statistic solutions

Sheet1Sayed Mansoor AfzaliID: 12195Prof. Karim SoroushStatistics IIDate: October 20, 2013Assignment 1Question 1.A recent study of the hourly wages of the maintenance crews for major airlines showed that the mean hourly salary was $ 16.5 with a standard deviation of $ 3.5. If we select a crew member at random, what is the probability the crew member earns?a. Between 16.5 and $ 20 per hourb. More than $ 20 per hourc. Less than $ 15 per hourd. Between $ 10 and 20 per hourAnswersa.As for this caseMean16.5SD3.5x (1)16.5x (2)20FormulaValuePercentage00%

134.13%

Hence, there is 34.13% chance that the selected crew members would earn b/w $16.5 and $20.b.As for this caseMean16.5SD3.5xMore than20FormulaValuePercentageRelevant Percentage134.13%50% - 34.13%15.87%Hence, there is 15.87% chance that the selected crew members would earn more than $20.

c.As for this caseMean16.5SD3.5xLess than15FormulaValuePercentageRelevant Percentage-0.4316.64%50% - 16.64%33.36%Hence, there is 33.36% chance that the selected crew members would earn less than $15.

d.As for this caseMean16.5SD3.5x (1)10x (2)20FormulaValuePercentageTotal-1.8646.86%80.99%

134.13%

Hence, there is 80.99% chance that the selected crew members would earn b/w $10 and $20.

Question 2.The mean of normal distribution is 400 pounds and the standard deviation is 10 poundsa. What is the area between 415 and 400 pounds?b. What is the area between the mean and 395 pounds?c. What is the probability of selecting a value at random and discovering that it has a value is less than 395 pounds?

Answersa.As for this caseMean400SD10x (1)400x (2)415FormulaValuePercentage00%

1.543.32%

Hence, the area between 415 and 400 pounds is equal to 43.32%.b.As for this caseMean400SD10x (1)400x (2)395FormulaValuePercentage00%

-0.519.15%

Hence, the area between the mean and 395 pounds is equal to 19.15%.

c.As for this caseMean400SD10xLess than395FormulaValuePercentageRelevant Percentage-0.5019.15%50% - 19.15%30.85%Hence, there is 30.85% probability that the randomly selected value would be less than 395.Question 3A normal distribution has a mean of 50 and standard deviation of 4 a. Compute the probability of value between 44 and 55b. Compute the probability of value greater than 55c. Compute the probability of value between 52 and 55 d. Determine the value of x below which 95 percent of the values will occur.Answersa.As for this caseMean50SD4x (1)44x (2)55

FormulaValuePercentageTotal-1.543.32%82.76%

1.2539.44%

Hence, the probability of value between 44 and 55 is equal to 82.76%.b.As for this caseMean50SD4xGreater than55FormulaValuePercentageRelevant Percentage1.2539.44%50% - 39.44%10.56%Hence, there is 10.56% probability that the value selected would be greater than 55.c.As for this caseMean50SD4x (1)52x (2)55FormulaValuePercentageTotal0.519.15%20.29%

1.2539.44%

Hence, the probability of value between 52 and 55 is equal to 20.29%.d.Mean50SD4Z1.65Formulax56.60

Hence, 95% of all data values will lie below x=56.60.Question 4A normal distribution has a mean of 80 and standard deviation of 14 a. Compute the probability of value between 75 and 90b.Compute the probability of value between 55 and 70 c. Determine the value of X above which 80 percent of the values will occur.Answersa.As for this caseMean80SD14x (1)75x (2)90


0.7126.11%

Hence, the probability of value between 75 and 90 is equal to 40.17%.b.As for this caseMean80SD14x (1)55x (2)70


-0.7126.11%

Hence, the probability of value between 55 and 70 is equal to 20.22%.c.Mean80SD14Z-0.84Formulax68.24

Hence, 80% of all data values will lie above x=72.72.

Question 5The mean expenditure of a family of four is 6120 afs. Assuming a standard deviation of 1500 and normally distributed.

a. What is the probability a single family of four spent over 6000 afs.b. What is the probability 100 families spent over 6000 on the average.Answera.As for this caseMean6120SD1500xMore than6000FormulaValuePercentageRelevant Percentage-0.083.19%50% + 3.19%53.19%Hence, there is 53.19% chance that a single family of four spent over 6000 afs.b.As for this casen100p53.19%q46.81%

FormulaMeanFormulaSD53.194.99

FormulaValuePercentage1.3641.31%

Hence, there is 41.31% chance that 100 familes of four spent over 6000 afs on average.Question 6.A researcher found that a sample of 100 with mean of 50.3 and standard error of 10.1 generated a confidence interval of 48.3204 to 52.2796. What level of confidence can be attributed to this interval?

