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Assignments2 solutions: Due by Midnight Sunday, September 30th, 2012(dropbox of week 2) (Chapters 4, 5, and 6): Total 70 points.

True/False (one point each)

Chapter 4

1. If events A and B are independent, then P(A/B) is always equal to zero.  FALSE In fact P(A/B) equals P(A) which is not zero unless A is an impossible event.

2. If events A and B are mutually exclusive, then P(A/B) is always equal to zero.  True This is obvious from the definition of Mutually exclusive events. If B occurs then A cannot occur.

3. Events whose union has the probability equal to 1 are called Exclusive Events. FALSE They are called Exhaustive Events

Chapter 5

4. If the probability of success is 0.4 and the number of trials in a binomial

distribution is 150, then its standard deviation is 36. FALSE σ= √(np(1-p))

=√(150*.4*.6) = 6

5. If a fair coin is tossed 100 times, then the variance of the random variable defined as the number of heads is exactly five. FALSE σ2 = np(1-p)= 100*.5*.5= 25. So the standard dev is 5 not the Variance.

6. If a fair coin is tossed 20 times then the probability of exactly 10 Tails is more than 15 percent. TRUE It is 17.62 percent

Chapter 67. The number of defective pencils in a lot of 1000 is an example of a continuous random variable. FALSE It is a result of counting- so discrete.

8. For a continuous distribution, P(X ≤ 100) = P(X<100). TRUE See my Instructions

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on this.

9. All continuous random variables are normally distributed. FALSE Continuous random variables can be highly skewed and non-normal. A normal random variable is a popular example of a continuous random variable, but a continuous r.v. need not be normal.

10. The mean of a standard normal distribution is always equal to 1. FALSE. Its mean is zero and variance (or std deviation) equal to 1.

11. Even if the sample size is more than 1000, we cannot always use the normal approximation to binomial. TRUE For example if p is .001 then np would be only 1 even if sample size is 1000.

12. For a binomial probability experiment, with n=150 and p=.1, we cannot use the normal approximation to the binomial distribution without continuity correction. FALSE

Multiple Choice (2 points each)

Multiple Choice(Chapter 4) 1. Two mutually exclusive events having positive probabilities are ______________ dependent. A. AlwaysB. SometimesC. Never

They are necessarily dependent because the occurrence of one (seriously) affects the probability of the other (makes it zero). Instructions on Ch 4 page 4

2.  If P(A)>0 and P(B)>0 and events A and B are independent, then: A. P(A)=P(B)B. P((A|B))=P(A)C. P(A B)=0

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D. P(A B)=P(A)/ P(B/A)E. Both A and C are correct

See My Instructions on Ch 4 page 5. Independence does not imply equality of probabilities. So the first choice is clearly wrong. The third choice applies to mutually exclusive events not independent events.. The fourth choice is also incorrect because there should be multiplication on the right hand side not division. So the correct answer is B.

Chapter 5

3. In a study conducted by UCLA, it was found that 25% of college freshmen support increased military spending. If 6 college freshmen are randomly selected, find the probability that: Exactly 3 support increased military spending A. 0.0330B. 0.1318C. 0.7844D.0 .9624(You can use Table on page 855 for n=6 and p=.25) or use computer to get the following:

Binomial distribution : n= 6 p= 0.25

    cumulativeX P(X) probability

00.1779

8 0.17798

10.3559

6 0.53394

20.2966

3 0.83057

30.1318

4 0.96240

40.0329

6 0.99536

50.0043

9 0.99976

60.0002

4 1.000001.0000

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4. A fair die is rolled 10 times. What is the probability that an even number (2, 4 or 6) will occur between 2 and 4 times (inclusive)? 

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A. 0.6123B. 0.1709C. 0.1611D. 0.3662E. 0.3223

Here n=10 and p=.5. We need P(2<=X<=4) or P(X=2)+ P(X=3) +P(X=4). From the Table on page 856 find these probabilities and add and you get the answer. Or you can use computer as follows:

Binomial distribution

10 n

0.5 p

    cumulative

X P(X) probability

0 0.00098 0.00098

1 0.00977 0.010742 0.04395 0.05469

3 0.11719 0.17188

4 0.20508 0.37695

5 0.24609 0.62305

6 0.20508 0.82813

7 0.11719 0.94531

8 0.04395 0.98926

9 0.00977 0.99902

10 0.00098 1.00000

1.00000

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0.043945

0.117188

0.205078

0.36621

5. Which one of the following statements is not a necessary assumption of the binomial distribution? A. Sampling is with replacementB. The experiment consists of n identical trialsC. The probability of success remains constant from trial to trialD. Trials are independent of each otherE. Each trial results in one of two mutually exclusive outcomes

6. If p=.6 and n=10, then the corresponding binomial distribution is A. Right skewedB. Left skewedC. SymmetricD. BimodalIf p > 0.5 it is left skewed for any number of trials. It is symmetric if p = 0.5 and right skewed if p < 0.5.

