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. The distribution follows the CLT, so we can conclude the sampling dist is fairly normal.

  2. P (xbar>83)= (Z>(83-80)/2) = P(Z>1.5) 1-P(z>1.5) = 1-0.9332 = 0.0668. Of 100 random samples of size n=49, roughly 7 samples would have a mean greater than 83.

  3. P(xbar<75.8) = P(Z<(75.8-80)/2) = P(Z<-2.1) = 0.0179 Of 100 random samples of size n=49, about 2 samples would have a mean less or equal to 75.8.

  4. P(78.3<xbar<85.1) P((78.3-80)/2<Z<(85.1-80/2)) = P(-0.85<Z<2.55) = 0.9946-0.1977 = 0.7969. Of 100 random samples of size n=49, roughly 79 of the means will fall between 78.3 and 85.1.

17

  1. The sample must be normally distributed. The sample size is not significant enough to assume normality. The sample distribution is normal, the sample mean is 64 and the sample SD is 4.907

  2. P(xbar<67.3) = P(Z<(67.3-64)/4.907) = P(Z<0.673) = 0.7486. Of 100 random sample distributions of size n=64, roughly 75 of the means will be less than 67.3.

  3. P(xbar>65.2) = P(Z>(65.2-64)/4.907) = P(Z>0.245) = 1-0.5948 = 0.4052. Of 100 random sample distributions of size n=64, roughly 40 of the means will be greater than or equal to 65.2.

19

  1. P(x<260) = P(Z<(260-266/16)) = P(Z<-0.375) = 0.3557. Of n=100 randomly selected pregnancies, about 35 pregnancies would last less than 260 days.

  2. mx=266 ox=16/√20=3.578

  3. P(xbar<260) = P(Z<(260-266)/3.578) = P(Z<-1.68) = 0.0465. Of 100 randomly selected samples of size n=20 about 5 samples will result in a mean gestation period of less than or equal to 260 days

  4. m=266 ox=16/√50 P(xbar<260) = P(Z<(260-266)/2.26) = P(Z<-2.65) = 0.004. Of 1000 randomly selected samples of size n=50 about 4 samples will result in a mean gestation period of less than or equal to 260 days.

  5. I would conclude that this is highly unlikely to occur, but is still possible.

  6. P(256<xbar<276) = P((256-266)/16/√15<Z<(276-266)/16/√15) = P(-2.42<Z<2.42) = 0.9920-0.0080 = 0.9840. Of 100 randomly selected samples of size n=15, roughly 98 of the mean gestation period lengths will fall within 10 days of the mean.

21

  1. P(x>95) = P(Z>(95-90)/10) = 1-0.6915 = 0.0385. There is a 0.0385 or 3.85% probability a randomly selected 2nd grader reads at above 95 words per minute.

  2. ox=10/√12=2.89 P(xbar>95) = P(Z>(95-90)/2.89) = P(Z>1.73) = 1-0.9582 = 0.0418. Of 100 randomly selected samples of size n=12, 4 would have a mean rate of greater than 95 words per minute.

  3. ox=10/√24 P(xbar>95) = P(Z>(95-90)/10/√24) = P(Z>2.45) = 1-0.9929 = 0.0071. Of 1000 simple random samples of size n=24, roughly 7 samples would have a mean rate of greater than 95 words per minute.

  4. Increasing the sample size results in less samples resulting in a mean greater than 95 because as n increases, ox dereases.

  5. This is not an unusual sample. P(xbar>92.8) = P(Z>(92.8-90)/(10/√20)) = P(Z>1.25) = 1-0.8944 = 0.1056. Of 100 simple random samples of size n=20, roughly 11 will result in mean score higher than 92.8.

  6. P(xbar>c) = 0.050 = P(Z>(c-90)/(10/√20)) = 0.050 z=1.645 c=93.7 wpm

23

  1. P(x>0) = P(Z>(0-0.007233)/0.04135) = 1-P(Z>-0.17) = 0.5675.Of n=100 randomly selected months, 57 would have a positive return.

  2. P(xbar>0) = P(Z>-0.007233/(0.04135/√12)) = 1-P(Z>-0.64) = 1-0.2709=0.7291. Of 100 srs of size n=12 months, 73 will result in mean positive return.

