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

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standard deviation of sampling distribution = 14/7=2 The sampling distribution is approximately normal with mean=80 and standard deviation=2.

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P(x-bar>83)=P(z>1.5)=1-0.9332=0.0668

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P(x-bar<=75.8)=P(z<-2.1)=0.0179

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P(78.3<x-bar<85.1)=P(-0.85<z<2.55)=0.9946-0.1977=0.7969

17

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The population distribution must be normal. If the population is normally distributed, the standard deviation of sampling distribution = 4.91. The sampling distibution is normally distributed with mean=64 and standard deviation=4.91.

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P(x-bar<67.3)=P(z<0.672)=0.7486

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P(x-bar>=65.2)=P(z>=0.244)=0.4052

19

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P(x<260)=P(z<-0.375)=0.3557

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The standard deviation of sampling distribution=3.58 The sampling distribution of the sample mean length of human pregnancies is approximately normal with mean=266 and standard deviation=3.58

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P(x-bar<=260)=P(z<=-1.68)=0.0465

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the standard deviation of sampling distribution with n=50 is 2.26 P(x-bar<=260)=P(z<=-2.65)=0.004

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The probability of randomly selecting a pregnancy with a mean gestation period of 260 days or less from a ramdom saple of 50 pregnancies is very low.

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the standard distribution of sampling distribution with n=15 is 4.13 P(256<x-bar<276)=P(-2.42<z<2.42)=1-2*0.0078=0.9844

21

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P(x>95)=P(z>0.5)=1-0.6915=0.3085

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standard deviation of sampling distribution=2.89 P(x-bar>95)=P(z>1.73)=1-0.9582=0.0418

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standard deviation of sampling distribution=2.04 P(x-bar>95)=P(z>2.45)=1-0.9929=0.0071

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The probability decreases as the sample size increases, because the standard deviation of sampling distribution decreases as the sample size increases.

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Assuming the reading program is ineffective, P(x-bar<92.8)=0.8944, meaning that if the reading program is ineffective, 89.44% of students cannot read faster than 92.8 words per minute. Since the sample mean is 92.8 wpm, the reading program is effective.

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Assume the standardization of that value=t. Since P(z>t)=0.05, P(z<t)=0.95. According to the table, t=1.645 x-bar=1.645*2.24+90=93.68 There is a 5% chance that the mean reading speed of a random sample of 20 second grade students will exceed 93.68

23

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P(x>0)=P(z>-0.17)=1-0.4325=0.5675

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With n=12, the standard deviation of sampling distribution=0.012 P(x-bar>0)=P(z>-0.60)=1-0.2743=0.7257

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with n=24, the standard deviation of sampling distribution=0.00844 P(x-bar>0)=P(z>-0.86)=1-0.1949=0.8051

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with n=36, the standard deviation of sampling distribution=0.00689 P(x-bar>0)=P(z>-1.05)=1-0.1469=0.8531

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As time horizon increases, the probability of mean monthly rate of return increases.

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

8.2

11

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Since 750.80.2=12>10, the sampling distribution is approximately normal, with mean=0.8 and standard deviation=0.046

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P(p-hat>=0.84)=P(z>=0.87)=1-0.8078=0.1922

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P(p-hat<=0.68)=P(z<=-2.61)=0.0045

12

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Since 2000.650.35=45.5 >10, the sampling distribution is approximately normal, with mean=0.65 and standard deviation=0.034

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P(p-hat>=0.68)=P(z>=0.88)=1-0.8106=0.1894

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P(p-hat<=0.59)=P(z<=-1.76)=0.0392

13

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Since 10000.350.65=227.5>10, the sampling distribution is approximately normal, with mean=0.35 and standard deviation=0.015

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P(P-hat>=0.39)=P(z>=2.67)=1-0.9962=0.0038

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P(p-hat<=0.32)=P(z<=-2)=0.0228

14

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Since 14600.420.58>10, the sampling distribution is approximately normal, with mean=0.42 and standard deviation=0.0129

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P(p-hat>=0.45)=P(z>=2.33)=1-0.9901=0.0099

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P(p-hat<=0.4)=P(z<=-1.55)=0.0606

15

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It’s qualitative, because people can only answer whether they can or cannot.

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P hat is a ramdom vairable because the probability of selecting a person who can order meal in a foreign language from a random sample varies.

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Since 2000.470.53=49.82>10, the sampling distribution is approximately normal, with mean=0.47 and standard deviation=0.035

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P(p hat>0.5)=P(z>0.86)=1-0.8051=0.1949

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P hat=80/200=0.4. P(p hat<=0.4)=0.0228. Since the probability of 80 or fewer Americans can order meals in a foreign language is 2.28% which is considerably low, it is unusual.

16

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It is qualitative, because people can only answer whether they are or not.

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P hat is a ramdom variable because the probability of selecting a satisfied American from a ramdom sample varies.

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Since 1000.820.18=14.4>10, the sampling distribution is approximately normal, with mean=0.82 and standard deviation=0.0384

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P(p hat>0.85)=P(z>0.78)=1-0.7823=0.2177

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P(p hat<=0.75)=P(z<=-1.82)=0.0344, which is low and unusual

17

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Since 5000.390.61>10, the sampling distribution is approximately normal, with mean=0.39 and standard deviation=0.0218

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P(p hat<0.38)=P(z<-0.46)=0.3228

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P(0.40

=0.42)=P(z>=1.38)=1-0.9162=0.0838, which is low and unusual