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. Type answer here. 14/7=2 the population of the sample size is unknown, but the sample size is big, therefore we can use the Central Limit Theorem to say that the sampling distribution is approximatly normal.

  2. Type answer here. 1-P(Z>1.50)=1-0.9332=0.0668
  3. Type answer here. P(Z<-2.10)=0.0179

  4. Type answer here. 0.9946-0.1977=0.7969 17

  5. Type answer here. the population must be normally distributed. if this is the case, then the sampling distribution is normal. the mean and standard deviation of the sampling destribution are 64 and 4.907.

  6. Type answer here. P(Z<0.67)=0.7486

  7. Type answer here. 1-P(Z<0.24)=0.4052 19

  8. Type answer here. P(X<260)=P(Z<0.38)=0.3520
  9. Type answer here. the length of human pregnancy is normally distributed, the sampling distribution is normal with it being 3.578
  10. Type answer here. 0.0465
  11. Type answer here. 0.0040
  12. Type answer here. WE COULD CONCLUDE THAT THE SAMPLE CAME FROM A POPULATION THAT IS LESS THAN 260 DAYS.

  13. Type answer here. 0.9922-0.0078=0.9844 21

  14. Type answer here. 1-0.6915=0.3085
  15. Type answer here. 1-0.9582=0.0418
  16. Type answer here. 1-0.9929=0.0071
  17. Type answer here. IT DECREASES THE PROBABILITY. BECAUSE THE STANDARD DEVIATION DECREASES AS N INCREASES.
  18. Type answer here. NOT UNUSUAL. 1-0.8944=0.1056 THE NEW READING PROGRAM IS NOT MORE EFFECTIVE THAN THE OLD PROGRAM.

  19. Type answer here. 93,7 WORDS PER MINUTE

23

  1. Type answer here. 1-0.4325=0.5675
  2. Type answer here. 1-0.2709=0.7291
  3. Type answer here. 1-0.1949=0.8051
  4. Type answer here. 1-0.1469=0.8531
  5. Type answer here. THE LIKELIHOOD OF EARNING A POSITIVE RATE OF RETURN 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. Type answer here. THE DISTRIBUTION OF P HAT IS APPROXIMATELY NORMAL, WITH THE MEAN=0.8 AND STANDARD DEVIATION IS 0.046.
  2. Type answer here. 1-0.8078=0.1922

  3. Type answer here. 0.0047 12

  4. Type answer here. THE DISTRIBUTION OF P HAT IS APPROXIMATELY NORMAL, WITH THE MEAN=0.65 AND STANDARD DEVIATION IS 0.0337
  5. Type answer here. 0.1867

  6. Type answer here. 0.0375 13

  7. Type answer here. THE DISTRIBUTION OF P HAT IS APPROXIMATELY NORMAL, WITH MEAN EQUALS TO 0.35 AND STANDARD DEVIATION EQUALS TO 0.015
  8. Type answer here. 1-0.9960=0.0040

  9. Type answer here. 0.0233 14

  10. Type answer here. THE DISTRIBUTION OF P HAT IS APPROXIMATELY NORMAL, WITH MEAN EQUALS TO 0.42, AND STANDARD DEVIATION EQUALS TO 0.0129
  11. Type answer here. 1-0.9901=0.0099

  12. Type answer here. 0.0606 15

  13. Type answer here. QUALITATIVE
  14. Type answer here. BECAUSE IT VARIES FROM SAMPLE TO SAMPLE. AND THE VARIABILIY IS THE INDIVIDUALS IN THE SAMPLE AND THEIR ABILITY TO ORDER A MEAL IN A FOREIGN LANGUAGE.
  15. Type answer here. THE DISTRIBUTION OF P HAT IS APPROXIMATELY NORMAL, WITH MEAN OF 0.47 AND STANDARD DEVIATION EQUALS 0.035.
  16. Type answer here. P HAT= 0.5 1-0.8023=0.1977

  17. Type answer here. P HAT=0.4 0.0239 16

  18. Type answer here. QUALITATIVE
  19. Type answer here. BECAUSE IT VARIES FROM SAMPLE TO SAMPLE. AND THE VARIABILITY OF THE SOURCE IS THE INDIVIDUALS IN THE SAMPLE AND THEIR ANSWER TO THE QUESTION
  20. Type answer here. THE DISTRIBUTION OF P HAT IS APPROXIMATELY NORMAL WITH MEAN OF 0.82, AND STANDARD DEVIATION OF 0.0384
  21. Type answer here. 0.2177

  22. Type answer here. 0.0344 17

  23. Type answer here. THE DISTRIBUTION IS APPROXIMATELY NORMAL, WITH MEAN EQUALS 0.39 AND STANDARD DEVIATION EQUALS 0.022
  24. Type answer here. 0.3228
  25. Type answer here. 0.3198

  26. Type answer here. 0.0838