Previous: 2.8 – Expected Value, Variance, Standard Deviation

## Problem

The length of time *X*, needed by students in a particular course to complete a 1 hour exam is a random variable with PDF given by

For the random variable *X*,

- Find the value
*k*that makes*f*(*x*) a probability density function (PDF) - Find the cumulative distribution function (CDF)
- Graph the PDF and the CDF
- Use the CDF to find
- Pr(
*X*≤ 0) - Pr(
*X*≤ 1) - Pr(
*X*≤ 2)

- Pr(
- find the probability that that a randomly selected student will finish the exam in less than half an hour
- Find the mean time needed to complete a 1 hour exam
- Find the variance and standard deviation of
*X*

## Solution

### Part 1

The given PDF must integrate to 1. Thus, we calculate

Therefore, *k* = 6/5.

### Part 2

The CDF, *F*(*x*), is area function of the PDF, obtained by integrating the PDF from negative infinity to an arbitrary value *x*.

If *x* is in the interval (-∞, 0), then

If *x* is in the interval [0, 1], then

If *x* is in the interval (1, ∞) then

Note that the PDF *f* is equal to zero for *x* > 1. The CDF is therefore given by

### Part 3

The PDF and CDF of *X* are shown below.

### Part 4

These probabilities can be calculated using the CDF:

Note that we could have evaluated these probabilities by using the PDF only, integrating the PDF over the desired event.

### Part 5

The probability that a student will complete the exam in less than half an hour is Pr(*X* < 0.5). Note that since Pr(*X* = 0.5) = 0, since *X* is a continuous random variable, we an equivalently calculate Pr(*x* ≤ 0.5). This is now precisely *F*(0.5):

### Part 6

The mean time to complete a 1 hour exam is the expected value of the random variable *X*. Consequently, we calculate

### Part 7

To find the variance of *X*, we use our alternate formula to calculate

Finally, we see that the standard deviation of *X* is

source: http://wiki.ubc.ca/Science:MATH105_Probability/Lesson_2_CRV/2.12_Example

Previous: 2.8 – Expected Value, Variance, Standard Deviation