7.1

31.

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

Interpretation 1. The probability that a randomly selected cell phone plan in the United States is less than $44 is 0.1587.

Interpretation 2. 15.87% of the cell phone plans in the United States are less than $44 per month.

32.

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

Interpretation 1. The probability that a randomly selected refrigerator lasts for more than 17 years is 0.1151.

Interpretation 2. 11.51% of refrigerators last more for than 17 years.

33.

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

Interpretation 1. The probability that a randomly selected full-term baby weighs more than 4410 grams is 0.0228.

Interpretation 2. 2.28% of full-term babies weigh more than 4410 grams.

34.

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

Interpretation 1. The probability that a randomly selected 10-year-old male is less than 46.5 inches tall is 0.0496.

Interpretation 2. 4.96% of all 10-year-old males are less than 46.5 inches tall.

35.

Interpretation 1. The probability that a randomly selected human pregnancy is more than 280 days is 0.1908.

Interpretation 2. 19.08% of human pregnancies last more than 280 days.

Interpretation 1. The probability that a randomly selected human pregnancy is between 230 and 260 days is 0.3416.

Interpretation 2. 34.16% of human pregnancies last between the 230 and 260 days.

36.

Interpretation 1. The probability that the gas tank on Elena’s Toyota Camry records more than 26 miles per gallon is 0.3309.

Interpretation 2. The gas tank on Elena’s Toyota Camry gets more than 26 miles per gallen 33.09% of the times it is filled.

Interpretation 1. The probability that the gas tank on Elena’s Toyota Camry records between 18 and 21 miles per gallon is 0.1107.

Interpretation 2. The gas tank on Elena’s Toyota Camry gets between 18 and 21 miles per gallen 11.07% of the times it is filled.

7.2

5.

  1. Area = 0.0071
  2. Area = 0.3336
  3. Area = 0.9115
  4. Area = 0.9998

7.

  1. Area = 0.9987
  2. Area = 0.9441
  3. Area = 0.0375
  4. Area = 0.0009

9.

  1. Area = 0.9586
  2. Area = 0.2088
  3. Area = 0.8479

11.

  1. Area = 0.0456
  2. Area = 0.0646
  3. Area = 0.5203

13. z = -1.28

15. z = 0.67

17. z1 = -2.57; z2 = 2.57

33. x = 40.62 is at the 9th percentile.

35. x = 56.16 is at the 81st percentile.

37.

  1. The probability that a randomly selected fertilized chicken egg hatches in less than 20 days is 0.1587.
  2. The probability that a randomly selected fertilized chicken egg takes over 22 days to hatch is 0.1587.
  3. The probability that a randomly selected fertilized chicken egg takes between 19 and 21 days to hatch is 0.4772.
  4. Yes, because the probability that a randomly selected fertilized chicken egg takes takes less than 18 days to hatch is 0.0013. Only about 1 in a 1000 eggs would take less than 18 days to hatch.

39.

  1. The probability that a randomly selected 18-ounce bag of Chips Ahoy! contains between 1000 and 1400 chocolate chips, inclusive, is 0.8658.
  2. The probability that a randomly selected 18-ounce bag of Chips Ahoy! that contains fewer than 1000 chocolate chips is 0.0132.
  3. 0.7019 of 18-ounce bags of Chips Ahoy! contain more than 1200 chocolate chips.
  4. 0.1239 of 18-ounce bags of Chips Ahoy! contain fewer than 1125 chocolate chips.
  5. An 18-ounce bag of Chips Ahoy! that contains 1485 chocolate chips is at the 96th percentile.
  6. An 18-ounce bag of Chips Ahoy! that contains 1050 chocolate chips is at the 4th percentile.

41.

  1. 0.4013 of pregancies last more than 270 days.
  2. 0.1587 of pregnancies last fewer than 250 days.
  3. 0.7590 of pregnancies last between 240 and 280 days.
  4. The probability that a randomly selected pregnancy lasts more than 280 days is 0.1894.
  5. The probability that a randomly selected pregnancy lasts fewer than 245 days is 0.0951.
  6. Yes, because the probability that a randomly selected pregnancy is less than 224 days is 0.0043. Only about 4 in 1000 births are very preterm.

43.

  1. 0.0764 of the rods have a length less than 24.9 cm.
  2. 0.0324 of the rods will be discarded.
  3. If 5,000 rods are manufactured in a day, the plant manager should expect to discard 162 rods.
  4. The plant managager should manufacture 11,804 rods.

45.

  1. In games where a team is favored by 12 or fewere points, the probability that the favored team wins by 5 or more points relative to the spread is 0.3228.
  2. In games where a team is favored by 12 or fewere points, the probability that the favored team loses by 2 or more points relative to the spread is 0.4286.
  3. The favored team is equally likely to win or lose relative to the spread. A mean of 0 implies that the spreads are accurate for games in which a team is favored by 12 or fewer points.

