251y0231 12/02/02 ECO251 QBA1
THIRD EXAM
NOVEMBER 20, 2001 Name: _____Key______
(Open this document in 'Print Layout' view!)Section Enrolled: (Circle) MWF 10 MWF 11 TR 11
Since there is no correct answer for the Take-home exam a possible solution appears below. Many errors you made are in ‘Things that you should never do on a Statistics Exam or Anyplace Else.’ Wake up and read it!
251x023a 11/16/01 ECO251 QBA1
THIRD EXAM
NOVEMBER 22-23, 2002
TAKE HOME SECTION
Name: Seymour Butz
Section Enrolled: (Circle) MWF 10 MWF 11 TR 11 TR 12:30
Social Security Number: 234567891
Throughout this exam show your work! Please indicate clearly what sections of the problem you are answering! If you are following a rule like , please state it! If you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why!
Part I. Do all the Following (10 Points) Show your work!
My Social Security Number is 265398248. If I write it as below, so that, for example, Minitab tells me that the sample mean is 5.222 and the sample standard deviation is 2.682. (Please don’t do these again!) Write your Social Security next to it in the same way and call it
1 2
2 6
3 5
4 3
5 9
6 8
7 2
8 4
9 8
Compute the following, showing the steps clearly as if you had done it by hand. Do not tell me “That is what my calculator said,” though you are welcome to use your calculator, Excel or Minitab to check your work:
1. The sample variance of (2).
2. The sample covariance between and (2)
3. The sample correlation between and (2)
4. Interpret the correlation (1)
Answers to questions 5) and 6) must be based on the mean, standard deviation, covariance and correlation that you found in questions 1-4. Do not recompute the answers after changing or !
5. If all the numbers in rise by 20, (so that if was [2, 3, 4, 5, 6, 7, 8, 9, 1], it is now [22, 23, 24, 25, 26, 27, 28, 29, 21]) what will the new mean and standard deviation of be? What will be the correlation between and be now? (1.5)
6. If, instead, all the numbers in are multiplied by 3.5, (so that if was [2, 3, 4, 5, 6, 7, 8, 9, 1], it is now [7, 10.5, 14, 17.5, 21, 24.5, 28, 31.5, 3.5]) what will the new mean and standard deviation of be? What will be the correlation between and be now? (1.5)
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Worksheet
13
1 2 4 2 4 4
2 6 36 3 9 18
3 5 25 4 16 20
4 3 9 5 25 15
5 9 81 6 36 54
6 8 64 7 49 56
7 2 4 8 64 16
8 4 16 9 81 36
9 8 64 1 1 8
47 303 45 265 227
13
, , , , and
Note that the and columns and their sums were not needed!
so
Solution: From above:
1. The sample variance of
2. The sample covariance between and
3. The sample correlation between and
4. Interpretation. The negative sign of the covariance and correlation indicates that and tend to move in opposite directions. The correlation squared is about .019. On a zero to one scale, this is extremely weak.
5. If all the numbers in rise by 20, what will the new mean and standard deviation of be? What will be the correlation between and be now?
means that works for sample means too, so that the sample mean rises by 20 to 25.0000.
And , so
. The standard deviation remains 2.73861.
( Also, , so )
and if and , . In this case and , and we started with a
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correlation of , so the correlation between the two is -0.136135. In short, adding 20 to has no effect.
6. If, instead, all the numbers in are multiplied by 3.5, what will the new mean and standard deviation of be? What will be the correlation between and be now? (1.5)
means that works for sample means too, so that the sample mean, which was 5.00000 is multiplied by 3.5 to become 17.5.
And , so if our original standard deviation was we find
(Also, , and so )
and if and , . , so the correlation between the two is unchanged at
.
Part II: Take your Social Security number again, let be the 2nd and 3rd digits of the number, be the 2nd, 3rd and 4th digits and and . ( For example, my Social Security number is 265398248, so , , ,
A jorcillator’s lifespan (failure time) can be represented by a continuous uniform distribution between and years (My jorcillator has a lifespan between 65 and 653 years).
1. What is the probability that it lasts between and years? (1)
2. What is the probability that it lasts between and 1000 years? (1)
3. What is the mean life of such a jorcillator ? (1)
4. What is the standard deviation of the life of such a jorcillator ? (1)
5. If I have five such jorcillators, what is the probability that at least one lasts between and 1000 years? (1)
Solution: Seymour’s Social Security number was 234567891, so that , , , and Our jorcillator has a lifespan from to years and we want . .
1. So . In the diagram below (not to scale), shade the area between 68 and 102.
0 34 68 102 345
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The height of the box is and the base of the shaded area is
102 – 68 = 34. So the area of the shaded area is .
Another way to do a problem of this type is to remember that for any continuous distribution, we can use differences between cumulative distributions, where the cumulative distribution is and and .
So
2. This time we want. Remember, and
In the diagram below, shade the area between 68 and 345 (since there is no area between 345 and 1000).
0 34 68 345 1000
The height of the box is and the base of the shaded area is
345 – 68 = 277. So the area of the box is .
If we use the cumulative distribution method, remember .
So .
