7.1

31.

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

Interpretation 1. 15.87 cell phone plans in the US cost less than 44$/minute.

Interpretation 2. The probability that a randomly selected cell phone plan in the USA is less than 44$/month is 0.1587.

32.

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

Interpretation 1. 11.51% of refrigerators last more than two years

Interpretation 2. The probability that a randomly selected refridgerator will last more than 17 years is .1151

33.

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

Interpretation 1. 2.28% of full term babies weigh more than 4410g.

Interpretation 2.The probability that a randomly selected baby will weigh more thant 4410g is 0.0228g.

34.

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

Interpretation 1. 4.95% of 10 year old males are less than 46.5 inches tall.

Interpretation 2. The probability that a randomly selected 10 year old male is less than 46.5 inches tall is .0495.

35.

Interpretation 1. The probability that a randomly selected gestation period will last longer than 280 days is 0.1908.

Interpretation 2. 19.08% of gestation periods last longer than 280 days.

Interpretation 1. The probability that a randomly selected gestation period will last between 230 and 260 days is 0.3416.

Interpretation 2. 34.16% of gestation periods last between 230 and 360 days.

36.

Interpretation 1. The probability that a randomly selected car will get more than 26 miles/gallon of gass is .3309

Interpretation 2. 33.09% of cars get more thant 26 miles/gallon of gas.

Interpretation 1. 11.07% of cars get between 18 and 21 miles/gallon of gas.

Interpretation 2. The probability that a randomly selected car will get between 19 and 21 miles/gallon of gas is .1107

7.2

5.

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

7.

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

9.

  1. .9586
  2. .2088
  3. .8479

11.

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

13. -1.28

15. .68

17. z1 = 2.58, z2 = 2.58

33. 40.9

35. 56.16

37.

  1. z = .1587
  2. z = .1587
  3. z = .4772
  4. z = .0013–> It would be unusual for an egg to hatch in less than 18 because there is less than a 1% chance that this would happen

39.

  1. z = .8658
  2. z = .0132
  3. z = .7019
  4. z = .123
  5. z = .9641 (96th percentile)
  6. z = .0359 (4th percentile)

41.

  1. z = .4013
  2. z = .1587
  3. z = .759
  4. z = .1894
  5. z = .0951
  6. z = .0838

43.

  1. z = .0764
  2. z = .0324
  3. 162
  4. 10335

45.

  1. z = .3228
  2. z = .4286
  3. The team is just as likely to win as it is to lose. The statement is accurate.

47.

  1. 20.9
  2. 21+(-1.96)(1)=19.04, 21+(1.96)(1)=22.96

56. He did relatively better on the SAT.

8.1

## Here is the syntax you can use to check the probabilities you look up are correct.

## Say you want to know the Pr(X < 5) and X is Normal with a mean of 12 and standard deviation 4

pnorm(5, mean = 12, sd = 4 )
## [1] 0.04005916

15.

  1. X is aproximately normal with µ = 80, σ = 20
  2. z = .0668
  3. z = .0179
  4. z = .7969

17.

  1. X is aproximately normal with µ = 64, σ = 4.9
  2. z = .7486
  3. z = .4052

19.

  1. z = .3520
  2. X is approximately normal with µ = 266, σ = 3.57
  3. z = .0475
  4. z = .0040
  5. The results would be unusual
  6. z = .9844

21.

  1. z = .3085
  2. z = .0418
  3. z = .0073
  4. Increasing the population increases the probability because there are more individuals that are more likely to fit under that results.
  5. The new reading program could have helped increase the students reading level, however, a reading level of 92.8 isn’t any higher than the standard deviation of 10, so it’s not increadibly surprising.
  6. 93.7

23.

  1. z = .5675
  2. z = .7291
  3. z = .8051
  4. z = .8531
  5. The longer you invest in the stock, the more likely it is the interest rate will be positive and the higher the interest rate will be.

Here is the syntax you can use to check your answers. (Forward and Backward)

Say you want to know the \(Pr (\hat{P} < .35)\) and \(\hat{P} \sim \mathcal{N}(.4,.07)\)

pnorm(.35, mean = .4, sd = .07 )
## [1] 0.2375253

Here is the syntax you can use to check if a “Backward” calcuation is corect.

Say you know the probability to the left of \(\hat{p}\) = .04 and you want to know what the appropriate \(\hat{p}\) is. You also know that \(\hat{P} \sim \mathcal{N}(.4,.07)\)

qnorm(.05, mean = 12, sd = 4)
## [1] 5.420585

Section 8.2

11.

  1. P is approximately normal with p = .8, σ = .046
  2. z = .2
  3. z = .0048

12.

  1. P is approximately normal with p = .35, σ = .015
  2. z = .0039
  3. z = .0228

13.

  1. P is approximately normal with p = .65, σ = .034
  2. z = .8106
  3. z = .0392

14.

  1. P is approximately normal with p = .42, σ = .0129
  2. z = 2.325
  3. z = .9394

15.

  1. P is approximately normal with p = .47, σ = .03529
  2. z = .1949
  3. z = .0228

16.

  1. P is approximately normal with p = .82, σ = .038
  2. z = .7852
  3. z = .0329

17.

  1. P is approximately normal with p = .39, σ = .218
  2. z = .0823
  3. z = .95413
  4. z = .091185