Acollege admissions officer for an mba program has determined that historically, candidates have undergraduate grade point averages (gpa) that are normally distributed with standard deviation 0.45. a random sample of twenty five applicants from the current year is taken, yielding a sample mean gpa of 2.90. find a 95% convidence interval for the population mean.

Step-by-step explanation:

1) 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)

represent the population standard deviation

n=25 represent the sample size  

We have the following distribution for the random variable:

And by the central theorem we know that the distribution for the sample mean is given by:

2) Confidence interval

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

  (1)

Since the Confidence is 0.95 or 95%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that

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

So on this case the 95% confidence interval would be given by (2.724;3.076)    

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1