578Assignment2 F14 Sol

8/10/2019 578Assignment2 F14 Sol

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Assignments2 solutions: Due by Midnight Monday October 13th, 2014

(drop box of week 2) (Chapters 5, 6, 7 and 8) Total 75 points.

True/False (1 point each)

Chapter 51. If the probability of success is 0.4 and the number of trials in a binomial

distribution is 150, then its variance is 6. FALSE 2= (np(1-p)) =(150*0.4*0.6) =36. But the standard deviation is 6.

2. If a fair coin is tossed 20 times then the probability of less than 10 Tails is less

than 0.4 (less than 40% chance). FALSE It is 41.19 percent

3. The probability that a person catches a cold during the cold and flu season is 0.3.

If 10 people are chosen at random, the standard deviation for the number ofpersons catching cold is 1.45. (Hint: convert the problem to a binomial distribution

problem). TRUEHere p = 0.3 and n=10. Therefore, variance = 10*0.3*0.7 = 2.1

and the standard deviation is sq. root of 2.1 = 1.45

Chapter 6

4. For any distribution, P(X 10) is greater than or equal to P(X < 10). FalseThiscan be true only for a discrete distribution. For a continuous distribution, the two

probabilities are equal.

5. All continuous random variables are normally distributed. FALSE Continuous

random variables can be highly skewed and non-normal. Even if it is symmetrical

it may not be normal but other distribution like t-distribution. A normal random

variable is a popular example of a continuous random variable, but a continuous

r.v. need not be normal.

6. The standard deviation of a standard normal distribution is always equal to

1. True. Its mean is zero and variance (or std deviation) equal to 1.

7. If the sample size is as large as 1000, we can safely use the normal

approximation to binomial even for small p. FALSE(Instructions on Ch6) : For

example if p is .001 then np would be only 1 even if sample size is 1000.


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Chapter 7

8. The standard deviation of the sampling distribution of sample proportions

increases as the sample size increases. FALSE

9. If the population is normally distributed then the sample mean may or may notbe normally distributed for small sample size. FALSE If the population is

normally distributed then the sample mean is also normally distributed even for

small sample size.(Instructions on Ch 7, property 4)

Chapter 8

10. First a confidence interval is constructed without using the finite population

correction factor. Then, for the same identical data, a confidence interval is

constructed using the finite population correction factor. The width of the interval

without the finite population correction factor is wider than the confidence interval

with the finite population correction factor. TRUE

11. When the population is normally distributed and the population standard

deviation is unknown, then for any sample size n, the sampling distribution

of is based on the t distribution. TRUE

12. When the level of confidence and sample standard deviation remain the same, a

confidence interval for a population mean based on a sample of n=100 will be

narrower than a confidence interval for a population mean based on a sample of n=

150. FALSE

13. When the level of confidence and the sample size remain the same, a

confidence interval for a population meanwill be narrower, when the sample

standard deviation s is large than when s is small. False

14. When the level of confidence and sample proportion p remain the same, a

confidence interval for a population proportion p based on a sample of n=100 will

be wider than a confidence interval for p based on a sample of n=400. TRUE

15. The sample mean and the sample proportion are unbiased estimators of the

corresponding population parameters. TRUEBut the Sample Standard Deviation

is not an unbiased estimator.


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Multiple Choice (2 points each)

Chapter 5

1. 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 at least 3 support increased military spending

A. 0.8306

B. 0.1318

C. 0.1694

D.0 .9624

Remember that at least 3 means 3 or more.(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. You can add the probabilities for 3 or more or get it from the complement

rule using cumulative probability for 2 or less: 1- 0.83057

cumulative

X P(X) probability

0 0.17798 0.177981 0.35596 0.533942 0.29663 0.83057

3 0.13184 0.96240

4 0.03296 0.99536

5 0.00439 0.999766 0.00024 1.00000

1.00000

2. A fair die is rolled 10 times. What is the probability that an even number (2, 4or 6) will occur less than 4 times?A. 0.1719B. 0.1172C. 0.6230D. 0.8281

E. 0.9453

Here n =10 and p = 0.5. We need P(X < 4) which is P(X 3)= [P(X= 0) + P(X =1) + P(X = 2) + P(X=3)] = 0.1719. From the Table on page 854 you get theanswer. Or you can use computer as follows:


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Binomial distribution10 n

0.5 p

cumulative

X P(X) probability

0 0.00098 0.000981 0.00977 0.010742 0.04395 0.054693 0.11719 0.171884 0.20508 0.376955 0.24609 0.623056 0.20508 0.828137 0.11719 0.945318 0.04395 0.98926

9 0.00977 0.9990210 0.00098 1.00000

1.00000

3. Which one of the following statements is not a necessary assumption of the

binomial distribution?

