# 4.12: Chapter Solution (Practice + Homework)

- Page ID
- 5559

**1**.

\(x\) | \(P(x)\) |
---|---|

0 | 0.12 |

1 | 0.18 |

2 | 0.30 |

3 | 0.15 |

4 | 0.10 |

5 | 0.10 |

6 | 0.05 |

**3**.

0.10 + 0.05 = 0.15

**5**.

1

**7**.

0.35 + 0.40 + 0.10 = 0.85

**9**.

1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45

**11**.

\(x\) | \(P(x)\) |
---|---|

0 | 0.03 |

1 | 0.04 |

2 | 0.08 |

3 | 0.85 |

**13**.

Let \(X =\) the number of events Javier volunteers for each month.

**15**.

\(x\) | \(P(x)\) |
---|---|

0 | 0.05 |

1 | 0.05 |

2 | 0.10 |

3 | 0.20 |

4 | 0.25 |

5 | 0.35 |

**17**.

1 – 0.05 = 0.95

**18**.

\(X =\) the number of business majors in the sample.

**19**.

2, 3, 4, 5, 6, 7, 8, 9

**20**.

\(X =\) the number that reply “yes”

**22**.

0, 1, 2, 3, 4, 5, 6, 7, 8

**24**.

5.7

**26**.

0.4151

**28**.

\(X =\) the number of freshmen selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.

**30**.

1,2,…

**32**.

1.4

**35**.

0, 1, 2, 3, 4, …

**37**.

0.0485

**39**.

0.0214

**41**.

\(X =\) the number of U.S. teens who die from motor vehicle injuries per day.

**43**.

0, 1, 2, 3, 4, ...

**45**.

No

**48**.

- \(X =\) the number of pages that advertise footwear
- 0, 1, 2, 3, ..., 20
- 3.03
- 1.5197

**50**.

- \(X =\) the number of Patriots picked
- 0, 1, 2, 3, 4
- Without replacement

**53**.

\(X =\) the number of patients calling in claiming to have the flu, who actually have the flu.

\(X = 0, 1, 2, ...25\)

**55**.

0.0165

**57**.

- \(X =\) the number of DVDs a Video to Go customer rents
- 0.12
- 0.11
- 0.77

**59**.

4. 4.43

**61**.

4

**63**.

- \(X =\) number of questions answered correctly
- \(X \sim B(32, 13)(32, 13)\)
- We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find \(P(x > 24)\). The event "more than 24" is the complement of "less than or equal to 24."
- \(P(x > 24) = 0\)
- The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero.

**65**.

- \(X =\) the number of college and universities that offer online offerings.
- \(0, 1, 2, …, 13\)
- \(X \sim B(13, 0.96)\)
- \(12.48\)
- \(0.0135\)
- \(P(x = 12) = 0.3186 P(x = 13) = 0.5882\) More likely to get 13.

**67**.

- \(X =\) the number of fencers who do not use the foil as their main weapon
- \(0, 1, 2, 3,... 25\)
- \(X \sim B(25,0.40)\)
- \(10\)
- \(0.0442\)
- The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.

**69**.

- \(X =\) the number of audits in a 20-year period
- \(0, 1, 2, …, 20\)
- \(X \sim B(20, 0.02)\)
- \(0.4\)
- \(0.6676\)
- \(0.0071\)

**71**.

- \(X =\) the number of matches
- \(0, 1, 2, 3\)
- In dollars: \(−1, 1, 2, 3\)
- \(\frac{1}{2}\)
- The answer is \(−0.0787\). You lose about eight cents, on average, per game.
- The house has the advantage.

**73**.

- \(X \sim B(15, 0.281)\)
Figure \(\PageIndex{4}\)

- \(\text{Mean }= \mu=n p=15(0.281)=4.215\)
- \(\text{Standard Deviation } =\sigma=\sqrt{n p q}=\sqrt{15(0.281)(0.719)}=1.7409\)

- \(P(x > 5)=1 – 0.7754 = 0.2246\)

\(P(x = 3) = 0.1927\)

\(P(x = 4) = 0.2259\)

It is more likely that four people are literate that three people are.

**75**.

- \(X =\) the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.
- \(X \sim G(0.40)\)
- \(2.5\)
- \(0.0187\)
- \(0.2304\)

**77**.

- \(X =\) the number of pages that advertise footwear
- \(X\) takes on the values 0, 1, 2, ..., 20
- \(X \sim B(20, \frac{29}{192})\)
- \(3.02\)
- No
- \(0.9997\)
- \(X =\) the number of pages we must survey until we find one that advertises footwear. \(X \sim G(\frac{29}{192})\)
- \(0.3881\)
- \(6.6207\) pages

**79**.

0, 1, 2, and 3

**81**.

- \(X \sim G(0.25)\)
- \(\text{Mean }=\mu=\frac{1}{p}=\frac{1}{0.25}=4\)
- \(\text{Standard Deviation }=\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.25}{0.25^{2}}} \approx 3.4641\)

- \(P(x = 10) = 0.0188\)
- \(P(x = 20) = 0.0011\)
- \(P(x \leq 5) = 0.7627\)

**82**.

- \(X \sim P(5.5); \mu = 5.5; \sigma=\sqrt{5.5} \approx 2.3452\)
- \(P(x \leq 6) \approx 0.6860\)
- There is a 15.7% probability that the law staff will receive more calls than they can handle.
- \(P(x > 8) = 1 – P(x \leq 8) \approx 1 – 0.8944 = 0.1056\)

**84**.

Let \(X =\) the number of defective bulbs in a string.

Using the Poisson distribution:

- \(\mu = np = 100(0.03) = 3\)
- \(X \sim P(3)\)
- \(P(x \leq 4) \approx 0.8153\)

Using the binomial distribution:

- \(X \sim B(100, 0.03)\)
- \(P(x \leq 4) = 0.8179\)

The Poisson approximation is very good—the difference between the probabilities is only \(0.0026\).

**86**.

- \(X =\) the number of children for a Spanish woman
- \(0, 1, 2, 3,...\)
- \(0.2299\)
- \(0.5679\)
- \(0.4321\)

**88**.

- \(X =\) the number of fortune cookies that have an extra fortune
- \(0, 1, 2, 3,... 144\)
- \(4.32\)
- \(0.0124\) or \(0.0133\)
- \(0.6300\) or \(0.6264\)
- As \(n\) gets larger, the probabilities get closer together.

**90**.

- \(X =\) the number of people audited in one year
- \(0, 1, 2, ..., 100\)
- \(2\)
- \(0.1353\)
- \(0.3233\)

**92**.

- \(X =\) the number of shell pieces in one cake
- \(0, 1, 2, 3,...\)
- \(1.5\)
- \(0.2231\)
- \(0.0001\)
- Yes

**94**.

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