Econ 103 tutorial 17
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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 103 tutorial 17

ECON 103Tutorial 17

Rob Pryce

www.robpryce.co.uk/teaching


Question 5 5a

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

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

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

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

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

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

Question 5.25

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


Question 5 25a

Question 5.25a

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


Question 5 25b

Question 5.25b

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


Question 5 25c

Question 5.25c

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


Question 5 39a

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

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

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

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

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

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

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

Any Questions

Email:[email protected]

Web:www.robpryce.co.uk/teaching

Office Hour:Thursday, 3pm, Charles Carter

C floor, near C07


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