Problem 38E

Determine the cumulative distribution function of the random variable in Exercise 3-16.

3-16.The sample space of a random experiment is {a, b, c, d,

e, f}, and each outcome is equally likely. A random variable is

defined as follows:

outcome |
a |
b |
c |
d |
e |
f |

x |
0 |
0 |
1.5 |
1.5 |
2 |
3 |

Determine the probability mass function of a. Use the probability mass function to determine the following probabilities:

Solution :

Step 1 of 1:

From the given information the sample space of a random experiment is {a, b, c, d, e, f}.

Then the table is given below.

outcome |
a |
b |
c |
d |
e |
f |

x |
0 |
0 |
1.5 |
1.5 |
2 |
3 |

Our goal is:

We need to determine the cumulative distribution function.

From the given information is the number of 0’s is 2.

Then the total number of variables is 6.

So P(x=0) is

P(x=0) =

P(x=0) = 0.333

Therefore, P(x=0) = 0.333.

The number of 1.5 is 2 and the total number of variables is 6.

P(x=1.5) =

P(x=1.5) = 0.333

Therefore, P(x=1.5) = 0.333.

The number of 2 is 1 and the total number of variables is 6.

P(x=2) =

P(x=2) = 0.167

Therefore, P(x=2) = 0.167.

The number of 3 is 1 and the total number of variables is 6.

P(x=2) =

P(x=2) = 0.167

Therefore, P(x=2) = 0.167.

From the available information, each outcome is equally likely.

Then the cumulative distribution function is

x |
0 |
1.5 |
2 |
3 |

P(X) = f(x) |
0.333 |
0.333 |
0.167 |
0.167 |

F(x) |
0.333 |
(0.333+0.333)= 0.666 |
(0.666+0.167)= 0.833 |
(0.833+0.167)=1 |

Then, F(x) is