5.6 Let HIV stand for a woman infected with the HIV virus and let black and non-black signify black and non-black women. Assuming a hypothetical population of 10,000,000, a contingency table is as follows:
| HIV | No HIV | Total | |
| Black | 6,625 | 1,293,375 | 1,300,000 |
| Non-black | 5,875 | 8,694,250 | 8,700,000 |
| Total | 12,500 | 9,987,500 | 10,000,000 |
The probability that a randomly selected black woman carried the HIV virus is 6,625/1,300,000 = 0.005096, which is nearly ten times larger than the probability that a randomly selected non-black woman carried the HIV virus: 5,875/8,700,000 = 0.000675.
5.12 Using the binomial distribution with p = 0.1 and n = 3, we have P[x = 3] = 0.001 and P[x = 2] = 0.027. Therefore, P[x greater than or equal to 2] = 0.028.
5.16 The Brewster coach is appealing to the fallacious law of averages, that victories must be offset by losses, and vice versa. Thus he concludes that Chatham's victories must be followed by losses; while Brewster's losses must be followed by victories. By this logic, every team would end up with an equal number of wins and losses. In reality, some teams are better than other teams and do win more than half their games, while other teams lose more than half of their games. A team that is in first place at the halfway point in a season is probability better than the other teams, and there is no reason why a first-place team should be destined to lose games in order to offset their wins. A team in first place has the best chance of winning the championship. (As it turned out, Chatham finished the season in first place, with a record of 24 wins, 19 losses and 1 tie, while Brewster finished with 21 wins, 22 losses, and 1 tie.)
5.28 After converting to standardized z value, we can look the probabilities up in Table 3 (or use computer software for the larger z values)
| IQ | z = (IQ - 100)/15 | Probability |
| 150 | 3.3333 | 0.00043 |
| 185 | 5.6667 | 0 .00000000746 |
| 210 | 7.3333 | 0.00000000000011 |
5.34 Using a normal approximation to a binomial distribution with p = 0.75 and n = 8,023, the probability that more than 6,021 would be yellow-seeded is
The probability that fewer than 6,013 would be yellow-seeded is
Therefore, the probability that the number of yellow-seeded plants would be between 6,013 and 6,021 is 1 - 0.461 - 0.456 = 0.083. (The awkwardness of deciding whether 6013 and 6021 are inside or outside the interval is an argument in favor of using a continuity correction--which is beyond the scope of this textbook.)
