12/31/99 251y9943 ECO251 QBA1 Name key
FINAL EXAM DECEMBER 16, 1999
Part I. Do all the Following (20 Points) Make Diagrams!
A.
1. Diagram! These are expected and will be provided in class for version 2.
2.
3.
4.
5. This is the point with a probability of .89 above it or .11 below it. It is the 11th percentile of
So . From the diagram the closest we can come to this
probability using the Normal table is So
Note that probabilities cannot be above 1 or below 0.
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B. . As usual, people made diagrams of with zero in the middle. Make up your mind! If you are diagramming , put the mean in the middle; if you are diagramming put zero in the middle.
1.
Remember .
2.
3.
4.
5. This is the point with a probability of .89 above it or .11 below it. It is the 11th percentile of .
Because it is below the 50th percentile, which is the mean for the Normal distribution, the point we
want will be below 1.5. On the previous page we found So
To check to see if this is correct:
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II. (4 points-2 point penalty for not trying .) Show your work!
A manufacturer of printheads investigates the amount of time before a printhead fails. The data for a sample of 8(in millions of characters)are below.
1.5
1.4
2.0
1.6
1.3
2.1
1.4
1.5
Compute the sample standard deviation, . Show your work. (4)
Solution:
Printhead lives
1 1.5 2.25
2 1.4 1.96
3 2.0 4.00
4 1.6 2.56
5 1.3 1.69
6 2.1 4.41
7 1.4 1.96
8 1.5 2.25
12.8 21.08
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III. Do at least 4 of the following 7 Problems (at least 12 each) (or do sections adding to at least 48 points - Anything extra you do helps, and grades wrap around) . Show your work!
1. Using the sample standard deviation from the sample of 8 on the previous page
a. Compute a 99% confidence interval for the mean time before a printhead fails. (4)
b. Compute a 99% confidence interval for the mean time before a printhead fails assuming that the sample of 8 was taken from a batch of 80 printheads. (4)
c. Compute a 99% confidence interval for the mean time before a printhead fails assuming that the population standard deviation is known to be 1.40 and the population is very large, but the sample is still of size 8. (4)
Solution: From the previous page , , , .
a) For a) and b) population variance is unknown, so we must use t instead of z. The general formula for a confidence interval is . Here the degrees of freedom are and since the confidence level is 99%, or . Also .
From the previous problem -
Putting this all together or 1.24 to 1.96. More formally,
b) The population size, , is less than 20 times the sample size so and or 1.25 to 1.95.
c) This time is known, so we can replace with . and or 0.325 to 2.875.
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2. The Freshman class at Chest Wester University has SAT Math scores that are normally distributed with a mean of 485 and a standard deviation of 70.
a. What is the probability that a class of 49 will have a mean score below 476? (3)
b. What is the probability that a person randomly picked from the Freshman class will have a score below 476? (2)
c. Assuming that the class is quite large, what is the probability that 10 people picked randomly from the Freshman class will all have a score below 476? (2)
d. What score would you have to have to be at the 95th percentile of this class? (3)
e. Create a symmetrical interval around 485 that contains 80% of the Freshman math scores. (2).
f. What is the probability that, if I pick 10 people from the Freshman class, 8 or more of them will have scores above 485? (2)
Solution: From the problem statement , i.e. and for the sample mean,
a)
b)
c) Binomial probability of 10 out of 10 when is
d) From the table . So
e) From the table . So or 395.26 to 574.74.
f) Since the probability of being above the mean is .5, this is a binomial problem with
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3. 20% of the patients at my hospital are dissatisfied with their experience. Last Monday 20 patients were discharged. All filled out surveys.
a. In this group of 20, what was the mean and variance of the number that were dissatisfied? (2)
b. What was the probability that over half of those discharged were dissatisfied? (2)
c. What was the probability that at least one was dissatisfied? (2)
d. If, instead, 65% of my patients were dissatisfied, what is the probability that at least half of the group of 20 would be dissatisfied? (2)
e. If 65% of my patients were dissatisfied, what is the chance that at least one of the group of 20 would be dissatisfied? (2)
f. If 65% of my patients were dissatisfied, and 200 were discharged in one day, what is the probability that at least half would be dissatisfied? (Don’t answer yet!)
(i) Can this problem be done using the Poisson distribution? Why? If it can be done using the Poisson distribution, do it! (2)
(ii) Can this problem be done using the Normal distribution? Why? If it can be done using the Normal distribution, do it! (3)
Solution: Binomial
a)
b)
c)
d) 10 or more successes when is 10 or fewer failures when the probability of
failure is .35.
e) The probability of one being satisfied is .35. The probability that all are satisfied is . So the probability that at least one is not satisfied is . To use the table to do this, reason that if at least one is not satisfied, not all are satisfied. Look up for .
f) (i) You cannot use the Poisson distribution since is below 500.
