After manufacture, computer disks are tested for errors. Let X be the number of errors detected on a randomly chosen disk. The following table presents values of the cumulative distribution function F(x) of X.

X |
F(X) |

0 |
0.41 |

1 |
0.72 |

2 |
0.83 |

3 |
0.95 |

4 |
1.00 |

a. What is the probability that two or fewer errors are detected?

b. What is the probability that more than three errors are detected?

c. What is the probability that exactly one error is detected?

d. What is the probability that no errors are detected?

e. What is the most probable number of errors to be detected?

Solution 8E

Step1 of 3:

We have computer disks are tested for errors After manufacture. Let us consider the random variable X it presents the number of errors detected on a randomly chosen disk. We have table that presents values of the cumulative distribution function F(x) of X.

X |
F(X) |

0 |
0.41 |

1 |
0.72 |

2 |
0.83 |

3 |
0.95 |

4 |
1 |

Our goal is:

a).We need to find What is the probability that two or fewer errors are detected?

b).We need to find What is the probability that more than three errors are detected?

c).We need to find What is the probability that exactly one error is detected?

d).We need to find What is the probability that no errors are detected?

e).We need to find What is the most probable number of errors to be detected?

Step2 of 3:

a).

Consider,

X |
F(X) |

0 |
0.41 |

1 |
0.72 |

2 |
0.83 |

3 |
0.95 |

4 |
1 |

The probability that two or fewer errors are detected is given by

Put x = 2 in above equation we get

In a above given given table for x = 2, F(2) = 0.83

Hence,

= 0.83

Therefore, 0.83.

b).

Consider,

X |
F(X) |

0 |
0.41 |

1 |
0.72 |

2 |
0.83 |

3 |
0.95 |

4 |
1 |

The probability that more than three errors are detected is given by

=

=

Put x = 3 in above equation we get

=

In a above given given table for x = 3, F(3) = 0.95

Hence,

= 1 - 0.95

= 0.05

Therefore, 0.05.

Step3 of 3:

c).

Consider,

X |
F(X) |

0 |
0.41 |

1 |
0.72 |

2 |
0.83 |

3 |
0.95 |

4 |
1 |

The probability that exactly one error is detected is given by

=

=

In a above given given table for x = 0, F(0) = 0.41 and for x = 1, F(1) = 0.72.

Hence,

= 0.72 - 0.41

= 0.31

Therefore, 0.31.

d).

Consider,

X |
F(X) |

0 |
0.41 |

1 |
0.72 |

2 |
0.83 |