47.

  1. The 17th percentile for incubation times of fertilized chicken eggs is 20 days.
  2. The incubation times that make up the middle 95% of fertilized chicken eggs is 19 to 23 days.

56. The SAT score is higher relative to the ACT score. The SAT score is higher than 84.6% of those who took the SAT. The ACT score is only higher than 83.2% of those who took the ACT.

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. The sampling distribution is approximately normal with a mean of 80 and a standard deviation of 2.
  2. 0.0668
  3. 0.0179
  4. 0.7969

17.

  1. In order to use the normal model to compute probabilities involving the sample mean, the population must be normally distributed. The sampling distribution is approximately normal with a mean of 64 and a standard deviation of 4.907.
  2. 0.7486
  3. 0.4052

19.

  1. The probability that a randomly selected pregnancy lasts less than 260 days is 0.3520.
  2. The sampling distribution is normal with a mean of 266 and a standard deviation of 3.578.
  3. The probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less is 0.0465.
  4. The probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less is 0.0040.
  5. If a random sample of 50 pregnancies resulted in a mean gestation period of 260 days or less, we might conclude that the sample represented a population whose mean gestation period is less than 266 days.
  6. The probability a random sample of size 15 will have a mean gestation period within 10 days of the mean is 0.9844.

21.

  1. The probability a randomly selected student will read more than 95 more words per minute is 0.3085.
  2. The probability that a random sample of 12 second grade students results in a mean reading rate of more than 95 words per minute is 0.0418.
  3. The probability that a random sample of 24 second grade students results in a mean reading rate of more than 95 words per minute is 0.0071.
  4. Increasing the sample size decreases the probability that a randomly selected student will read more than 95 words per minute.
  5. We might conclude that the new reading program is not improving the reading speed of the students.
  6. There is a 5% chance that the mean reading speed of a random sample of 20 second grade students will exceed 93.9 words per minute.

23.

  1. The probability that a randomly selected month has a positive return rate is 0.5675.
  2. The probability that the mean monthly rate of return will be positive in the next 12 months is 0.7291.
  3. The probability that the mean monthly rate of return will be positive in the next 24 months is 0.8051.
  4. The probability that the mean monthly rate of return will be positive in the next 36 months is 0.8531.
  5. The likelihood of earning a positive rate of return on stocks increases as the investment time horizon increases.

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. The sampling distribution is approximately normal with a mean of 0.8 and standard deviation of 0.046.
  2. The probability of obtaining x = 63 or more individuals with the characteristic is 0.1922.
  3. The probability of obtaining x = 51 or fewer individuals with the characteristic is 0.0047.

12.

  1. The sampling distribution is approximately normal with a mean of 0.65 and standard deviation of 0.034.
  2. The probability of obtaining x = 136 or more individuals with the characteristic is 0.1894.
  3. The probability of obtaining x = 118 or fewer individuals with the characteristic is 0.0392.

13.

  1. The sampling distribution is approximately normal with a mean of 0.35 and standard deviation of 0.015.
  2. The probability of obtaining x = 390 or more individuals with the characteristic is 0.0040.
  3. The probability of obtaining x = 320 or fewer individuals with the characteristic is 0.0233.

14.

  1. The sampling distribution is approximately normal with a mean of 0.42 and standard deviation of 0.0129.
  2. The probability of obtaining x = 657 or more individuals with the characteristic is 0.0102.
  3. The probability of obtaining x = 584 or fewer individuals with the characteristic is 0.0606.

15.

  1. The sampling distrubtion of the proportion of Americans who can order a meal in a foreign language is approximately normal with a mean of 0.47 and standard deviation of 0.035.
  2. 0.1977 is the probability the proportion of Americans who can order a meal in a foreign language is greater than 0.5.
  3. It would be unusual that, in a survey of 200 Americans, 80 or fewer Americans can order a meal in a foreign language because the probability is 0.0239.

16.

  1. The sampling distrubtion of the proportion of Americans who are satisfied with the way things are going in their life is approximately normal with a mean of 0.82 and standard deviation of 0.038.
  2. 0.2174 is the probability that the proportion who are satisfied with the way things are going in their life exceeds 0.85.
  3. It would be unusual for a servey of 100 Americans to reveal that 75 or fewer are satisfied with the way things are going in their life because the probability is 0.0342.

17.

  1. The sampling distrubtion of the proportion of adult Americans who believe marriage is obselete is approximately normal with a mean of 0.39 and standard deviation of 0.022.
  2. 0.3228 is the probability that in a random sample of 500 adult Americans less than 38% believe that marriage is obsolete.
  3. 0.3198 is the probability that in a random sample of 500 adult Americans between 40% and 45% believe that marriage is obsolete.
  4. It would not be unusual for a random sample of 500 adult Americans to result in 210 or more who believe marriage is obsolete because the probability is 0.0838.