Here is some code that will allow you to check your answers. For example if \(\bar{X} \sim \mathcal{N}(5,2)\) and you want to know what the \(Pr(\bar{X} < 3)\) is, you can use the following code. Meaning just change the mean and sd to fit the problem you are #working on.

pnorm(2, mean = 5, sd = 2)
## [1] 0.0668072

8.1

15

  1. Since μx=μ=80, σx=14/√49=2, and n=49, we can conclude that the sampling distribution of x is about normal

  2. P(x>83)=1-0.9332=0.0668, it means that if we randomly select 100 samples out of n=49 of a population with μ=80 and σ=40, 6.68, or about 7 of the samples would have a mean greater than 83.

  3. P(x≤75.8)=0.0179, it means that if we randomly select 100 samples from n=49 of a population with μ=80 and σ=40, 1.79, or about 2 of the samples would have a mean less than or equal to 75.8.

  4. P(78.3<x<85.1)=0.9946-0.1977=0.7969, it means that if we randomly select 100 samples from n=49 of a population with μ=80 and σ=40, 79.69 or about 80 of the samples would have a mean that is between 78.3 and 85.1.

17

  1. The population must be normally distributed, which means the sampling distribution of x is also normal.μx=64 and σ=17/√12=4.9

  2. P(x<67.3)=0.7486, which means that if we randomly select 100 samples from n=12 of a population with μ=64, σ=17, 74.86 or about 75 of the samples would have a mean less than 67.3.

  3. P(x≥65.2)=1-0.5948=0.4052, which means that if we randomly select 100 samples from n=12 of a population with μ=64, σ=17, 40.52 or about 41 of the samples would have a mean greater than or equal to 65.2.

19

  1. P(x<260)=0.3520, if randomly select from n=100 of a popolation, about 35 of the pregnacies would less than 260 days.

  2. The sampling distribution of x is normal since the length of human pregnancies is approximately normal, and μ=266, σ=16/√20=3.58.

  3. P(x≤260)=0.0465, with σ=16/√20, if randomly select 100 samples from n=20 of a population, 4.65 or about 5 of the samples would have a mean gestation period of 260 days or less.

  4. P(x≤260)=0.0040, with σ=16/√50, if randomly select 1000 samples from n=50 of a population, about 4 of the samples would have a mean gestation period of 260 days or less.

  5. The result is unusual so the sample might came from a population somewhat mean gestation period is less than 266 days.

  6. P(256≤x≤276)=0.9922-0.0078=0.9844, with σ=16/√15, if randomly select 100 samples from n=15 of a population, about 98 of the samples would have a mean gestation period between 256 and 276 days, include the two ends.

21

  1. P(x>95)=1-0.6915=0.3085, if randomly select a sample from n=100 from the population, 30.85 or about 31 of the students would read more than 95 words/minute.

  2. σ=10/√12, P(x>95)=1-0.9582=0.0418, if randomly select 100 samples from n=12 of the population, about 4 of the students would have a mean reading rate more than 95 words/minute.

  3. σ=10/√24, P(x>95)=1-0.9929-0.0071, if randomly select 1000 samples from n=24 of the population, about 7 of the students would have a mean reading rate that is more than 95 words/minute.

  4. Increasing the sample size would decreasing the probability of x>95, because σ would decreases as n increases.

  5. This result is not unusual and is not more effective than the old one, σ=10/√20, P(x>92.8)=1-0.8944=0.1056, if randomly select 100 samples from n=20 of the population, 10.56 or about 11 of the students would have a mean reading rate greater than 92.8 word/minute.

  6. P(x>N)=0.05, N=93.7 words/minute

23

  1. P(x>0)=1-0.4325=0.5675, if randomly select a sample from n=100 months, 56.75 or about 57 of the samples would have a positive rate of return.

  2. σ=0.04135/√12, P(x>0)=1-0.2709=0.7291, if randomly select 100 samples from n=12 months, 72.91 or about 73 of the samples would have a mean monthly rate that is positive.

  3. σ=0.04135/√24, P(x>0)=1-0.1949=0.8051, if randomly select 100 samples from n=24 months, 80.51 or about 81 of the samples would have a mean monthly rate that is positive.