(ii) You can use the Normal distribution because and . Without continuity correction . Or with the continuity correction,
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4. observations of the velocity that at which a ball is thrown and the distance that it is hit are reported below.(Ab p9-34)
Note that all the sums are computed for you. Do not compute any other sums when doing this problem. Find the covariance and correlation between the velocity of the pitch and the distance the ball goes. (7)
a) Interpret the correlation. What does this mean about the likelihood that a given hit will be a
home run? (2)
b) A sample is taken in another stadium. It is identical to the numbers shown here except all the distances are 5 feet longer. What would the new covariance and correlation be? (3)
326 20 106276 400 6520
334 30 111556 900 10020
339 40 114921 1600 13560
346 50 119716 2500 17300
356 60 126736 3600 21360
362 70 131044 4900 25340
369 80 136161 6400 29520
377 90 142129 8100 33930
387 100 149769 10000 38700
3196 540 1138308 38400 196250
Solution:
, ()
, ()
It is still true that a correlation cannot be above 1 or below –1!
a) The positive sign of , the sample correlation, indicates that x and y tend to move closely
together. If we square , we get approximately .996, which on a zero to one scale indicates a
very strong relationship. This means that the faster the pitch, the more likely that a hit will be a
home run.
b) We are leaving alone, but replacing by . From the syllabus supplement and the
outline if and so that and , and ,
. (Unchanged)
(Unchanged)
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5. a) Two events and are independent. , . Find the following.
(i) Can and be mutually exclusive? Why? (1)
(ii) (1)
(iii) (1)
(iv)(2)
b) Now assume that is independent of , and , is independent of , and ,
and is independent of , and . , , that , and are mutually
exclusive and collectively exhaustive relative to one another, and that , and are mutually
exclusive and collectively exhaustive relative to one another. Fill in the joint probability table below.(3)
c) Now assume that a machine has two components. is the probability that component 1 fails in
period 1, is the probability that component 1 fails in period 2 and is the probability that
component 1 fails in period 3. is the probability that component 2 fails in period 1, is the
probability that component 2 fails in period 2 and is the probability that component 2 fails in
period 3.
Assume that the machine works as long as any one component works. (i) If event occurs,
During which period will the machine go down? (1) What about event ? (1) Find the probability
that the machine goes down in period 1(1), period 2 (2) and period 3 (2). Use the joint probability table
that you created in b).
Solution: a) Two events and are independent. , .
(i) and cannot be mutually exclusive? Briefly, no two possible events can be both. If and are mutually exclusive but see (ii) below.
(ii) If and are independent .
(iii) If and are independent, by definition
(iv)
b) Because , and are mutually exclusive and collectively exhaustive relative to one another, and and , . Similarly, if , and are mutually exclusive and collectively exhaustive relative to one another and and , then . Because of the independence assumption, the numbers in the joint probability table will be the product of the marginal probabilities. .
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c) Because the machine works as long as the longest lived component works, the event downs the machine in period 1, but the event downs the machine in period 3. So the various joint events down the machine as follows:
Event Period Probability
1 .12
2 .15
3 .03
2 .16
2 .20
3 .04
3 .12
3 .15
3 .03
So
or
or
Note that, since no component lasts beyond period 3, these three probabilities must add to one.
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6.
a. A professor has a pool of 20 multiple choice questions involving discrete probability
distributions. Five of these questions involve the Hypergeometric distribution. Assume that he picks 7 questions at random to put on a quiz. What is the probability that at least one concerns the Hypergeometric distribution? (3)
b. What is the mean and variance of the number of questions that concern the Hypergeometric distribution on the above quiz? (2)
c. What would the answer to the above be if the pool of questions contained over 120 questions but the proportion that involve the Hypergeometric distribution was the same as in a)? Do not use the same distribution as you used in a).(2)
d. A student has not studied for the above seven-question quiz. Each multiple-choice question has four answers and she picks one randomly. What is the chance that she will get at least four right? (2)
e. A student is extremely well-prepared for an exam. Nevertheless she knows that since she is tired she will make an average of .1 mistake per page. The exam is 8 pages. What is the probability that she will turn in a perfect exam? (2)
f. In the exam in e, the professor takes off 2 points for each mistake. What is the variance of the number of points he will take off on her exam? (2)
Solution: a) Hypergeometric .
b)
c) Since there was some vagueness in this question, either of the two following answers were
accepted – 3 points was given for an answer to both. Binomial The rule of
thumb for replacing the Hypergeometric by the binomial is that the population should be 20 times
the sample.
(i) From the binomial table,
(ii)
d)
e) Poisson
f) Poisson Since .
Note that it is still true that probabilities cannot be above 1 or below 0.
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7. A soft drink machine dispenses an amount of beverage that is distributed according to the continuous Uniform distribution between 6.7 and 7.3 oz.
a. What is the mean and variance of the number of ounces dispensed by the machine. (2)
b. What is the probability that a given drink will contain more than 7.1 oz.? (2)
c. What is the probability that a given drink will contain between 6.5 and 7 oz? (1)
d. Using the results of a) and your knowledge of the distribution of the sample mean, if the machine is tested by being used 100 times, what is the mean and variance of , the average amount of soft drink dispensed? (2)
e. What is the probability that the sample mean in d) will exceed 7.02 oz.? (3)
f. What is the probability that the total amount of liquid dispensed in 100 uses will exceed 700 oz.? (2) 710 oz.? (2)