7. A fair die is rolled 36 times. What is the standard deviation of the even number (2, 4 or 6) outcomes? A. 18B. 6C. 4D . 3

The probability of 2 or 4 or 6 in any trial is 0.5 and there are 36 trials. Therefore,

σ=√(36*.5*.5) = 3 using the formula given in the instructions for binomial distribution.

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8. A fair die is rolled 10 times. What is the probability that an odd number (1, 3 or 5) will occur less than 4 times? A . 0.1719B. 0.1172C. 0.2051D. 0.3770E. 0.0439 Add P(X=0)+P(X=1)+P(X=2)+P(X=3) from table on page 856 or can use excel to find the cumulative probability

9. In the most recent election, 20% of all eligible college students voted. If a random sample of 20 students were surveyed: Find the probability that at least five students voted in the election. A. 0.1746B. 0.2182C . 0.3704D.0.6296

P(X>=5)= 1-P(X<=4)= 1- 0.629648 (using excel’s Binomdist) or using table and adding the probabilities up to 4 and subtracting from 1.

Chapter 6

10. The fill weight of a certain brand of adult cereal is normally distributed with a mean of 910 grams and a standard deviation of 5 grams. If we select one box of cereal at random from this population, what is the probability that it will weigh less than 904 grams?

A. 0.3849B. 0.7698C. 0.2302D. 0.8849E . 0.1151

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normal distribution P(low

er) P(upp

er) z.1151 .8849 -1.20

11. The area under the normal curve between z=0 and z=1 is ________________ the area under the normal curve between z=1 and z=2. A. Equal toB. Less thanC. Greater thanD. A, B or C above dependent on the value of the meanE. A, B or C above dependent on the value of the standard deviation

Remember the height of the curve declines as we move farther from the mean value 0. Therefore, the area under the curve (or probability will decrease for the same interval length as we move farther from the mean.

12. If the wages of workers for a given company are normally distributed with a mean of $15 per hour, then the proportion of the workers earning more than $13 per hour

A. Is greater than the proportion of workers earning less than $18 per hourB. Is less than 50%C. Is less than the proportion of workers earning more than the mean wage $15D. Is greater than the proportion earning less than $13 per hourE. Cannot tell without knowing the standard deviationNote that $13 is to the left of the mean $15 and the distribution is normal (symmetric). Therefore, the area to its right (or proportion) is greater than the area to its left.

13. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue. The historical records of the company show that over the past 8 years 14 percent of the accounts are delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What is the probability that at least 30 accounts will be classified as delinquent? 

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A. 31.86%B. 18.14%C .  81.86%D. 63.72%E. 75.84%

Here nπ = 250*(.14) = 35, n(1-π) = 215 and nπ(1-π) = 30.1 all >=10. So normal approximation without continuity correction can be performed. You can work with X or proportion to get the same answer. Let us work with proportions. The mean of the sample proportion (or its expected value) is 0.14 (the population proportion). The std error is √{π(1-π)/n} = 0.0219 and given value p = 30/250 = 0.12Therefore Z = (.12 - .14)/.0219 = - 0.91 (rounded to two decimal places).P(X>=30) = P(Z>= -0.91) = 1- 0.1814 = 0.8186 from table or MegaStat

Normal distribution

P(lower) P(upp

er) z

.8186 .1814 0.91

.1814 .8186 -0.91

14. Which of the following statements is not a property of the normal probability distribution? A. The normal distribution is symmetricB. Exactly 90% of all possible observed values of the random variable x are within plus or minus two standard deviations of the population meanC. The mean, median and mode are equalD. The area under the normal curve to the right of the mean is equal to the area under the normal curve to the left of the meanE. All of the above answers are properties of the normal distributionChoice B is selected because 95.44% corresponds to two standard deviations not 90%.