3. The formula for the mean is , and . So .
4. The formula for the variance is . So . So the standard deviation of the life of the jorcillator is ?
5. If I have five such jorcillators, the probability that at least one lasts between and 1000 years? (1)
We already found that . Because the probability of lasting is a constant and we have tries, we use the Binomial distribution. , with
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Check your answer! and . is the density – not a probability.
1)
2)
3)
4) , so
5)
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Part III. Do the following problems. ( Do at least 35 points ). Show your work! Choose the problems you do carefully, since most people cannot finish this exam. You may do parts of problems!
1. (13) The following table represents the joint probability of and .
3 / 5 / 91 / ___ / 0 / 0
/ 5 / 0 / .20 / 0
6 / 0 / 0 / .65
a. Fill in the missing number. (1)
b. Are and independent? Why?(2)
c. Compute , the covariance of and , and interpret it. (3)
d. Compute , the correlation of and , and interpret it. (3)
e. Find the distribution of . (1)
f. Using only the results of a)-d), find the mean and variance of .(3)
Solution: a) the missing number must be .15 because the numbers inside the table must add to 1.
b) Check for independence: First you need to find and . Look at the upper left hand probability below. Its value is .15 and it represents . If and were independent , we would have . Since this is not true, we can say that and are not independent. A faster way to do this is to note that the zeroes are not parts of rows or columns of zeroes.
c) (iii) Compute .
To summarize , , , , and
. Positive, so and move together.
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d) Compute , the correlation of and ,
and
So that . The sign has been interpreted above. Since the correlation squared is about .7, on a zero to one scale this is only moderately strong.
e) This is an easy one, since there are so many zeroes.
If we run down the columns of the table and forget about the zeroes:
3 / 1 / 4 / .155 / 5 / 10 / .20
9 / 6 / 15 / .65
f) Using only the results of a)-d), find the mean and variance of .(3)
and
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2. (16) Find the following probabilities. (2 each)
a. Assume that has the standard Normal distribution. Make diagrams.
(i)
0
0 2.01
Shade the entire area below 2.01.
(ii)
0 1.02 3.41
Shade the area between 1.02 and 3.41.
(iii)
-2.34 -1.63 0
Shade the area between -2.34 and -1.63.
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b. Assume that has the binomial distribution. Normal diagrams won’t help!
(i)
(ii) . Since 5 successes in 12 tries correspond to 7 failures, 10 successes
in 12 tries correspond to 2 failures and the probability of failure is 1 - .55 = .45 ,
(iii)
(iv) Do (iii) without using a table. (3 points)
Note that , with
(v) Find the mean and standard deviation for the variable in (iii)
,
so
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3. (13) A firm is hiring two executives from a group of 10 executives that is 30% female. Assume that the individuals are chosen at random (actually without regard to gender).
a. What is the probability that both of the individuals selected are women? (2)
b. What is the probability that at least one woman is selected? (2) How many of you know that ‘at
least one’ does not mean ‘exactly one.’
c. What is the mean and standard deviation of the distribution of the number of women
selected. (2)
d. If the pool (population) is larger than 10, but the firm still is hiring only two people
and the population is still 30% female, how large would the pool have to be before we could
use the binomial distribution to solve these problems? What is the probability that at
least one woman is selected using the binomial distribution. (3)
e. Assume that the pool is much, much larger than 10. How many people would the firm have
to be hiring before we could use the Poisson distribution? What is the probability that at least
one woman is hired now? (Answers without calculations will not be accepted for full
credit.) (4)
Solution: Hypergeometric Distribution with Remember .
and .
a. Exactly 2 women:
b. At least one woman:
.
c. . so so
d. The condition for using the binomial distribution is that the population size is above 20 times the sample size. This would mean a population above 20 times 2 or 40. It would still be true that and so that we could use the Binomial table which says . Thus
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e. We still have , and the condition to replace the Binomial distribution with the Poisson is If we solve for we find , so the sample would have be of size 151, and The outline says , so and
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4. (22) As you all know by now, a Jorcillator has two components, a phillinx and a flubberall.
a. Assume that, in an average week 10 of my Jorcillators need repair. Using your table tell me
the probability that (i) at least one, (ii) more than 20 and (iii) more than 30 need repair. (3)
b. What is the standard deviation of the number that need repair? (1)
c. Assume instead that in each of the first 200 weeks the Flubberall has a constant 10% chance
of failing, what is the chance that the Flubberall fails in the third week? (1.5)
d. What is the chance that it fails between the 3rd and the 10th week? (2)
e. If represents the week in which the Flubberall fails, what is the mean and standard
deviation of ? (1.5)
f. What is the chance that the Flubberall lasts beyond the 10th week? What about the 20th? (2)
g. Assume that the Jorcillator needs both components to operate, that the Phillinx follows the
same distribution as the Flubberall and that their lives are independent, what is the chance
that the Jorcillator does not last beyond the 10th week? (4)
h. For the grand slam, what is the chance that the Jorcillator dies after the tenth, but not after the
20th week? (6)
Solution: a) As I posted on the notice board last week, “If you are looking for numbers of successes when the number of tries is given and the average number of successes per unit time or space is given, you want the Poisson distribution. In this case, the mean (parameter) is so:
(i) (ii)
(iii)