Statistical Analysis for Management
Autor: Daniel Vega • September 20, 2017 • Exam • 1,007 Words (5 Pages) • 764 Views
The Fox School of Business
Executive MBA Program
NAME: xxxx
Show all your work here. Underline or highlight your final answer.
- 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 |
- Find the expected value of the exchange rate.
The first step is to verify that the sum of the probabilities is equals to 1.
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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
- 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 |
- What does the standard deviation represent?
Represents how far the values are from the media.
- 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 |
- 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.
- 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 |
- 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 |
- 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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