Stat 2300 International: Midterm Answers and Calculations for Fall 2006, Exams of Business Statistics

The answers and calculations for the midterm exam of stat 2300 international for the fall 2006 semester. It covers various statistical concepts such as variables, sampling, probability distributions, and confidence intervals.

Typology: Exams

Pre 2010

Uploaded on 07/31/2009

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Stat 2300 International, Fall 2006– Sample Midterm Answers
Friday, October 20, 2006
. 1. d. variable.
2. b. sampling without replacement.
3. d. Statistical Inference
4. a. Outliers
5. c. three
6. d. Z-score
7. a. Independent events
8. b. mutually exclusive.
9. b. P(A | B) = P(A)
10. b. Poisson random variable
11. b. (n) (p).
12. a. its outcomes are countable.
13. b. zero, one.
14. d. Poisson and exponential
15. d. below the mean.
16. c. for any sample size.
17. b. increases, normal
18. d. .09 and 1.34
19. b. the t distribution.
20. b. size is large.
21. z = (33.2-32)/.8 = 1.5; value from table = 0.4332;
Probability = 0.5 + 0.4332 = 0.9332 = 93.32%
22. (i) 9.6 (ii) 10 (iii) 10
23. [mean = 10] (i) 10 (ii) 11.6 (iii) about 3.41
24. Probability = .95*.92 = .874 = 87.4%
25. [P(B) = .4, P(M) = .5, P(B|M) = .2]
P(BM) = P(M)*P(B|M) = .5*.2 = .1 = 10%
26. Expected Outcome = .25*(-40,000) + .7*10,000 + .05*70,000 = $500
27. Probability = 10! / (10! * 0!) * .9^10 * .1^0 = .3487 = 34.87%
28. 1 – P(x = 0) = 1 – e^(-2) * 2^0 / 0! = 1- .1353 = 0.8647 = 86.47%
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Stat 2300 International, Fall 2006– Sample Midterm Answers Friday, October 20, 2006 .

  1. d. variable.
  2. b. sampling without replacement.
  3. d. Statistical Inference
  4. a. Outliers
  5. c. three
  6. d. Z-score
  7. a. Independent events
  8. b. mutually exclusive.
  9. b. P(A | B) = P(A)
  10. b. Poisson random variable
  11. b. ( n ) ( p ).
  12. a. its outcomes are countable.
  13. b. zero, one.
  14. d. Poisson and exponential
  15. d. below the mean.
  16. c. for any sample size.
  17. b. increases, normal
  18. d. .09 and 1.
  19. b. the t distribution.
  20. b. size is large.
  21. z = (33.2-32)/.8 = 1.5; value from table = 0.4332; Probability = 0.5 + 0.4332 = 0.9332 = 93.32%
  22. (i) 9.6 (ii) 10 (iii) 10
  23. [mean = 10] (i) 10 (ii) 11.6 (iii) about 3.
  24. Probability = .95*.92 = .874 = 87.4%
  25. [P(B) = .4, P(M) = .5, P(B|M) = .2] P(B∩M) = P(M)P(B|M) = .5.2 = .1 = 10%
  26. Expected Outcome = .25(-40,000) + .710,000 + .05*70,000 = $
  27. Probability = 10! / (10! * 0!) * .9^10 * .1^0 = .3487 = 34.87%
  28. 1 – P(x = 0) = 1 – e^(-2) * 2^0 / 0! = 1- .1353 = 0.8647 = 86.47%
  1. P(6.5 <= x <= 7.5) = P((6.5-8)/.5 <= z <= (7.5-8)/.5) = P(-3 <= z <= -1) = .1587 - .0013 = .1574 = 15.74%
  2. mu = np = 200.4 = 80, sigma = sqrt(np(1-p)) = sqrt(200.4*.6) = sqrt(48) =
    P(x <= 95) = P(x <= 95.5) = P(z <= (95.5-80)/6.9282) = P(z <= 2.24) = .9875 = 98.75%
  3. meanxbar = mean = 50, sigmaxbar = sigma/sqrt(n) = 1 / sqrt(100) = .1; P(xbar > 50.2) = P(z > (50.2-50)/.1) = P(z > 2) = 1 - .9772 = .0228 = 2.28%
  4. for 98%CI: z_alpha/2 = 2.33; CI: [.1 +/- z_alpha/2 * sqrt(.1*.9/299)] = [.1 +/- 2.33 * .01735] = [.0596, .1404]
  5. Mean 27. Confidence Level(95.0%) 7. 95% CI: 20.22047036 35.
  6. (i) Mean 102. (ii) Median 99 (iii) Standard Deviation 59.