Question

The probability of obtaining a defective 10-year old widget is
68.9%. For our purposes, the random variable will be the number of
items that must be tested before finding the first defective
10-year old widget. Thus, this procedure yields a geometric
distribution.

Use some form of technology like Excel or StatDisk to find the
probability distribution.

*(Report answers accurate to 4 decimal places.)*

k |
P(X = k) |
---|---|

1 | |

2 | |

3 | |

4 | |

5 | |

6 or greater |

Answer #1

Solution:

For the given geometric distribution, we are given

p = 0.689

q = 1 – p = 1 – 0.689 = 0.311

P(X=k) = p*q^(k – 1)

P(X=1) = 0.689*0.311^(6 - 1) = 0.00200457

P(X=2) = 0.689*0.311^(6 - 2) = 0.006445562

P(X=3) = 0.689*0.311^(6 - 3) = 0.020725279

P(X=4) = 0.689*0.311^(6 - 4) = 0.066640769

P(X=5) = 0.689*0.311^(6 - 5) = 0.214279

P(X≤6) = 1 – P(X≤5) = 1 – (0.00200457 + 0.006445562 + 0.020725279 + 0.066640769 + 0.214279)

P(X≤6) = 1 - 0.31009518

P(X≤6) = 0.68990482

The table is given as below:

k |
P(X=k) |

1 |
0.0020 |

2 |
0.0064 |

3 |
0.0207 |

4 |
0.0667 |

5 |
0.2143 |

6 or more |
0.6899 |

Total |
1.0000 |

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