7.1

31.

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

Interpretation 1. 15.87% of the cell phone plans are less than $44 per month.

Interpretation 2. The probability is 0.1587 that a randomly selected cell phone plan is cheaper than $44 per month.

32.

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

Interpretation 1. 11.5% of refrigerators will have lives more than 17 years.

Interpretation 2. The probability is 0.1151 that the lives of refrigerators of a randomly selected group is more than 17 years.

33.

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

Interpretation 1. 2.28% of all full-term babies have a birth weight of more than 4,410 grams.

Interpretation 2.The probability is 0.0228 that the birth weight of a randomly chosen full-term baby is more than 4,410 grams.

34.

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

Interpretation 1. 4.96% of the heights of 10 year old males is less than 46.5

Interpretation 2. The probability is 0.0496 that a randomly selected 10 year old males’ height is less than 46.5 per month.

35.

Interpretation 1. The proportion of human pregnancies that last more than 280 days is 0.1908

Interpretation 2. The probability that a randomly selected human pregancy lasts more than 280 days is 0.1908

Interpretation 1.The proporation of human pregnancies that last between 230 and 260 days is 0.3416

Interpretation 2. The probability that a randomly selected human pregnancy lasts between 230 and 260 days is 0.3416

36.

Interpretation 1. The proportion of times Elena’s fills up her gas tank that last more than 26 miles is 0.3309

Interpretation 2. The probability that a randomly selected car is chosen to see how long many times they need to fill up lasts more than 26 miles is 0.3309

Interpretation 1. The proporation of gas tanks that last between 18 and 21 miles is 0.1107

Interpretation 2. The probability that a randomly selected car lasts more than 26 miles is 0.3309

7.2

5.

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

7.

  1. 0.9987
  2. 0.9441
  3. 0.0375

  4. 0.0009 9.

  5. 0.9586
  6. 0.2088

  7. 0.8479

11.

  1. 0.0456
  2. 0.2088
  3. 0.5203

13. z= -1.28

15. z= 0.67

17. z1= -2.575; z2= 2.575

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

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

37.

  1. P (X < 20)= 0.1587
  2. P (x > 22)= 0.1587
  3. P (19 less than or equal to X less than or equal to 21)= 0.4772
  4. Yes, P (X < 18)= 0.0013. About 1 egg in 1000 hatches in less than 18 days.

39.

  1. P (1000 less than or equal to X less than or equal to 1400)= 0.8658
  2. P (X < 1000)= 0.0132
  3. 0.7019 of the bags have more than 1200 chocolate chips
  4. 0.1230 of the bags have fewer than 1125 chocolate chips
  5. A bag that contains 1475 chocolate chips is at the 96th percentile
  6. A bag that contains 1050 chocolate chips is at the 4th perecentile

41.

  1. 0.4013 of pregnancies 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. P (X > 280)= 0.1894
  5. P (X less than or equal to 245)= 0.0951
  6. Yes, 0.0043 of births are very preterm. So about 4 births in 1000 births are very pattern.

43.

  1. 0.0764 of the rods have a length of les than 24.9 cm.
  2. 0.0324 of the rods will be discarded
  3. The plant manager expects to discard 162 of the 5000 rods manufactured.
  4. To meet the order, the plant manager should manufacture 11,804.

45.

  1. P (X greater than or equal to 5)= 0.3228
  2. P (X less than or equal to -2)= 0.4286
  3. The favored team is equally likely to win or lose relative to the spread. Yes, a mean of 0 implies the spreads are accurate.

47.

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

56. Reporting the probability as <0.0001 accurately describes that they event is not highly likely, but again not certain. Reporting the probability as 0 might be incorrectly interrupted to mean that the event will never occur even though it has a chance of occurring.

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 bar is approximately normal with average x bar, sigma x bar= 2
  2. P (x bar > 83)= 0.0668. If we take 100 simple random samples of size n= 49 from a population with average =80 and sigma= 14, then about 7 of the samples will result in a mean that is greater than 83.
  3. P (x bar less than or equal to 75.8)= 0.0179. If we take 100 simple random samples of size n=49 from a population with average=80 and sigma= 14, then about 2 of the samples will result in a mean that is less than or equal to 75.8
  4. P (78.3 < x bar < 85.1)= 0.7969. If we take 100 simple random samples of size n=49 from a population with average= 80 and sigma= 14, then about 80 of the samples will result in a mean that is between 78.3 and 85.1

17.

  1. The population must be normally distributed to compute probabilities involving the sample mean. If the population is normally distributed, then the sampling distribution of x bar is also normally distributed with average x bar= 64 and sigma x bar is 17/ sqaure root of 12= 4.907
  2. P (x bar < 67.3)= 0.7486. If we take 100 simple random samples of size n= 12 from a population that is normally distributed with average= 64 and sigma= 17, then about 75 of the samples will result in a mean that is less than 67.3
  3. P (x bar greater than or equal to 65.2)= 0.4052. If we take 100 simple random samples of size n= 12 from a population that is normally distributed with average= 64 and sigma= 17, then 41 of the samples will result in a mean that is greater than or equal to 65.2

19.

