1. A study is being conducted by a group of insurance brokers to analyze whether or not pet owners who do not have pet insurance would opt for medical treatment for their pet if the cost of the treatment exceeded $5,000. A random stratified sample of 195 pet owners without pet insurance was collected, and out of those sampled, 37 said they would opt for treatment when presented with the choice if the treatment would cost over $5,000 in total. Calculate the margin of error for this result at the 20% level of significance, and construct the corresponding confidence interval for the true proportion of pet owners that do not have pet insurance who would be willing to pay over $5,000 for medical treatment for their pet.

Standard Normal Distribution Table

E==

Round to four decimal places if necessary

Enter 0 if normal approximation cannot be used

Round to four decimal places if necessary

Enter 0 if normal approximation cannot be used

2. The average lifespan of a CleanFreek dishwasher is normally distributed with a mean of 8 years and a standard deviation of 1.7 years. Standard Normal Distribution Table

a. What is the probability that a CleanFreek dishwasher will last longer than 12 years? P(X>12)=(>12)=

b. What is the probability that a CleanFreek dishwasher will last fewer than 8 years? P(X<8)=(<8)=

c. What length warranty should be established on the dishwashers so that no more than 1.7% of the units will need to be replaced under warranty? x==

years

Round to 2 decimal places.

3. The mean annual salary of a sample of 425 office managers is $55,570 with a standard deviation of $9,110. Calculate the margin of error and construct the 95% confidence interval for the true population mean annual salary for office managers.

We may assume that the sample standard deviation s is an accurate approximation of the population standard deviation (i.e., s ), given that the sample size is so large (n > 200).

Standard Normal Distribution Table

E==

Round to the nearest dollar

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Round to the nearest dollar

4. In a growing industry, the mean number of hours of productivity lost by employees per week due to online social media engagement is 9 hours, with a standard deviation of 2.1 hours.

Note: Assume the population data is normally distributed.

Standard Normal Distribution Table

a. What is the probability that an employee will lose more than 11 hours of productivity due to online social media engagement?

P(X>11)=(>11)=

Round to four decimal places if necessary

b. What is the probability that 9 employees will lose more than 10 hours of productivity due to online social media engagement?

P(X>10)=(X>10)=

Round to four decimal places if necessary

5. Waiters at a restaurant chain earn an average of $256 per shift (regular pay + tips) with a standard deviation of $33. For random samples of 37 shifts at the restaurant, within what range of dollar values will their sample mean earnings fall, with 98% probability?

Standard Normal Distribution Table

Range:

to

Round to the nearest cent