ECON 103 Tutorial 17 - PowerPoint PPT Presentation

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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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ECON 103Tutorial 17

Rob Pryce

www.robpryce.co.uk/teaching

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

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

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

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

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

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.

Question 5.25

Return on stocks followed normal distribution – mean 12.2% and standard deviation 7.2%

Question 5.25a

Return on stocks followed normal distribution – mean 12.2% and standard deviation 7.2%

Question 5.25b

Return on stocks followed normal distribution – mean 12.2% and standard deviation 7.2%

Question 5.25c

Return on stocks followed normal distribution – mean 12.2% and standard deviation 7.2%

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

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

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

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?

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?

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

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?

Any Questions

Email:r.pryce@lancaster.ac.uk

Web:www.robpryce.co.uk/teaching

Office Hour:Thursday, 3pm, Charles Carter

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