Noise levels at 5 airports were measured in decibels yielding the following data: 152,154,139,124,120 Construct the 80% confidence interval for the mean noise level at such locations. Assume the population is approximately normal. Step 1 of 4 : Calculate the sample mean for the given sample data. Round your answer to one decimal place.

So on this case the 80% confidence interval would be given by (127.108;148.492)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

(2)  

(3)  

The mean calculated for this case is

The sample deviation calculated

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.80 or 80%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.1,4)".And we see that

Now we have everything in order to replace into formula (1):

So on this case the 80% confidence interval would be given by (127.108;148.492)    

its d. the first coordintaes are 10,32

step-by-step explanation:

i think the answer would be c