A. Trials are independent of each other

B. The experiment consists of n identical trials

C. The probability of success remains constant from trial to trialD. Sampling is with replacement

E. Each trial results in one of two mutually exclusive outcomes

4. If p = 0.55 and n =15, then the corresponding binomial distribution is

A. Right skewed

B. Left skewed

C. Symmetric

D. BimodalIf p = 0.5 it is symmetric for any number of trials. If p is larger than 0.5, then the

distribution is left skewed. If p is smaller than 0.5, then it is right skewed.

5. In the most recent election, 20% of all eligible college students voted. If a

random sample of 20 students were surveyed:


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Find the probability that less than five students voted in the election.

A. 0.1746

B. 0.2182

C. 0.3704

D. 0.6296

P(X < 5)= P(X 4) = 0.629648 (using Excels Binomdist) or using table andadding the probabilities up to 4.

Chapter 6

6. The area under the normal curve between z = 1 and z = 2 is ________________the area under the normal curve between z =2 and z = 3.A. Greater thanB. Less thanC. Equal toD. 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 meanvalue 0. Therefore, the area under the curve (or probability will decrease for thesame interval length as we move farther from the mean.

7. The internal auditing staff of a local manufacturing company performs a sampleaudit each quarter to estimate the proportion of accounts that are delinquent morethan 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 auditingstaff randomly selected 250 customer accounts. What is the probability that at least30 accounts will be classified as delinquent?A. 31.86%B. 18.14%C. 81.86%D. 63.72%E. 75.84%

Here n = 250*(0.14) = 35, n(1-) = 215 and n(1-) = 30.1 all 10. So, normalapproximation without continuity correction can be performed. You can work with

X or proportion to get the same answer. Let us work with X first. Mean n = 35.Variance = n(1-) = 30. Standard deviation (or standard error) = 30 = 5.4863.


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The Z score of 30 = (30-35)/5.4863 = -0.91 (rounded to two decimal places).

Now I will show you that the same result is obtained working with proportions.

The mean of the sample proportion (or its expected value) is 0.14 (the population

proportion). The standard error is {(1-)/n} = 0.0219 and given value p = 30/250= 0.12

Therefore Z = (0.12 - 0.14)/0.0219 = - 0.91 (rounded to two decimal places).

Thus, P(X 30) = P(Z -0.91) = 1- 0.1814 from table or MegaStat

Normal distribution

P(lower) P(upper) z

0.8186 0.1814 0.91

0.1814 0.8186 -0.91

Chapter 7

8. A manufacturing company measures the weight of boxes before shipping them

to the customers. If the box weights have a population mean and standard deviation

of 90 lbs. and 24 lbs. respectively, then based on a sample size of 36 boxes, theprobability that the average weight of the boxes will be less than 94 lbs. is:

A. 34.13%

B. 84.13%

C. 15.87%

D. 56.36%

E. 16.87%

The = 24/36 = 4. Therefore the Z score is (94-90)/4 = 1.0

P(Z


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staff randomly selected 250 customer accounts. What is the probability that more

than 40 accounts will be classified as delinquent?

A. 42.07%

B. 92.07%

C. 90.15%D. 40.15%

E. 7.93%

Here = 0.13, n = 250, and 1- = 0.87. We have n= 32.5 and n(1- ) = 217.5,both greater than 10, and n(1- ) = 28.275 > 10. So, normal approximationwithout continuity correction is appropriate. The standard error of p = p=

= 0.0213

Next, 40 accounts out of 250 in proportion is 40/250 = 0.16. The question

refers to more than 40. Therefore, the question is P(p 0.16)?

Using the standardization process with p= 0.13 (the population proportion)

and p= 0.0213, we have P(Z

) = P(Z 1.41) = 1- 0.9207 = 0.0793

from the table or MegaStat.

