A leading agricultural company must maintain strict control over the size, weight, and number of seeds they package for sale to customers. An SRS of 81 one-pound scoops of seeds was collected as part of a Six Sigma quality improvement effort within the company. The number of seeds in each scoop is given in the excel file “seedcount”.

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1. A leading agricultural company must maintain strict control over the size, weight, and number of seeds they package for sale to customers. An SRS of 81 one-pound scoops of seeds was collected as part of a Six Sigma quality improvement effort within the company. The number of seeds in each scoop is given in the excel file “seedcount”.

a) Find the mean, the standard deviation, and the standard error of the mean for this sample.

Mean: 128055/81=1850.925926

Standard deviation: 154.62

Standard error of the mean: standard deviation/ n= 154.62/81=1.9089

b) Calculate the 90% confidence intervals for the mean number of seeds in a one-pound scoop.

90% confidence interval: x±1.645*154.62/9= (1552.335, 1609.516)

c) Calculate the 95% confidence intervals for the mean number of seeds in a one-pound scoop.

95% confidence interval: x±1.96*154.62/9= (1546.736, 1615.115)

d) Do these data provide evidence that the average number of seeds in a one-pound scoop is greater than 1550? Using a significance level of 5%, state your hypotheses, the test statistic, the P-value, and your conclusion.

Hypotheses: H0: μ≤1550 Ha:μ> 1550

Z- test: 1850.92-μ0 / 154.62/9 =1.80

P- value: P (z>1.80)=1-(z<1.80)=1-0.9641=0.0359 <significance level of 5%

Conclusion: At the significance level of 5%, the average number of seeds in a one-pound scoop is greater than 1550.

2. In a city (outside North America) a traffic officer knows from experience that 30% of drivers do not have a driver’s license or have an expired one. This officer stops 10 cars looking to obtain bribes to pay for his lunch.

a) Given that the officer needs at least one bribe to have enough money for his lunch, what is the probability that the officer cannot get money for his lunch (i.e., all the drivers have a legal driver’s license)?

N=10

P(x=0)=10C0 *p0 *(1-p)10-0

=0.710

=0.0282

b) What is the probability that 4 or less than 4 of the drivers do not have a driver’s license or have an expired one?

p=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)

=0.849

c) What are the mean (i.e., expected value) and the standard deviation of the random variable? In one sentence, explain how to interpret the value of the mean.

Mean:10*0.3=3

Standard deviation: 10*0.3*0.710-5 =1.449

Normally, there are three drivers have no license or have an expired one in average 10 drivers.

d) What is the probability that the number of drivers that do not have a driver’s license (or have an expired one) is more than the mean?

P=P(x=4)+….+p(x=10)

=0.3504

3. Let X be a random variable distributed as Normal [mean=m=5, Var(X)=s2=4]. Find the probabilities of the following events:

Text Description automatically generated

Note that |X-5| is the absolute value.

ⅰ:P (X ≤6)= P (6-m / s)= p (Z<1/2)=0.6915

ⅱ: P (X>4)=1-P(X≤4)=1-P (4-m / s)=1- p(z<-0.5)=1-0.3085=0.6915

ⅲ P( ∣X-5∣>1)=1-( p(-1≤x-5≤1))=1-(p(x≤6)-p(x≤4))=1-(0.6915-0.3085)

=1-0.383

=0.617

 

4. Scores on the SAT Math Reasoning test are approximately Normal with mean 515 points and standard deviation 116 points. The scores on the Verbal Reasoning test are also approximately Normal but with mean 502 and standard deviation 112. The correlation between Math Reasoning and Verbal Reasoning scores is about ρ = 0.70. Define the random variable X as Math Reasoning score and the random variable Y as Verbal Reasoning score.

(a) Determine the mean and standard deviation for the combined score (X + Y).

Mean for (X+Y)=515+502=1017

Standard deviation for (X+Y)= √1162+1122+0.7*116*112 =210.21

(b) Given that the Math and Verbal scores are approximately Normal, the combined score will also be approximately Normal. Determine the probability that a randomly chosen student will have a combined score of at least 1250.

 

m= 1017 s= 210.21

Z= 1250-1017/ 210.21= 1.1084

P(x>1250)= P(Z>1.1084)=1- P(z<1.1084)= 0.1338

 

5. When trying to start a new relationship, people want to make a favorable impression. Sometimes they will even stretch the truth a bit when disclosing information about themselves. A study of deception in online dating studied the accuracy of the information given in their online dating profiles by 110 online daters. The study found that 35 of 55 men and 40 of 55 women lied about their weight. Construct a two-way table. The data is in the Excel file “Assig 1 Q5”.

Applying the chi-square analysis, test whether the variables “Gender” and “Lying” are independent. State null and alternative hypotheses, report the test statistic, the P-value and summarize the results of your hypothesis test.

 

6. It is common for governments to gather data on family structure in the population. Information on family structure distribution is important because of the known associations between family structure and child well-being, behavior, and general success in life. Additionally, information on family structure in the population has business implications for such things as day care for employees and location of work (for example, from home or at organization). Here are data from 2006 on Canadian families with at least one child:

http://ebooks.bfwpub.com/psbe3e/tables/5_T_UN_10.gif

(a)  What is the probability that a randomly chosen child belongs to a “Father-only family” and is the only child at home?

(b) If you know that the family chosen is a “common-law couple”, what is the probability that the family has three or more children?

(c) If you know that the family chosen is a “married couple”, what is the probability that the family has three or more children?

(d) Are the events “Father-only family” and “one child at home” independent? How do you know?

 

7. Use this information to answer the following 3 questions. Suppose X~U[0,4] (that is, X is a continuous random variable that follows the Uniform probability distribution taking values between 0 and 4). The probability density function graph is given below:

1/4

 

 

 

 

0 4

Find the probabilities of the following events:

a) P(< 0.5)

b) P(X > 1.5)

c) P(2 < X 3)

 

8. Problems 1 in the Appendix C of your textbook (page 744)

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