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15. According to a hospital administrator, historical records over the past 10 years have shown that 20% of the major surgery patients are dissatisfied with after-surgery care in the hospital. A scientific poll based on 400 hospital patients has just been conducted. What is the probability that at least 70 patients will not be satisfied with the after-surgery care?

A . 89.44% B. 39.44%C. 10.56%D. 78.88%E. 84.49%

Here nπ = 400*(.20) = 80, n(1-π) = 320 and nπ(1-π) = 64.1 all >=10. So normal approximation without continuity correction can be performed. You can work with X or proportion to get the same answer. Let us work with proportions. The mean of the sample proportion (or its expected value) is 0.20 (the population proportion). The std error is √{π(1-π)/n} = 0.02 and given value p = 70/400 = 0.175. Therefore Z = (.175 - .20)/.02 = - 1.25P(X>=70) = P(Z>= -1.25) = 1- 0.1056 = 0.8944 from table or MegaStat

Essay Type (each question carries 4 points)

Chapter 4

1.  At a college, 60 percent of the students are female and 30 percent of the students receive a grade of C. About 42 percent of the students are female and not C students. Use this contingency table.

P(lower) P(upper) z

.1056 .8944 -1.25

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Gender\Grade C Not C --- CFemale (F) 0.42 0.60Male (M)

0.30If a C student is selected at random, what is the probability the student is male?

Gender\Grade C Not C --- CFemale (F) 0.18 0.42 0.60Male (M) 0.12 0.28 0.40

0.30 0.70 1.00

P(M/C) = P (M∧C)P(C ) = 0.12

0.30 = 0.40

Some of you just answered 0.12, but that is the joint probability of male and C, not the conditional probability of male given C. What the question is asking is the probability of "male" given “ C”.

2. About 30% of the customers of a store buy Cigars and 35 % of customers buy Beer. About 10% buy Cigars but do no Beer. The contingency table is given below.

Beer No Beer

Cigars 0.10 0.30

No cigar

0.35

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Determine the probability that a customer will buy at least one of these items: either Cigar or Beer (or both).

Answer: First we complete the contingency table as follows:

Beer No Beer Total

Cigars 0.20 0.10 0.30

No cigar 0.15 0.55 0.70

Total 0.35 0.65 1.00

Now, P(C and B) = P(C) + P(B) – P(C and B) = 0.30 + 0.35- .20 = 0.45 which is also equal to 1-P(neither) = 1- 0.55 from the table. So, you can use either rule of addition or rule of complements here.

3.  Four employees who work as drive-through attendees at a local fast food restaurant are being evaluated. As a part of quality improvement initiative and employee evaluation these workers were observed over three days. One of the statistics collected is the proportion of time employee forgets to include a napkin in the bag. Related information is given in the table below.

Worker % of Dinners Packed % of forgetting Napkin

Joe 0.30 0.05

Jan 0.20 0.02

Cheryl 0.15 0.12

Clay 0.35 0.08

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You just purchased a dinner and found that there is no napkin in your bag, what is the probability that Cheryl has prepared your order?  

Answer: First note that the last column in the above table gives conditional probabilities. For example 6% is the probability of forgetting napkin given that Joe packed the dinner or P(No napkin/Joe). In the question we are given that No napkin has occurred and asked to find the probability of Cheryl in light of this result. So here we are asked a reverse conditionality than the one given in the contingency table. According to the Instructions for Chapter 4, this requires Bayesian rule. Therefore,

P(Cheryl/ No napkin) = 0.018/0.063 = 0.2857 or 28.57%

The numerator is:

P(Cheryl and No napkin)=P(Dinner packed by Cheryl)*P(No napkin given that Cheryl packed Dinner) = 0.15*0.12 = 0.018

The denominator is:

P(No napkin)= P(Joe and no napkin)+ P(Jan and No napkin)+ P(Cheryl and No napkin)+ P(Clay and no napkin) = 0.015 + 0.006 + 0.018 + 0.024 = 0.063 as shown in the table below (everything converted to decimals instead of percentage (working with percentages can lead to errors, especially when you have to multiply the probabilities).

