{1} = \frac{3}{k+1}

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