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Statistical Analysis for Management

Autor:   •  September 20, 2017  •  Exam  •  1,007 Words (5 Pages)  •  764 Views

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The Fox School of Business

Executive MBA Program

NAME: xxxx

Show all your work here. Underline or highlight your final answer.

  1. A company orders components from Japan. When the products are delivered, it must convert dollars into yen to pay the producer. When the next order is delivered, it believes the exchange rate will take on the following values with shown probabilities.

Exchange Rate (yen per dollar)

Probability

100

0.1

110

0.2

120

0.4

130

0.2

140

0.1

  1. Find the expected value of the exchange rate.

The first step is to verify that the sum of the probabilities is equals to 1.

[pic 3]

We take the expected exchange rate and multiplies it by the probability and the sum all of them.

Exchange Rate (yen per dollar)

Probability

100

0,1

10

110

0,2

22

120

0,4

48

130

0,2

26

140

0,1

14

Expected Value

120

Expected Value = 120

  1. Find the standard deviation of the exchange rate.

Exchange Rate (yen per dollar)

Probability

x^2*Prob

100

0,1

10

1000

110

0,2

22

2420

120

0,4

48

5760

130

0,2

26

3380

140

0,1

14

1960

Expected Value

120

14520

Expected Value

120

Variance

120

Standard Deviation

10,95445115

  1. What does the standard deviation represent?

Represents how far the values are from the media.

  1. Find the probability that the exchange rate will be higher than 120 yen/dollar.

Probability is:

 

130

0,2

140

0,1

Total

0,3


  1. A typical incoming phone call to your catalog sales force results in a mean order of $30 with a standard deviation of $10. You may assume a Normal distribution.

  1. Find the probability that one call will result in an order of more than $35 tomorrow.

Normal Distribution

Mean

 $         30

Standard Deviation

 $         10

P(X>$35)

P((X-30)/1>($35-$30)/$10)

0,5

P(Z>0,5)

0,30854

  1. Consider a random sample of 25 phone calls. Find the probability that the average order will be more than $35 tomorrow.

P(Average>$35)

P((X-30)/1>((35-30)/10/Sqr(25)))

P(Average>((35-30)/10/Sqr(25)))

P(Average>$35)

0,00621

  1. Why are the probabilities computed in a) and b) different?

Because a) is the probability occurrence for just one event and the b) the probability of the average of the occurrences.

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