Question

It is estimated that 80% of a grapefruit crop is good; the other
20% have rotten centers that cannot be detected unless the

grapefruit are cut open. The grapefruit are sold in sacks of 12.
Let r be the number of good grapefruit in a sack.

(a) Make a histogram of the probability distribution of r.

(b) What is the probability of getting no more than one bad
grapefruit in a sack? (Round your answer to three decimal

places.)

What is the probability of getting at least one good grapefruit?
(Round your answer to three decimal places.)

(c) What is the expected number of good grapefruit in a sack?
(Round your answer to one decimal place.)

grapefruit

(d) What is the standard deviation of the r probability
distribution? (Round your answer to two decimal places.)

grapefruit

Answer #1

Part a)

X | P ( X ) |

0 | 0 |

1 | 0.0000 |

2 | 0.0000 |

3 | 0.0001 |

4 | 0.0005 |

5 | 0.0033 |

6 | 0.0155 |

7 | 0.0532 |

8 | 0.1329 |

9 | 0.2362 |

10 | 0.2835 |

11 | 0.2062 |

12 | 0.0687 |

**Histogram**

Part b)

X ~ B ( n = 12 , P = 0.2 )

**P ( X <= 1 ) = 0.275**

X ~ B ( n = 12 , P = 0.8 )

P ( X >= 1 ) = 1 - P ( X = 0 ) = 1.00 i.e 100%

Part c)

Mean = n * P = ( 12 * 0.8 ) = 9.6

Part d)

Variance = n * P * Q = ( 12 * 0.8 * 0.2 ) = 1.92

Standard deviation = (variance) = (1.92) = 1.39

It is estimated that 75% of a grapefruit crop is good; the other
25% have rotten centers that cannot be detected unless the
grapefruit are cut open. The grapefruit are sold in sacks of 11.
Let r be the number of good grapefruit in a sack.
(a) Make a histogram of the probability distribution of
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(b) What is the probability of getting no more than one bad
grapefruit in a sack? (Round your answer to three decimal
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