WorkerProportion of Dinners Packed

Proportion of forgetting Napkin Col. 2*Col.3

Joe 0.25 0.06 0.015Jan 0.30 0.02 0.006Cheryl 0.15 0.12 0.018Clay 0.30 0.08 0.024

0.063

This formula is also called the Bayesian rule for probability revision based on the results of an experiment. Here the prior probability of Cheryl (which is 15%) has been revised upward to 28.57% (called the revised or posterior probability) after noticing that the dinner had no napkins, because Cheryl is

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the most forgetful one. If the question were for Jan the posterior probability would be lower than the prior probability because she has a very low chance of forgetting napkin (only 2%).

Chapter 5

4. In a study conducted for the State Department of Education, 35% of the teachers who left teaching did so because they were laid off. Assume that we randomly select 15 teachers who have recently left their profession. Find the probability that at least 5 of them were laid off.

We can use Binomial Distribution (because there are only two possible outcomes for each teacher who left their profession (either laid off or not laid off). Here n = 15 and p= .35. Question asks P(X ≥ 5) (remember that “at least 5” means 5 or more). This can be easily found from its complement as = 1- P(X ≤ 4). Either using Excel Binomdist formula or the table at the back of the book P(X ≤ 4) = 0.3520. Therefore, P(X ≥ 5) = 1-.3520 = 0.6480 or 64.8%

Alternatively, using MegaStat:

Binomial distribution

15 n0.35 p

    cumulativeX P(X) probability0 0.00156 0.001561 0.01262 0.014182 0.04756 0.061733 0.11096 0.172704 0.17925 0.351945 0.21234 0.564286 0.19056 0.754847 0.13193 0.886778 0.07104 0.957819 0.02975 0.98756

10 0.00961 0.9971711 0.00235 0.9995212 0.00042 0.9999413 0.00005 1.0000014 0.00000 1.0000015 0.00000 1.00000

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1.00000

There is a small difference in the fifth decimal place compared to book table. You can also use Excel Binomdist(4,15,0.35,1) to get the probability P(X≤ 4) equal to 0.351943.

5. The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 25%. If 12 calculators are selected at random, what is the probability that more than 5 of the calculators will be defective?

Here n =12, p = 0.25 and P(X > 5) = 1- P(X ≤ 5).

Now this can be solved using excel or MegaStat or simply by looking at the binomial Table at the end of the book for n = 12 and p= .25:

Using Excel Binomdist (5, 12, 0.25, 1) we get P(X ≤ 5) = 0.945598 or 0.9456 (rounded). Therefore, P(X < 6) = 1- 0.9456 = 0.0544 or 5.44% chance.

Binomial distribution12 n

0.25 p    cumulativeX P(X) probability0 0.03168 0.031681 0.12671 0.158382 0.23229 0.390683 0.25810 0.648784 0.19358 0.842365 0.10324 0.945606 0.04015 0.985757 0.01147 0.997228 0.00239 0.999619 0.00035 0.99996

10 0.00004 1.0000011 0.00000 1.0000012 0.00000 1.00000

Chapter 6

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6. If x is a binomial random variable where n = 60 and p = .2, find the probability that x is less than or equal to 14 using the normal approximation to the binomial.

Answer: Need continuity correction because np(1-p) = 9.6 < 10. Following my instructions on continuity correction, less than or equal to 14 becomes less than or equal to 14.5 after continuity correction. ). The standard deviation is ((.2)(.8)(60)) = 3.0984 and the mean is np = 12. Therefore,Z = (14.5 – 12)/3.0984 = 2.5/3.0984 = 0.8069 or 0.81 rounded to two decimal places (for the Normal Table).P(Z 0.81) from normal table = 0.7910

Using computer and the four decimal value 0.8069 we get

Normal distribution

P(lower)

P(upper) z

.7901 .2099 0.81

There is a small difference because we have used Z value with four decimal places.

Some of you did not do the continuity correction and lost one point.

7. The weight of a product is normally distributed with a standard deviation of .5 ounces. What should the average weight be if the production manager wants no more than 1% of the products to weigh more than 10 ounces?

Answer: The Z value which has 1% area to the right (or 99% area to the left) is Z = 2.326 using the Excel function “Normsinv” for probability 0.99.

Using this we calculate the required µ which will give the specified probability or area for the specified X equal to 9 ounces. This is obtained by using the relation between X, Z, µ and σ shown in my Instructions (rearranging):P(Z ≥ 2.326) = 0.01μx = X- Zσ = 10 - (2.326)(.5) = 8.8370

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