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Solutions to Midterm Exam 2
Problem 1. It known that when the earnings of the Alpha Corporation go up, its stock price usually goes up. In fact, when earnings go up in one quarter, 65% of the time the stock price goes up the next quarter. When their earnings go down in a quarter, 23% of the time the stock price goes up during the next quarter. If the percent changes in earnings for a quarter are normally distributed with mean 2% and standard deviation 1.5%, what is the probability that the stock price will go up in any given quarter?
The first thing to determine here is the probability that earnings go up in a given quarter. Let X be the percent change in earnings.
|[pic] |[pic] |
| |[pic] |
| |[pic] |
| |[pic] |
Therefore, there is a 90.82% chance that earnings will go up in the next quarter. Now, make a contingency table, letting E be the event that earnings go up S the event that the stock goes up. We need three facts:
• There is no chance that either the stock or the earnings will stay exactly the same.
• [pic]
• [pic].
| |[pic] |[pic] | |
|[pic] |0.5903 |0.3179 |0.9082 |
|[pic] |0.0211 |0.0707 |0.0918 |
| |0.6114 |0.3886 |1.0000 |
The answer is 61.14%.
Problem 2. The Downhill Manufacturing Company produces snowboards. The average life of their product is 12 years. They calculate that there is a 1/5 chance that a snowboard will last more than 15 years.
a. (2 points) A snowboard is considered defective if its life is less than 7 years. If the distribution of the useful lives of snowboards is normal, what is the chance that a randomly selected snowboard is defective?
First, we need to determine the standard deviation of snowboard life, using what we are given:
|[pic] |[pic] (from the table) |
Therefore:
|[pic] |[pic] |
| |[pic] |
| |[pic] |
|[pic] |[pic] |
|[pic] |[pic] |
| |[pic] |
The probability of lasting less than 7 years can now be found:
|[pic] |[pic] |
| |[pic] |
| |[pic] |
| |[pic] |
b. (3 points) Downhill Manufacturing sells these snowboards in shipping containers —called pallets — that contain 80 snowboards each. Again, assuming the distribution of their life is normal, what is the probability that in a single pallet there are 12 or fewer defective snowboards? Assume independence between snowboards packed together.
This is a binomial question. Let Y = the number of defectives in a pallet. Then Y is binomially distributed with n = 80 trials and with probability of defective p = 0.0808. Therefore, the probability that there are 12 or less defectives is P(Y < 12). The expected number of defectives is [pic] The standard deviation of defectives is [pic].
We can use the Central Limit Theorem (with the continuity correction) to approximate the exact answer, using 12.5 as our cut-off value:
|[pic] |[pic] |
| |[pic] |
| |[pic] |
Problem 3. (2 points) A 55-year-old woman buys a 1-year term life insurance policy for an annual premium of $58. Under the policy, the insurance company must pay her family $1,000 if she dies during the year. If the average annual number of deaths for women in this age group is 11 deaths per 1,000 women, what is the expected value and standard deviation of the net income for the insurance company from this policy?
The probability of this woman dying during the next year is 11/1000 = 0.011, so the gain for the insurance company can be represented by a random variable X. The possible values of X are:
|Outcome |Payoff (X) |Probability |
|Woman Dies |[pic] |0.011 |
|Woman Doesn't Die |[pic] |0.989 |
Therefore
|[pic] |[pic] |
| |[pic] |
The insurance company makes $47 on average for a policy of this type.
Standard deviation:
|Payoff |Expected |Error |Error^2 |Weighted Error^2 |
|$(942.00) |$47.00 |$(989.00) |$978,121.00 |$10,759.33 |
|$58.00 |$47.00 |$11.00 |$121.00 |$119.67 |
| | | |Sum |$10,879.00 |
| | | |StDev |$104.30 |
Problem 4. (2 points) The heights of adult men are normally distributed with mean 176 cm and standard deviation 7 cm. How tall should a door be if it is desired that 96% of the men going through it do not hit their head? (Assume they do not crouch down!)
Let X be the height of an adult man. Then X is normally distributed with mean 176 cm and standard deviation 7 cm. We are looking for some height h such that P(X < h) =0.96. From the Z table we can infer that the number h must be 1.75 standard deviations above the mean. Therefore,
|[pic] |[pic] |
| |[pic] |
| |[pic] |
Note: in fact, the probability associated with 1.75 standard deviations is slightly less than 0.96, as you can see in the table. However, it is not necessary to interpolate between Z-values on the exam. Use the Z-value that is closest to the desired probability.
Problem 5. The manager of an internet retailing operation selling a single product has found that people who visit the web site buy the product 26% of the time, and that the behavior of customers seems to be independent. Assume that exactly 8 potential customers visit the site every day.
