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ECON 103 Tutorial 17

ECON 103 Tutorial 17. Rob Pryce www.robpryce.co.uk/teaching. Question 5.5a. An analyst has available 2 forecasts F 1 and F 2 of earnings per share. Wants a compromised forecast, with XF 1 + (1 – X)F 2 Probability function is random variable. Question 5.5b.

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ECON 103 Tutorial 17

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  1. ECON 103Tutorial 17 Rob Pryce www.robpryce.co.uk/teaching

  2. Question 5.5a An analyst has available 2 forecasts F1 and F2 of earnings per share. Wants a compromised forecast, with XF1 + (1 – X)F2 Probability function is random variable

  3. Question 5.5b An analyst has available 2 forecasts F1 and F2 of earnings per share. Wants a compromised forecast, with XF1 + (1 – X)F2 Probability function is random variable

  4. Question 5.5c An analyst has available 2 forecasts F1 and F2 of earnings per share. Wants a compromised forecast, with XF1 + (1 – X)F2 Probability function is random variable

  5. Question 5.5d An analyst has available 2 forecasts F1 and F2 of earnings per share. Wants a compromised forecast, with XF1 + (1 – X)F2 Probability function is random variable

  6. Question 5.5e An analyst has available 2 forecasts F1 and F2 of earnings per share. Wants a compromised forecast, with XF1 + (1 – X)F2 Probability function is random variable

  7. Question 5.15 Employees are paid $60 plus 20% of the money their calls generate. Amount of money generated is random variable – mean $700 and standard deviation $130 Find the mean and standard deviation of total pay.

  8. Question 5.25 Return on stocks followed normal distribution – mean 12.2% and standard deviation 7.2%

  9. Question 5.25a Return on stocks followed normal distribution – mean 12.2% and standard deviation 7.2%

  10. Question 5.25b Return on stocks followed normal distribution – mean 12.2% and standard deviation 7.2%

  11. Question 5.25c Return on stocks followed normal distribution – mean 12.2% and standard deviation 7.2%

  12. Question 5.39a Given a random sample size of n=900 from a binomial probability distribution with P=0.50 Find the probability that the number of successes is greater than 500

  13. Question 5.39b Given a random sample size of n=900 from a binomial probability distribution with P=0.50 Find the probability that the number of successes is fewer than 430

  14. Question 5.39c Given a random sample size of n=900 from a binomial probability distribution with P=0.50 Find the probability that the number of successes is between 440 and 480

  15. Question 5.39d Given a random sample size of n=900 from a binomial probability distribution with P=0.50 With probability 0.10, the number of successes is fewer than how many?

  16. Question 5.39e Given a random sample size of n=900 from a binomial probability distribution with P=0.50 With probability 0.08, the number of successes is greater than how many?

  17. Question 5.61 A random variable X is normally distributed – mean 100 and variance 100 A random variable Y is normally distributed – mean 200 and variance 400. X and Y have correlation coefficient of 0.5 Find the mean and variance of the random variable W = 5X + 4Y

  18. Question 5.67 Furnace will reduce energy costs by X X random variable – mean $200 and standard deviation $60 Find the mean and standard deviation of the total reduction over 5 years What assumptions?

  19. Any Questions Email: r.pryce@lancaster.ac.uk Web: www.robpryce.co.uk/teaching Office Hour: Thursday, 3pm, Charles Carter C floor, near C07

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