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Canine Crunchies Inc. (CCI) sells large bags of dog food to warehouse clubs. CCI uses an automatic filling process to fill bags. Weights of the filled bags are approximately normally distributed with a mean of 50 kilograms and a standard deviation of 1.25 kilograms.

What is the probability that a filled bag will weigh less than 49.5 kilograms?

Z = (49.5 -50)/ 1.25 = -0.4

P ( X < 49.5) = 0.3446 or 34.46%

Therefore there is a 34.46 percent chance that a filled bag would weigh less than 49.5Kg

What is the probability that a that a sampled filled bag will weigh between 48.5 and 51kilograms

Z score for 48.5 Kg = (48.5 – 50) / 1.25 = -1.2

Z score for 51Kg = (51-50) / 1.25 = 0.8

P (X < 51) – P (X< 48.5) = 0.7881 – 0.1151 = 0.673 or 67.3%

Therefore, there is a 67.3% chance that a randomly sampled a filled bag will weigh between 48.5 and 51 kilograms.

What is the minimum weight a bag of dog food could be and remain in the top 15% of all bags filled?

1-0.15 =.85

Z= 1.04

1.04 = (X -50.0)/1.25

X = 51.3kg therefore the bag must weigh a minimum of 51.3 kilograms.

d. CCI is unable to adjust the mean of the filling process. However, it is able to adjust the standard

deviation of the filling process. What would the standard deviation need to be so that no more than 2% of all filled bags weigh more than 52 kilograms?

1-0.02 =0.98

Z = 2.06

SD = 2/2.06 = 0.9708Kg

The Baily Hill Bicycle Shop sells mountain bikes and offers a maintenance program. The manager has found the average repair bill during the maintenance program’s first year to be $15.30 with a standard deviation of $7.00.

What is the probability a random sample of 40 customers will have a repair cost exceeding $16.00?

P [ 16 – 15.30/7. 40)

1-P(X > 16) = 0.265

Therefore the chance that a a random sample of 40 would have a repair cost of more than 16 is 0.265%

What is the probability the mean repair cost for a sample of 100 customers will be between $15.10 and $15.80?

P (15.1-15.3/7 100)

1-p (Z< 0.28)

P(X < 15.1) = 0.39

P(15.8- 15.3/7. 100).

1-p (Z<0.71)

P(X < 15.8) = 0.761

P(15.1 < X< 15.8) = 0.761 – 0.39 = 0.371

Therefore the probability that the mean repair cost for a sample of 100 will be between 15.1 and 15.8 is 0.371 or 37.1 percent.

The manager has decided to offer a Spring Special. He is aware of the mean and the standard deviation for repair bills last year. Therefore, he has decided to randomly select and repair the first 50 bicycles for $14.000 each. He notes this is not even 1 standard deviation below the mean price to make such repairs. He asks your advice. Is this a risky thing to do? Based upon the probability of a repair bill being $14.00 or less, what would you recommend? Discuss.

P ( Z< 14 - 15.3 / 7 50) = P(Z< - 1.31)

1-P(Z<1.31)

P(X<14)= 0.096

Based on the above calculations the manager will suffer losses and this is very risky, based on this conclusion I would not recommend this course of action.

In a study conducted by American Express, corporate clients were surveyed to determine the extent to which hotel room rates quoted by central reservation systems differ from the rates negotiated by the companies. The study found that the mean overcharge by hotels was $11.35 per night. Suppose a follow-up study was done in which a random sample of 30 corporate hotel bookings was analyzed. Only those cases where an error occurred were included in the study. The following data show the amounts by which the quoted rate differs from the negotiated rate. Positive values indicate an overcharge and negative values indicate an undercharge.

$15.45 $24.81 $6.02 $14.00 $25.60 $8.29

-$17.34 -$5.72 $11.61 $3.48 $18.91 $7.14

$6.64 $12.48 $6.31 -$4.85 $5.72 $12.72

$5.23 $4.57 $15.84 $2.09 -$4.56 $3.00

$23.60 $30.86 $9.25 $0.93 $20.73 $12.45

Compute a 95% confidence interval estimate for the mean error in hotel charges. Interpret the confidence interval estimate.

( 9.17 + (2.04 ) 10.4143/ 30)

= (5.28, 13.04)

Based on this information we can conclude that at a 95% confidante level the population mean of the error in hotel charges will lie between $5.28 and $13.04.

Based on the interval computed in apart a, do these sample data tend to support the results of the America Express study? Explain.

Yes, we can conclude based on the information gathered in part A from computing the interval estimate the results of the study conducted are supported by the findings in part a as $11.35 falls with in the range of the interval.

The director of state agency believes that the average starting salary for clerical employees in the state is less than $30,000 per year. To test her hypothesis, she has collected a simple random sample of 100 starting clerical salaries from across the state and found that the sample mean is $29,750.

State the appropriate null and alternative hypotheses.

HO => 30000

Ha = <30000

Assuming the population standard deviation is known to be $2,500 and the significance level for the test is to be 0.05, what is the critical value

(stated in dollars)?

Critical value = – 1.645

Critical value(dollars)= 30000 + (-1.645)[2500/ 100]

= 30000 + (- 4112.5/10)

= 30000 – 411.25

= $29588.75

Referring to your answer in part b, what conclusion should be reached with respect to the null hypothesis?

Because X is more than Xa we fail to reject Ho. As such, there is not enough evidence to conclude that the average starting salary is less than $30000.

Referring to your answer in part c, which of the two statistical errors might have been made in this case? Explain.

Based on the results in the previous question that we can not reject Ho, we would classify this as a type 2 error.

A recent article in The Wall Street Journal entitled” As Identity Theft Moves Online, Crime Rings Mimic Big Business” states that 39% of the consumer scam complaints by American consumers are about identity theft. Suppose a random sample of 90 complaints is obtained. Of these complaints, 40 were regarding identity theft. Based on these sample data, what conclusion should be reached about the statement made in the Wall Street Journal? (Test using a = 0.10).

HO P= 0.39

Ha P= 0.39

N= 90 P= .44

Rejection zone = z +1.645

Test statistic = 0.44- 0.39/ 0.0514 = 0.97

Conclusion: due to the fact that the test statistic does not lie within the rejection zone we can not reject Ho and the evidence supports the statements made by the Wall Street Journal.

Quantitative Grading Rubric (10%)

Criteria Excellent

80-100% Good

50-79% Requires Improvement

<50% Points

Use and Interpretation of formulas, graphs, tables Uses and develops appropriate model and interprets quantitative information with clearly defined variables, units or representations Uses and develops appropriate model and interprets quantitative information Uses and develops appropriate model and interprets quantitative information with few or many errors /2

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