1.A health insurance broker records the monthly extended health and dental insurance premiums for 18 non-smoking white-collar clients between the ages of 31 and 40 (measured in $):

108	57	141	105	63	139	48	85	138
100	139	137	100	58	107	117	114	148

T-Distribution Table

a.Calculate the sample mean and standard deviation.

x=x=

Round to the nearest cent

s=s=

Round to the nearest cent

b.Construct a 95% confidence interval for the mean monthly insurance premium for all non-smoking white-collar clients between the ages of 31 and 40.

< <<<

Round to the nearest cent

Use the Central Limit Theorem to find the probability of the indicated event, assuming that the distribution of the population data is unknown.

The amount of time students spend checking their phones during a lecture period has a mean of 9 minutes and standard deviation of 1.5 minutes. What is the probability that a sample of 30 students will spend an average of at most 9.7 minutes checking their phones during a lecture period?

Standard Normal Distribution Table

P(X<9.7)=P(X<9.7)=

In a growing industry, the mean number of hours of productivity lost by employees per week due to online social media engagement is 8 hours, with a standard deviation of 1.9 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 10 hours of productivity due to online social media engagement?

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

Round to four decimal places if necessary

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

P(X>9)=P(X>9)=

Round to four decimal places if necessary

An e-marketing consultant is hired to determine how long customers spend, on average, on a company's website. If the standard deviation is known from previous studies to be approximately 5.8 minutes, determine the number of customers that needs to be sampled in order to determine the mean length of time spent on the company's website, accurate to within a margin of error of 0.7 minutes, at the 5% level of significance.

Standard Normal Distribution Table

Number of Customers =

Roundupto the next whole number

Using two different confidence levels, a researcher constructs two confidence intervals for a population mean based on a sample of sizen=24.n=24.The standard error of the mean is 5.1031. Assume the population is normally distributed and the population standard deviation is unknown.

The confidence intervals are:

Interval 1:85.342<<106.45885.342<<106.458

Interval 2:83.142<<108.65883.142<<108.658

1.Determine the following:

a.The sample mean:

x=x=

Round to one decimal place

b.The margin of error, ME:

ME for Interval 1 =

ME for Interval 2 =

Round to three decimal places

c.The sample standard deviation:

s=s=

Round to the nearest integer

2.At what confidence level was each confidence interval constructed? Choose the closest answer.

a.What is the confidence level used in Interval 1?

(click to select) 80% 90% 95% 98% 99%

b.What is the confidence level used in Interval 2?

(click to select) 80% 90% 95% 98% 99%

A random sample of 92 LED 50"-TVs have an average retail price of $399, with a standard deviation of $64. Calculate the margin of error and construct a 95% confidence interval for the population mean retail price for LED 50"-TVs.

T-Distribution Table

E=E=

Round to the nearest dollar

< <<<

Round to the nearest dollar