  4. σ=0.04135/√36, P(x>0)=1-0.1469=0.8531, if randomly select 100 samples from n=36 months, about 85 of the samples will have a mean monthly rate that is positive.

  5. The likelihood of earning a positive rate of return would increases as the investment time horizon increases.

Similarily you can use the above code to determine the \(Pr(\hat{P} < \hat{p})\)

8.2

11

  1. The distribution of p is about normal with μp=0.8 and σp=√0.8(1-0.8)/75=0.046.

  2. P(p≥0.84)=1-0.8078=0.1922, if randomly select 100 samples from n=75 of a population, about 19 out of 100 samples would have 63 or more individuals with the characteristic.

  3. P(p≤0.68)=0.0047, if randomly select 1000 samples, 4.7 or about 5 would have 51 or fewer individuals with the characteristic.

12

  1. The distribution of p is about normal, with μp=0.65 and σp=√0.65(1-0.65)/200=0.034.

  2. P(p≥0.68)=1-0.8133=0.1867, if randomly select 100 samples, 18.67 or about 19 out of 100 would have 136 or more individuals with the characteristic.

  3. P(p≤0.59)=0.0375, 3.75 or about 4 out of the 100 random samples would have 118 or fewer individuals with the characteristic.

13

  1. The distribution of p is approximately normal, with μp=0.35 and σp=√0.35(1-0.35)/1000=0.015.

  2. p=390/1000=0.39, P(x≥390)=1-0.9960=0.0040, about 4 out of 1000 random samples would have 390 or more individuals with the characteristic.

  3. p=320/1000=0/32, P(x≤320)=0.0233, about 2 out of 100 random samples would have 320 or fewer individuals with the characteristic.

14

  1. The distribution is approximately normal, with μp=0.42 and σp=√0.42(1-0.42)/1460=0.013.

  2. p=657/1460=0.45, P(x≥657)=1-0.9898=0.0102, about 1 out of 100 random samples would have 657 or more individuals with the characteristic.

  3. p=584/1460=0.4, P(x≤584)=0.0606, about 6 out of 100 random samples would have 584 or fewer individuals with the characteristic.

15

  1. Qualitative because it has two possible answers, can order with foreign language or not.

  2. p is a random variable because it varies from sample to sample, and the source of the variability is individuals from the sample and their ability to order with foreign language.

  3. The distribution of p is about normal, with μp=0.47 and σp=√0.47(1-0.47)/200=0.035.

  4. p=100/200=0.5, P(p>0.5)=1-0.8023=0.1977, 19.77 or about 20 out of 100 random samples would have more than half can order a meal in a foreign language.

  5. p=80/200=0.4, P(x≤80)=0.0239, about 2 out of 100 random samples would have 80 or fewer can order a meal in a foreign language, which is unusual.

16

  1. Qualitative because it has two possible answers, satisfied or not.

  2. p is arandom variable because it varies from sample to sample, and the source of the variability is individuals in the sample and the percentage of whom are satisfied.

  3. The distribution of p is about normal, with μp=0.82 and σp=√0.82(1-0.82)/100=0.038.

  4. p=85/100=0.85, P(x≥85)=1-0.7823=0.2177, 21.77 or about 22 out of 100 random samples would have at least 85 of whom are satisified with their lives.

  5. p=75/100=0.75, P(x≤75)=0.0344, about 3 out of 100 random samples would have 75 or fewer whom are satisfied with their lives, which is unusual.

17

  1. The distribution of p is about normal, with μp=0.39 and σp=√0.39(1-0.39)/500=0.022.

  2. P(p<0.38)=0.3228, about 32 out of 100 random samples would have less than 38% who believe marriage is obsolete.

  3. P(0.40<p<0.45)=0.9970-0.6772=0.3198, 31.98 or about 32 out of 100 random samples would have between 40% and 45% of who believe that marriage is obsolete.

  4. P(x≥210)=1-0.9162=0.0838, about 8 out of 100 random samples would have at least 210 of who believe that marriage is obsolete, which is unusual.