AnswerAs for this casen100p50.00%q50.00%FormulaMeanFormulaSD505.00

Lower LimitFormulaValuePercentage-0.3413.31%

Upper LimitFormulaValuePercentage0.4617.72%

Total31.03%Hence, the confidence level that can be attributed to this interval is 31.03%.Question 7.A recent study of 100 people in Kabul found 27 were obese. Find the 90% confidence interval of the population proportion of the individuals living in Kabul who are obese.

AnswerAs for this casen100p0.27q0.73Z = 90%1.64FormulaMeanFormulaSD274.44

Lower LimitFormulaValue26.3

Upper LimitFormulaValue27.7

So, with confidence level of 90% the population proportion is between 26.3 and 27.7.Question 8.A survey of 90 families showed that 40 owned at least one car. Find 95 % confidence interval of the true proportion of the families who own at least one car.

AnswerAs for this casen90p0.44q0.56Z = 95%1.96FormulaMeanFormulaSD404.71

Lower LimitFormulaValue39.0


So, with confidence level of 95% the population proportion is between 39 and 41.Question 9.A company wishes to estimate the average time to service a customer.a. Assume that a sample of 100 customers yielded a mean of 37 minutes with a standard error of 5.2 minutes. Construct and interpret the 99 percent of interval.

b. Calculate the 99 percent interval under the same conditions if the sample size is 500.Answera.As for this casen100Mean37.00SD5.20Z = 99%2.58Lower LimitFormulaValue35.7


So, with confidence level of 99% the interval is between 35.7 and 38.3b.As for this casen500Mean37.00SD5.20Z = 99%2.58Lower LimitFormulaValue36.4


So, with confidence level of 99% the interval is between 36.4 and 37.6Question 10.The owner of small business wishes to estimate the average time required to complete a certain job task. He must ensure that he is 90 percent confident that that the error is less than 0.5 minutes. The standard deviation is known to be 3.2 minutes. How many observations of completion times must he take?

AnswerAs for this caseError0.5n?SD3.20Z = 90%1.64nFormulaValue110.2

So, with confidence level of 90% the owner must take 110 observations.Question 11.The dean of private university wants an estimate of the number of the out of state students enrolled. She must be 95 percent confident that the error is less than 3 percent. How large a sample must she take? If the sample reveals a proportion of 31 percent out of stators and there are 12,414 students. How many students do you estimate come from other states.

AnswerAs for this caseError0.03p0.31q0.69Z = 95%1.96nFormulaValue913

The total number of students coming from other states are 913.Question 12A restaurant owner wishes to find the 99% confidence interval of the true mean cost of a day martini. How large should the sample be if she wishes to be accurate within $ 10? A previous study shows that the standard deviation of the price was $ 12.

AnswerAs for this caseError10n?SD12.00Z = 99%2.58nFormulaValue10

So, with confidence level of 99% the owner must take 10 observations.Question 13A health care professional wishes to estimate the birth weights of infants. How large a sample must be selected if she desires to be 90% confident that the true mean is within 6 ounces of the sample mean? The standard deviation of the birth weights is known to be 8 ounces.

AnswerAs for this caseError6n?SD8Z = 90%1.64nFormulaValue5

So, with confidence level of 90% the health care professional must take 5 observations.Question 14The following data represent a random sample of 9 marks (out of 10) on a statistics quiz. The marks are normally distributed with a standard deviation of 2. Estimate the population mean with 90% confidence.

AnswerxZ = 90%1.64SD =2Population Mean =?79Sample Mean7FormulaValue56.89483109Total62Population MeanFormulaLower LimitUpper Limit5.807.98

Hence, with 90% confidence Level the population mean is between the interval 5.80 and 7.98.Question 15.How many rounds of golf do physicians (who play golf) play per year? A survey of 12 physicians revealed the following numbers.

Answerxn=12Z = 90%1.6439Standard Deviation411681FormulaValue1728915.4113310893713691832415225Sample Mean17289FormulaValue1214422.83298415126012748862

Population MeanFormulaLower LimitUpper Limit15.5530.12

Hence, physicians play on average from 15.55 to 30.12 rounds of gold per year

Question 16.Among the most exciting aspect of a university professors life is departmental meeting where such critical issues as the color the walls will be painted and who gets a new desk are decided. A sample of 19 professors was asked how many hours per year are devoted to these meetings .The responses are listed here .Assuming that the variable is approximately normally distributed. Estimate the mean number of hours spent at departmental meetings by all professors. Use a confidence level of 90%.

Answerxn=19Z = 90%14196Standard Deviation17289FormulaValue396.06361728939864416Sample Mean20400FormulaValue152259.6874998100Error CalculationDegree of Freedom525FormulaValueFormulaValue111210.10n - 1181522518324Total Critical Value8641.7344161842438

Population MeanFormulaLower LimitUpper Limit7.2812.09

Hence, with confidence level of 90%, professors spend around 7.28 - 12.09 hours in meetings.Question 17.The number of cars sold annually by used car salespeople is normally distributed with a standard deviation of 15. A random sample of 15 salespeople was taken and the number of cars each sold is listed here. Find the 95% confidence interval estimate of population mean. Interpret the interval estimate.