  3. ox=0.04135/√24 P(xbar>0) = P(Z>0.007233/(0.04135/√24)) = 1-P(Z>-0.86) = 1-0.1949 = 0.8051. Of 100 srs of size n=24, 81 of the samples will have a positive mean monthly return rate.

  4. ox=0.04135/√36 P(xbar>0) = P(Z>0.007233/(0.04135/√36)) = 1-P(Z>-1.05) = 1-0.1469 = 0.8531. Of 100 srs of size n=36, 85 of the samples will have a positive mean monthly return rate.

  5. The likelihood of obtaining a positive monthly return rate increases as the time horizon increases.

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

8.2

11

  1. The dist p is approx normal. op-hat = √((0.8x0.2)/75) = 0.046

  2. P(phat>0.84) = P(Z>(0.84-0.8)/√(0.046)) = 1-P(Z<0.87) = 1-0.8078=0.1922. There is a 19% chance that in 100 samples of size n=75 will result in 63 or more individuals with the characteristic.

  3. P(phat<0.68) = P(Z<(0.68-0.8)/√(0.046)) = P(Z<-2.60) = 0.0047. Out of 1000 randomly selected samples of size n=75, 5 will result in 51 or less individuals with the characteristic.

12

  1. The dist is approx normal. op-hat = √((0.65x0.35)/200) = 0.034

  2. P(phat>0.68) = P(Z>(0.68-0.65)/√(0.034)) = 1-P(Z<0.16) = 1-0.5636=0.4364. There is a 43% chance that in 100 samples of size n=200 will result in 136 or more individuals with the characteristic.

  3. P(phat<0.59) = P(Z<(0.59-0.65)/√(0.034)) = P(Z<-0.33) = 0.3707. There is a 37% chance that in 100 samples of size n=200 will result in 118 or more individuals with the characteristic.

13

  1. n=1000, less than 5% of pop size. np(1-p)=227.5>10. Dist is approx normal. mp=.35 op=0.015

  2. p-hat = 390/1000 = 0.39 P(p-hat>390) = P(z>(.39-.35)/(√.53x.65/1000)) = 1-P(Z<2.65) = 0.0040.Out of 1000 randomly selected samples of size n=1000, 4 will result in 390 or more individuals with the characteristic.

  3. p-hat = 320/1000 = 0.32 P(p-hat>320) = P(z>(.32-.35)/(√.53x.65/1000)) = P(Z<-1.99) = 0.0233.Out of 100 randomly selected samples of size n=1000, 2 will result in 320 or less individuals with the characteristic.

14

  1. n=1460, less than 1% of pop size. np(1-p)=227.5>10. Dist is approx normal. mp=.35 op=0.015

15

  1. The responses are qualitative because the answers are not numerical.

  2. It is a random variable because it varies between each sample. The source of the variability is each participant’s ability to order in a foreign language

  3. n=200, less than 5% of pop size. np(1-p)=200(0.47)(0.53)= 49.82>10. Variability is approx normal with mp=0.47 op=0.035

  4. phat=0.5 P(phat>0.5)=P(Z>.03/(0.035))

  5. phat=.4 P(x<80)=P(phat<0.4) P(Z<-0.07/0.035) = P(Z<-1.98) = 0.0239 Out of 1000 randomly selected samples of size n=200, 2 will result in 80 or less individuals can order in a foreign language.

16

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17

  1. n=500 is less than 5% of pop size np(1-p)=500(.39)(.61) = 118.9>10. Dist of phat is approx normal, with mp=0.39 and op=0.022

  2. P(phat<0.38) = P(Z<-0.01/0.022) = P(Z<-0.46) = 0.3228 Out of 100 randomly selected samples of size n=500, 2 will result in 38% or less individuals who believe marriage is obsolete.

  3. P(.4<phat<.45) = P((0.01/0.022)<Z(0.06/0.022)) = P(0.46<z<2.75) = 0.9970-0.6772 = 0.3198 Out of 100 randomly selected samples of size n=500, 32 will have been between 40% and 45% of individuals who believe marriage is obsolete.

  4. P(X>210) = P(phat>0.42) = 1-P(phat<0.42) = 1-P(Z<1.38) = 1-0.9162 = 0.0838 Out of 100 randomly selected samples of size n=500, 8 will result in 210 or more individuals who believe marriage is obsolete.