Chapter 8

10. The width of a confidence interval will be:

A. Wider for 90% confidence than for 95% confidence

B. Narrower for 95% confidence than for 99% confidence

C. Wider for a sample size of 100 than for a sample size of 50

D. Wider when the sample standard deviation (s) is small than when s is large

11. When the level of confidence and sample size remain the same, a confidence

interval for a population proportion p will be __________ when p(1-p) is smaller

than when p(1-p) is larger.A.Narrower

B. Wider

C. Neither A nor B, they will be the same

D. Cannot tell from the information given


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12. In a manufacturing process a random sample of 9 bolts manufactured has a

mean length of 3 inches with a standard deviation of .3 inches. What is the 95%

confidence interval for the true mean length of the bolt?

A. 2.804 to 3.196

B. 2.308 to 3.692C.2.769 to 3.231

D. 2.412 to 3.588

E. 2.814 to 3.186

Since the population standard deviation is not given (only sample std. dev given)

and the sample is small, t distribution is appropriate. Using MegaStat (you could

use t Table too)

Confidence interval - mean

95% confidence level

3 mean

0.3 std. dev.

9 n

2.306 t (df = 8)

0.231 half-width

3.231 upper confidence limit

2.769 lower confidence limit


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13. The internal auditing staff of a local manufacturing company performs asample audit each quarter to estimate the proportion of accounts that are delinquent(more than 90 days overdue). For this quarter, the auditing staff randomly selected400 customer accounts and found that 80 of these accounts were delinquent. Whatis the 95% confidence interval for the proportion of all delinquent customeraccounts at this manufacturing company?A. .171 to .229B. .199 to .201C. .167 to .233D. .149 to .252E. .161 to .239

Confidence interval -

proportion


0.2 proportion

400 n

1.960 z

0.039 half-width

0.239 upper confidence limit

0.161 lower confidence limit


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14. The internal auditing staff of a local manufacturing company performs a

sample audit each quarter to estimate the proportion of accounts that are current

(between 0 and 60 days after billing). The historical records show that over the past

8 years 70 percent of the accounts have been current. Determine the sample size

needed in order to be 99% confident that the sample proportion of the currentcustomer accounts is within .03 of the true proportion of all current accounts for

this company.

A. 1842

B. 897

C. 1549

D. 632

E. 1267

You can use the formula given in my instructions or book or MegaStat. We alwaysround up the required sample size to the whole number because the formula gives

the minimum.

Sample size - proportion

0.03 E, error tolerance

0.7 estimated population proportion


2.576 z

1548.143 sample size

1549 rounded up

Essay Type (each question carries 4 points)

Chapter 51. In a study conducted for the State Department of Education, 30% of the teachers

who left teaching did so because they were laid off. Assume that we randomly

select 12 teachers who have recently left their profession. Find the probability that

at least 7 of them were laid off.


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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= 12 and p = 0.3. Question is P(X 7) which can be easily foundfrom its complement as= 1-P(X 6). Either using Excel Binomdist formula or

the table at the back of the book P(X 6) = 0.9614. Therefore, P(X 7) = 1-0.9614 = 0.0386 or 3.86%

Alternatively, using MegaStat:

Binomial

distribution

12 n

0.3 p

cumulative

X P(X) probabili ty

0 0.01384 0.01384

1 0.07118 0.08503

2 0.16779 0.25282

3 0.23970 0.49252

4 0.23114 0.72366

5 0.15850 0.88215

6 0.07925 0.96140

7 0.02911 0.99051

8 0.00780 0.998319 0.00149 0.99979

10 0.00019 0.99998

11 0.00001 1.00000

12 0.00000 1.00000

1.00000

3.600 expected value

2.520 variance

1.587

standard

deviation

2. The J.O. Supplies Company buys calculators from a Korean supplier. The

probability of a defective calculator is 20%. If 15 calculators are selected at

random, what is the probability that 5 or more of the calculators will be defective?


