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===Answer 1=== | ===Answer 1=== | ||
| − | + | To compute the probability <math>P(1\leq X \leq 2)</math>, we can integrate the probability density function over the interval as such: | |
| + | |||
| + | <math>\int_1^2 \! f_X(x) \, \mathrm{d}x.</math> | ||
| + | |||
| + | Plugging in <math>f_X(x)</math>, we get: | ||
| + | |||
| + | <math>\int_1^2 \! k+\frac{1}{10}x \, \mathrm{d}x.</math> | ||
| + | |||
| + | Solving the integral we obtain: | ||
| + | |||
| + | <math>k+\frac{3}{20}</math> | ||
| + | |||
| + | 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: | ||
| + | |||
| + | <math>\int_0^2 \! k+\frac{1}{10}x \, \mathrm{d}x. = 1</math> | ||
| + | |||
| + | Computing the integral gives us: | ||
| + | |||
| + | <math>2k + \frac{4}{20} = 1</math> | ||
| + | |||
| + | And solving for k: | ||
| + | |||
| + | <math>k = \frac{2}{5}</math> | ||
| + | |||
| + | Plugging in k into our previous result we get a probability of: | ||
| + | |||
| + | <math>\frac{2}{5} + \frac{3}{20} = \frac{11}{20} = 55%</math> | ||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
Write it here | Write it here | ||
Revision as of 06:33, 2 March 2013
Contents
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) $.
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% $
Answer 2
Write it here
Answer 3
Write it here.
