Question

A local bottling company has determined the number of machine breakdowns per month and their respective probabilities as shown below.

Number of Breakdowns |
Probability |

0 | .18 |

1 | .32 |

2 | .25 |

3 | .17 |

4 | .08 |

1. The expected number of machine breakdowns per month is:

2. The probability of at least 3 breakdowns per month is:

3. The standard deviation of machine breakdowns per month is:

Answer #1

(1)

From the given data, the following Table is calculated:

Number of breakdowns (x) | Probability (p) | xp | x^{2} p |

0 | 0.18 | 0 | 0 |

1 | 0.32 | 0.32 | 0.32 |

2 | 0.25 | 0.50 | 1.00 |

3 | 0.17 | 0.51 | 1.53 |

4 | 0.08 | 0.32 | 1.28 |

Total = | 1.65 | 4.13 |

So,

Expected number of machine breakdown per month = E(X) = 1.65

So,

Answer is:

**1.65**

(2) The probability of at least 3 breakdowns pr month = 0.17 + 0.08 = 0.25

So,

Answer is:

**0.25**

(3)

Variance= E(X^{2}) - (E(X))^{2} = 4.13 -
1.65^{2} = 1.4075

So,

Standard Deviation =

So,

Answer is:

**1.1864**

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