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Here n =15, p = 0.20 and P(X >= 5) = 1 - P(X 4). Now this can be solvedusing excel or MegaStat or simply by looking at the binomial Table at the end

of the book for n = 15 and p= .20:

Using Excel Binomdist (4, 15, 0.20, 1) we get P(X 4) = 0.8358. Therefore, P(X>= 5) = 1- 0.8358 = 0.1642

Alternatively, using MegaStat:

Binomial

distribution

15 n

0.2 p

cumulative

X P(X) probabili ty0 0.03518 0.03518

1 0.13194 0.16713

2 0.23090 0.39802

3 0.25014 0.64816

4 0.18760 0.83577

5 0.10318 0.93895

6 0.04299 0.98194

7 0.01382 0.99576

8 0.00345 0.99922

9 0.00067 0.99989

10 0.00010 0.99999

11 0.00001 1.00000

12 0.00000 1.00000

13 0.00000 1.00000

14 0.00000 1.00000

15 0.00000 1.00000

1.00000

3.000 expected value

2.400 variance

1.549

standard

deviation


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Chapter 6

3. If x is a binomial random variable where n = 100 and p = 0.1, find the

probability that x is less than or equal to 12 using the normal approximation to

the binomial.

Answer: Need continuity correction because np(1-p) = 9 < 10. Following myinstructions on continuity correction, less than or equal to12 becomes less

than or equal to12.5 after continuity correction. ). The standard deviation is

((0.1)(0.9)(100)) = 3. Therefore,

Z = (12.510)/((0.1)(0.9)(100)) = 2.5/3 = 0.83P(Z 0.83) = 0.7967

normal distribution

P(lower) P(upper) z

.7967 .2033 0.83

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

4. The weight of a product is normally distributed with a standard deviation of 1.5

ounces. What should the average weight be if the production manager wants no

more than 5% of the products to weigh more than 15 ounces?

Answer: The Z value which has 5% area to the right (or 95% area to the left)

is Z = 1.645. Using this we calculate the required which will give the

specified probability or area for the specified X equal to 15 ounces. This is

obtained by using the relation between X, Z, and as shown in myInstructions (and rearranging):

P(Z 1.645) = 0.05x= X- Z = 15 - (1.645)(1.5) = 12.53

Chapter 7

5. A PGA (Professional Golf Association) tournament organizer is attempting to

determine whether hole (pin) placement has a significant impact on the average

number of strokes for the 13thhole on a given golf course. Historically, the pin has

been placed in the front right corner of the green and the historical mean number of


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strokes for the hole has been 4.25, with a standard deviation of 1.4 strokes. On a

particular day during the most recent golf tournament, the organizer placed the

hole (pin) in the back left corner of the green. 49 golfers played the hole with the

new placement on that day. Determine the probability of the sample average

number of strokes being greater than 4.35.

Here = 4.25, = 1.4, and n= 49. So, = 1.4/49= 0.2. Therefore,

Z = (4.35-4.25)/0.2 = 0.5

P(>= 4.35) = P(Z >= 0.5) = 1 - 0.6915 = 30.85% chance.

6. The population of lengths of aluminum-coated steel sheets is normally

distributed with a mean of 30.05 inches and a standard deviation of 0.32 inches. A

sample of 16 metal sheets is randomly selected from a batch. What is theprobability that the average length of a sheet is between 29.95 and 30.25 inches

long?

The Standard deviation of the sample mean = 0.32/16 = 0.08. Therefore,P{(29.9530.05)/0.08 Z (30.2530.05)/0.08} = P(-1.25 Z 2.5) = 0.9938- 0.1056 = 0.8882 or 88.82%.

Chapter 8

7. A small town has a population of 20,000 people. Among these 1,000 regularly

visit a popular local bar. A sample of 100 people who visit the bar is surveyed for

their annual expenditures in the bar. It is found that on average each person who

regularly visits the bar spends about $2400 per year in the bar with a standard

deviation of $200. Construct a 99 percent confidence interval around the mean

annual expenditure in the bar.

This problem needs finite population correctionbecause the population from

which sample is taken is only 1000 not 20,000(that number was given to trickyou) and is Notat least 20 times the sample of 100. The finite population

correction factor (given in my Instructions) is

(N-n)/(N-1) or 900/999 in this case or 0.901. Its square root is 0.949. So,

=

= (200/10)*0.949 = 18.98; For 99% confidence,

2.576*18.98 = 48.89. Therefore, the interval is 2400 48.89 or between 2351.11


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and 2448.89

8. The production manager for the XYZ manufacturing company is concerned that

the customer orders are being shipped late. He asked one of his planners to check

the timeliness of shipments for the orders. The planner randomly selected 900orders and found that 180 orders were shipped late. Construct the 95% confidence

interval for the proportion of orders shipped late.

(180/900) 1.96 [(0.2*0.8)/900] =0.1739 to 0.2261

578Assignment2 F14 Sol

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