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===Answer 1===
 
===Answer 1===
Write it here
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To compute the probability <math>P(1\leq X \leq 2)</math>, we can integrate the probability density function over the interval as such:
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<math>\int_1^2 \! f_X(x) \, \mathrm{d}x.</math>
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Plugging in <math>f_X(x)</math>, we get:
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<math>\int_1^2 \! k+\frac{1}{10}x \, \mathrm{d}x.</math>
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Solving the integral we obtain:
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<math>k+\frac{3}{20}</math>
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If we wanted to solve for the constant k, we could setup another integral over the entire function and set it equal to 1 like so:
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<math>\int_0^2 \! k+\frac{1}{10}x \, \mathrm{d}x. = 1</math>
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Computing the integral gives us:
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<math>2k + \frac{4}{20} = 1</math>
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And solving for k:
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<math>k = \frac{2}{5}</math>
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Plugging in k into our previous result we get a probability of:
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<math>\frac{2}{5} + \frac{3}{20} = \frac{11}{20} = 55%</math>
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:<span style="color:purple">Instructor's comment: Can anybody propose a more "compact" solution? -pm </span>
 
===Answer 2===
 
===Answer 2===
 
Write it here
 
Write it here

Latest revision as of 04:58, 4 March 2013

Practice Problem: normalizing the probability mass function of a continuous random variable


A random variable X has the following probability density function:

$ f_X (x) = \left\{ \begin{array}{ll} k+\frac{1}{10}x, & \text{ if } 0\leq x \leq 2,\\ 0, & \text{ else}, \end{array} \right. $

where k is a constant. Compute $ P(1\leq X \leq 2) $.


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You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!


Answer 1

To compute the probability $ P(1\leq X \leq 2) $, we can integrate the probability density function over the interval as such:

$ \int_1^2 \! f_X(x) \, \mathrm{d}x. $

Plugging in $ f_X(x) $, we get:

$ \int_1^2 \! k+\frac{1}{10}x \, \mathrm{d}x. $

Solving the integral we obtain:

$ k+\frac{3}{20} $

If we wanted to solve for the constant k, we could setup another integral over the entire function and set it equal to 1 like so:

$ \int_0^2 \! k+\frac{1}{10}x \, \mathrm{d}x. = 1 $

Computing the integral gives us:

$ 2k + \frac{4}{20} = 1 $

And solving for k:

$ k = \frac{2}{5} $

Plugging in k into our previous result we get a probability of:

$ \frac{2}{5} + \frac{3}{20} = \frac{11}{20} = 55% $

Instructor's comment: Can anybody propose a more "compact" solution? -pm

Answer 2

Write it here

Answer 3

Write it here.


Back to ECE302 Spring 2013 Prof. Boutin

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