  1. P (X < 260)= 0.3520. If we randomly select 100 human pregnancies, then about 35 of the pregnancies will last less than 260 days.
  2. The sampling distribution of x bar is normal with average x bar= 266 and sigma x bar= 16/ square root of 20= 3.578
  3. P (x bar less than or equal to 260)= 0.0465. If we take 1000 simple random samples of n= 20 human pregnancies, then about 5 of the samples will result in a mean gestation period of 260 days or less.
  4. P (x bar less than or equal to 260)= 0.0040. If we take 1000 simple random samples of size n= 50 human pregnancies, then about 4 of the samples will result in a mean gestation period of 260 days or less.
  5. This result would be unusual, so the sample likely came from a population whose mean gestation period is less than 266 days. (f)P (x bar greater than or equal to 256 and less than or equal to 276)=0.9844. Ifwe take 100 simple random samples of n=15 human pregnancies, then about 98 of the samples will result in a mean gestation period between 256 and 276 days, inclusive.

21.

  1. P (X > 95)=0.3085. If we select a simple random sample of n=100 second grade students, then about 31 of the students will read more than 95 words per minute.
  2. P (x bar > 95)=0.0418. If we take 100 simple random samples of size n=12 second grade students, then about 4 of the samples will result in a mean reading rate that is more than 95 words per minute.
  3. P (x bar > 95)=0.0071. If we take 1000 simple random samples of size n= 24 second grade students, then about 7 of the samples will result in a mean reading rate that is more than 95 words per minute.
  4. Increasing the sample size decreases P (x bar > 95). This happens because sigma x bar decreases as n increases.
  5. A mean reading rate of 92.8 wpm is not unusual since P (x bar greater than or equal to 92.8)=0.1056. This means that the new reading program is not abundantly more effective than the old program.
  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. P (X > 0)=0.5675. If we select a simple random sample of n=100 months, then about 57 of the months will have positive rates of return.
  2. P (X bar > 0)=0.7291. If we take 100 simple random samples of sizw n= 12 months, then about 73 of the samples will result in a mean monthly rate of return that is positive.
  3. P (x bar > 0)=0.8531. If we take 100 simple random samples of size n=24 months, then about 81 of the samples will result in a mean monthly rate of return that is positive.
  4. P (x bar > 0)=0.8531. If we take 100 simple random samples of size n=36 months, then about 85 of the samples will result in a mean monthly rate of return that is positive.
  5. The likelihood of earning a positive rate of return 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 of p hat is approximately normal with average (u) p hat=0.8 and sigma p hat = square root of 0.8 (0.2)/ 75= 0.046
  2. P (p hat greater than or equal to 0.84)=0.1922. About 19 out of 100 random samples of size n=75 will result in 63 or more individuals (that is 84% or more) with the characteristic.
  3. P (p hat less than or equal to 0.68)= 0.0047. About 5 out of 1000 random samples of size n=75 will result in 51 or fewer individuals (that is 68% or less) with the characteristic.

12.

  1. The sampling distribution of p hat is approximately normal with the average (u) p hat= 0.65 and sigma p hat= square root of 0.65(.35)/200=0.034
  2. P (p hat greater than or equal to 0.68)=0.882
  3. P (p hat less than or equal to 0.59)= -1.76

13.

  1. The sampling distribution of p hat is approximately normal with avergae (u) p hat= 0.35 and sigma p hat= sqaure root of 0.35 (0.65)/ 1000= 0.015

  2. P (p hat greater than or equal to 0.39)=0.0040. About 4 out of 1000 random samples of size n=1000 will result in 390 or more individuals (that is 39% or more) with the characteristic.

  3. P (p hat less than or equal to 0.32)= 0.0233. About 2 out of 100 random samples of size n=1000 will result in 320 or fewer individuals (that is 32% or less) with the characteristic.

14.

  1. The sampling distribution of p hat is approximately normal with average (u) p hat=0.42 and sigma p hat= square root of 0.42(.58)/1,500,000
  2. P (p hat is greater than or equal to 657)=162.93
  3. P (p hat is less than of equal to 584)=144.81

15.

  1. The sampling distribution of p hat is approximately normal with average (u) p hat= 0.47 and sigma p hat= square root of 0.47(0.53)/200= 0.035
  2. P (p hat > 0.5)=0.1977. About 20 out of 100 random samples of size n=200 Americans will result in more than 100 individuals (that is more than 50%) who can order a meal in a foregin language.
  3. P (p hat less than or equal to 0.4)= 0.0239. About 2 out of 100 random samples of size n=200 Americans will result in 80 or fewer individuals (that is 40% or less) who can order a meal in a foreign language. This result in unusual.

16.

  1. The sampling distribution of p hat is approximately normal with average (u) p hat= 0.82 and sigma p hat= square root of 0.82(0.18)/100= 0.038
  2. P (p hat is greater than or equal to 0.85)=0.789
  3. P (p hat is less than or equal to 75)=-1.842

17.

  1. The sampling distribution of p hat is approximately normal with average (u) p hat= 0.39 and sigma p hat= square root of 0.39(0.61)/500= 0.022
  2. P (p hat < 0.38)=0.3228. About 32 out of 100 random samples of size n=500 adult Americans will result in fewer than 190 individuals (that is less than 38%) who believe that marriage is obsolete.
  3. P (0.40 < p hat < 0.45)= 0.3198. About 32 out of 100 random samples of size n=500 adult Americans will result in between 200 and 225 individuals (that is between 40% and 45%) who believe marriage is obsolete.
  4. P (p hat greater than or equal to 0.42)=0.0838. Aboout 8 out of 100 random samples of size n=500 adult Americans will reuslt in 210 or more individuals (that is 42% or more) who believe that marriage is obsolete. This result is not unusual.