Question

6.

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution. |

a. |
What is the probability Linda Lahey, company president, received
exactly one non-work-related e-mail between 4 p.m. and 5 p.m.
yesterday? |

Probability |

b. |
What is the probability she received five or more
non-work-related e-mails during the same period? |

Probability |

c. |
What is the probability she did not receive any non-work-related
e-mails during the period? |

Probability |

7.

A study of the checkout lines at the Safeway Supermarket in the South Strand area revealed that between 4 and 7 p.m. on weekdays there is an average of four customers waiting in line. What is the probability that you visit Safeway today during this period and find: |

a. |
No customers are waiting? (Round your answer to 4
decimal places.) |

Probability |

b. |
Four customers are waiting? (Round your answer to 4
decimal places.) |

Probability |

c. |
Four or fewer are waiting? (Round your answer to 4
decimal places.) |

Probability |

d. |
Four or more are waiting? (Round your answer to 4
decimal places.) |

Probability |

Answer #1

6)λ =2

a)

probability = | P(X=1)= |
{e^{-λ}*λ^{x}/x!}= |
0.2707 |

b)

probability = | P(X>=5)= | 1-P(X<=4)= |
1-∑_{x=0}^{x-1 }
{e^{-λ}*λ^{x}/x!}= |
0.0527 |

c)

probability = | P(X=0)= |
{e^{-λ}*λ^{x}/x!}= |
0.1353 |

7)λ =4

a)

probability = | P(X=0)= |
{e^{-λ}*λ^{x}/x!}= |
0.0183 |

b)

probability = | P(X=4)= |
{e^{-λ}*λ^{x}/x!}= |
0.1954 |

c)

probability = | P(X<=4)= |
∑_{x=0}^{x }
{e^{-λ}*λ^{x}/x!}= |
0.6288 |

d)

probability = | P(X>=4)= | 1-P(X<=3)= |
1-∑_{x=0}^{x-1 }
{e^{-λ}*λ^{x}/x!}= |
0.5665 |

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