7.1

31.

shadenorm(mu = 62, sig = 18, below = 44, col = "blue", dens = 200)

Interpretation 1. 15.87% of the plans are less than 44$

Interpretation 2. The probability of randomly selecting a plan less than 44$ is 15.87%

32.

shadenorm(mu = 14, sig = 2.5, above = 17, col = "blue", dens = 200)

Interpretation 1. 11.51% of fridges last longer than 17 years.

Interpretation 2. The probability of randomly selecting a fridge that lasts longer than 17 years is 11.51%.

33.

shadenorm(mu = 3400, sig = 505, above = 4410, col = "blue", dens=200)

Interpretation 1. 2.28% of babies weigh more than 4410 grams at birth.

Interpretation 2. The probability of having a baby that weighs more than 4410 grams at birth is 2.28%.

34.

shadenorm(mu = 55.9, sig = 5.7, below = 46.5, col = "blue", dens=200)

Interpretation 1. 4.96% of 10-year-olds are less than 46.5" tall.

Interpretation 2. The probability of randomly selecting a 10-year-old shorter than 46.5" is 4.96%.

35.

Interpretation 1. 19.08% of pregnancies last longer than 280 days.

Interpretation 2. The probability of having a pregnancy longer than 280 days is 19.08%.

Interpretation 1. 34.16% of pregnancies last between 230 and 260 days.

Interpretation 2. The probability of having a pregnancy between 230 and 260 days is 34.16%.

36.

Interpretation 1. 33.09% of the time the care performs better than 26mpg.

Interpretation 2. The probability of the car performing at 26mpg or better is 33.09%.

Interpretation 1. 11.07% of the time the car performs between 18 and 21 miles per gallon.

Interpretation 2. The probability of the car performing between 18 and 21 miles per gallon is 11.07%.

7.2

5.

  1. .0071
  2. .3336
  3. .9115
  4. .9998

7.

  1. .9987
  2. .9441
  3. .0375
  4. .0009

9.

  1. .0414
  2. .2088
  3. .8479

11.

  1. .0456
  2. .0646
  3. .5203

13. -1.28 was the closest I could find (.1003)

15. .67 was the closest I could find (.2514)

17. -2.575 and 2.575

33. -1.34

35. .87

37.

  1. .1587
  2. .0228
  3. .4772
  4. It would be unusual in the sense that it is not very probable, clocking in at only .0013.

39.

  1. .8658
  2. .0132
  3. .2981
  4. .1230
  5. 96th
  6. 3rd

41.

  1. .4013
  2. .1587
  3. .5122
  4. .1894
  5. .0951
  6. Yes, they occur in only .43% of cases

43.

  1. .0764
  2. .0324
  3. 162
  4. I’m not sure I understand this question right. That said, in the first batch of 10k, one would expect 1528 rods to be incorrectly sized to fit the order. If the factory added another 1528 rods to production, another 234 (233.48) rods would be incorrectly sized. After adding 234, another 36 (35.76) would be incorrectly sized. After adding 36, another 6 (5.5) would be incorrectly sized. After adding 6, another 1 (.92) would be incorrectly sized. After adding 1, there is no longer a probability worthy of calculation that the rod would be incorrectly sized. So, the factory should produce 11,805 rods to be sure they have 10k sized correctly, and because probability is unpredictable, I would suggest they just be safe and make 12k.

45.

  1. .3228
  2. .4286
  3. This means that on average, the spread accurately predicts the result of the game correctly. In context, the mean being 0 suggests that on average, the distance between the predicted result and the actual result, is 0. By virtue of the explanation being the answer, yes, this would mean that spreads (of 12 points favored or fewer) are accurate.

47.

  1. 23.12 days
  2. 95% of chicken incubations last between 19.04 days and 22.96

56. In the hypothetical scenario, the ACT score is marginally better than its SAT counterpart. The z-score for a score of 26 on the ACT is -.94; the z-score for a score of 1240 on the SAT is -1.02. On the graph, the ACT score is slightly greater than the SAT score.