**1.** Marilyn's (correct) answer: "I know it sounds strange, but more families with four children have three of one sex and one of the other sex than any other combination. The chances of having all girls or all boys is 1 in 8; the chances of having two of each are 3 in 8; and the chances of having three girls and one boy (or three boys and one girl) are 4 in 8."

**2.** Five dice, since the central limit theorem tells us that the probability
distribution approaches a normal distribution as the number of dice m being
cumulated increases: m = 5 is larger than m = 1. As the number of experiments
increases, the empirical histogram is increasingly likely to be close to the
theoretical probability distribution. The empirical distribution for 1 dice
will approach the theoretical distribution, which is not normal but uniform--with
each of the numbers 1-6 having a 1/6 probability of occurring.

**3.** Here is a contingency table with a total of 27,000 women who give birth at age 35, of whom (1/270)27,000 = 100 will suffer from Down's syndrome:

Positive Reading | Negative Reading | Total | |

Down's | 89 | 11 | 100 |

No Down's | 6,725 | 20,175 | 26,900 |

Total | 6,814 | 20,186 | 27,000 |

Of the 6,814 cases with positive readings, a stunning 6,725/6,814 = 0.987 are false positives. (The false negative rate is 11/ 20,186 = 0.0005; the main benefit of this blood test is that it would screen out many women from an amniotic-fluid test that risks a miscarriage. For those who get a positive blood-test result, a follow up amniotic-fluid test would be necessary.)

**4.** It is true that a random sample gives every member of the population an equal chance of being picked, but this does not guarantee that the sample will accurately represent the population or that attitudes, demographics, and geography will have the same proportions in the sample as in the population. By its very nature, a random sample is subject to sampling error, the uncontrollable chance that some groups or characteristics will be under- or over-represented.

**5.** Here are the relative frequencies and the histogram heights:

Previous | Number of | Relative | Histogram |

Sentences | Persons | Frequency | Height |

0 | 2,562 | 2,562/28,974 = 0.08842 | 0.08842 |

1-2 | 2,808 | 2,808/28,974 = 0.09691 | 0.09691/2 = 0.04846 |

3-5 | 4,911 | 4,911/28,974 = 0.15950 | 0.15950/3 = 0.05650 |

6-10 | 7,977 | 7,977/28,974 = 0.27532 | 0.2753/5 = 0.05506 |

11-15 | 10,716 | 10,716/28,974 = 0.36985 | 0.36985/5 = 0.07397 |

Total | 28,974 |

Here is a bar graph:

A bar graph gives undue prominence to those with a large number of previous sentences.

**6.** There are two ways they can drive down the correct road: Mr. Jones can give correct directions and his wife follows them, or he can give wrong directions and she does the opposite. The probabilities are

In general, if P is the probability of following her husband's directions, the probability of choosing the correct road is

which is maximized at 0.7 by P = 0; that is; always doing the opposite of what he recommends.

**7.** Converting to Z-values, corporate stock's Z = (0 - 15)/20 = -0.75 and Treasury bond's Z = (0 - 6)/9 = -0.67. Bonds are more likely to have a negative return.

**8.** The probability that an accepted student will attend is p = 0.5 and, if n students are accepted, then the expected size of the freshman class is pn = 0.5n. For this expected size to be 400, n must be 800 (0.5n = 400 implies n = 400/0.5 = 800).

The actual number who will attend if 800 are accepted follows the binomial distribution (assuming acceptance decisions are independent of one another).

Using a normal approximation,