Imagine that the manager is on an incentive plan that pays him $300 for any day in which the site generates three or more sales; otherwise his pay is $100 per day.
a. (2 points) What is the probability of him earning the $300 on any random day?
The quickest way to solve this is to see that there are three ways in which he would fail to earn the $300, namely if the site makes 0, 1, or 2 sales. If we add up the probabilities of these three events and subtract the total from 100%, we will have the answer.
The number of sales made (S) is binomially distributed with n = 8 and p = 0.26. The manager will get the $300 if S ( 3. Therefore:
|[pic] |[pic] |
| |[pic] |
| |[pic] |
| |[pic] |
b. (3 points) What is the expected value of his pay on any random day?
There is a 34.65% chance that he gets $300, and a 65.35% chance that he gets $100, so therefore:
|[pic] |[pic] |
| |[pic] |
| |[pic] |
c. (2 points) The manager is offered his choice of two alternative incentive schemes, by which he will receive either (a) no base salary, but a commission of $75 per sale, or (b) a fixed salary of $160 per day, or (c) the original plan outlined above. Which plan should he select if he wants to maximize the expected value his of earnings? (Circle one.)
a) a commission of $75 per sale
b) a fixed salary of $160 per day
c) the original plan
In order to evaluate option (a), we need to know the expected number of sales per day:
[pic]
Therefore, under that plan his expected earnings would be 2.08 * $75 = $156
The original plan has the highest expected value.
Problem 6. A salesman for a company has annual sales that are normally distributed with a mean of $1,200,000 and a standard deviation of $300,000. He earns a base salary of $35,000, and a commission of 4.5% of sales. Let ( be the correlation between his sales and his income. Assume that sales are never negative.
a. (2 points) What would you expect the sign of ( to be (positive or negative)?
Clearly every time the salesman increases sales he sees an increase in his income, so ( must be positive.
b. (2 points) Find (.
Every sales dollar brings in a fixed amount of additional income, therefore ( = +1. If you imagine a graph with sales on the X axis and the salesman's income on the Y axis, you should see that all the possible points fall on a straight line. Knowing the value of one of the two variables in a given year gives us perfect knowledge about the value of the other variable in the same year.
Problem 7. The Clinton administration advocated that every state require high school students to pass exit exams to graduate; twenty-six states already required or had approved these tests. But in a federal lawsuit in San Antonio, lawyers argued that the Texas Assessment of Academic Skills (TAAS) was unfair to thousands of minority students. The lawyers noted that these minorities made up 40 percent of Texas high school seniors but 80 percent of those who failed to pass the test by the end of 12th grade.
Out of a total of 212,773 12th graders in the class of 1999, 5,461 students had not passed the TAAS by the end of the school year.
a. (2 points) Based on 1999 data, what would you estimate to be the probability that a randomly selected Texas high school senior would fail to pass the TAAS by the end of the 12th grade?
Define the following events:
|Event |Symbol |Probability |
|Student Fails the TAAS |[pic] | |
|Student is a Minority |[pic] |0.40 |
|Minority, Given Failed |[pic] |0.80 |
|[pic] |[pic] |
| |[pic] |
| |[pic] |
b. (2 points) Given that a student is in the minority group, what would you estimate to be the probability that he/she would fail to pass the TAAS by the end of the 12th grade?
We could approach this in several ways. One possibility is to construct a 2 x 2 "box" diagram, using the following steps:
|[pic] |[pic] |(given) |
|[pic] |[pic] |(given) |
|[pic] |[pic] |(from Part a) |
|[pic] |[pic] |
| |[pic] |
| |[pic] |
|[pic] |[pic] |
| |[pic] |
| |[pic] |
|[pic] |[pic] |
| |[pic] |
| |[pic] |
[pic] can be calculated in several ways, by subtraction. The finished box:
| |[pic] |[pic] | |
|[pic] |0.02053 |0.37947 |0.4 |
|[pic] |0.00513 |0.59487 |0.6 |
| |0.02567 |0.97433 | |
Now, to solve (b):
|[pic] |[pic] |
| |[pic] |
| |[pic] |
c. (3 points) If a group of 1000 randomly selected Texas high school seniors took the TAAS, what would you estimate to be the probability that 40 or more of them would fail?
It is important to recognize that this is a binomial question. We need to find out [pic], given that X is binomial with [pic] and [pic] (from Part a). Having recognized this, the next thing is to realize that we can apply the Central Limit Theorem and avoid doing many discrete binomial calculations.
|[pic] |[pic] |[pic] |[pic] |
|[pic] |[pic] |[pic] |[pic] |
Applying the continuity correction, our cut-off value should be 39.5. Therefore:
|[pic] |[pic] |
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