Answerxn=15Z = 95%796241SD =1543184958336466435610110201633969796241331089Sample Mean583364FormulaValue71504168.6060360010110201745476Error CalculationDegree of Freedom553025FormulaValueFormulaValue8877440.05n - 114102975761Total Critical Value2.145Population MeanFormulaLower LimitUpper Limit60.2976.91

Hence, with confidence level of 95%, car saler sell 60-77 cars annually on average.

Question 18.It is known that the amount of time needed to change the oil on a car is normally distributed the amount of time to complete a random sample of 10 oil changes was recorded and listed here. Compute the 99% confidence interval estimate of the mean of the population.

Answerxn=10Z = 99%11121SD =5.21010016256Sample Mean15225FormulaValue1832416.90121442562520400Error CalculationDegree of Freedom18324FormulaValueFormulaValue245760.01n - 191693095Total Critical Value3.25Population MeanFormulaLower LimitUpper Limit11.6022.20

Hence, with confidence level of 99%, the true mean of the population lies between 11.6 - 22.2.Question 19.Because of different sales ability ,experience and deviation ,the incomes of real estate agents vary considerably. Suppose that in a large city the annual income is normally distributed with a standard deviation of $15,000. A random sample of 16 real estate agents was asked to report their annual income (in $1,000).The responses are listed here .Determine the 99% confidence interval estimate of the mean annual income of all real estate agents in the city .

Answerxn=16Z = 99%654225SD=15.094883657324911112321836889613721502500735329684624Sample Mean806400FormulaValue93864975.6384705611312769411681Error CalculationDegree of Freedom603600FormulaValueFormulaValue7759290.01n - 115121097778Total Critical Value2.947Population MeanFormulaLower LimitUpper Limit64.5786.68

Hence, with confidence level of 99%, the true mean of the population lies b/w 64.57 - 86.68.

Question 20.The operation manager of a plan making cellular telephones has proposed rearranging the production process to be more efficient. She wants to estimate the assemble the telephone using the new arrangement. She believes that the population standard deviation is 15 seconds. How large a sample workers should she take to estimate the mean assembly time to within 2 seconds with 95% confidence?

AnswerSD =15Z =95%Error =2nFormulaValue216

So, with a confidence level of 95%, she must take 216 observations.Question 21.A medical researcher wants to investigate the amount of time it takes for patients headache pain to be relived after taking a new prescription painkiller .She plans to use statistical methods to estimate the mean of population of relief times .She believes that population is normally distributed with a standard deviation of 20 minutes. How large a sample should take to estimate the mean time within 1 minute with 90%confidence?

AnswerSD =20Z =90%Error =1nFormulaValue1076

So, with a confidence level of 90%, the medical researcher should take 1,076 observations.Question 22. The operation manager of a large plant would like to estimate the average amount of time workers take to assemble a new electronic component .After observing a number of workers assembling similar devices, she guesses that the standard deviation is 6 minutes. How large a sample of workers should she take if she wishes to estimate the mean assembly time to within 20 seconds? Assume that the confidence level is to be 99%.

AnswerSD =6Z =99%Error =0.33nFormulaValue2157

So, with a confidence level of 99%, she should take 2,157 observations.Question 23.a. Determine the sample size necessary to estimate a population mean to within 90% confidence given that the population standard deviation is 10.

b. Suppose that the sample mean was calculated as 150. Estimate the population mean with 90% confidence.

Answera.SD =10Z =90%nFormulaValue269

So, with a confidence level of 90%, the sample size should be 269.b.Mean =150Population MeanFormulaLower LimitUpper Limit149151

So, with a confidence level of 90%, the population mean falls between 149 and 151.Question 24.a. A statistics practitioner would like to estimate a population mean within 50 units with 99% confidence and population standard deviation is 250 .what sample size should be used?b. Redo Part a changing the standard deviation to 50.c. Redo Part a using a 95% confidence level.d. Redo Part a wherein we wish to estimate the population mean within 10 units.

Answera.SD =250Z =99%Error =50nFormulaValue166

b.SD =50Z =99%Error =50nFormulaValue7

c.SD =250Z =95%Error =50nFormulaValue96

d.SD =250Z =99%Error =10nFormulaValue4160

Question 25.a. Determine the sample size required to estimate a population mean to within 10 units given that the population standard deviation is 50. A confidence level would be 99% .

b. Repeat Part a changing the standard deviation to 100 c. Redo Part a using a 95% confidence leveld. Repeat Part a wherein we wish to estimate the population mean to within 20 units.

Answera.SD =50Z =99%Error =10nFormulaValue166

b.SD =100Z =99%Error =10nFormulaValue666

c.SD =50Z =95%Error =10nFormulaValue96

d.SD =50Z =99%Error =20nFormulaValue42