An evaluator wishes to make a statement about the emotional maturity of the freshmen population, so she decides to sample the population and administer an emotional maturity test. Putting aside any concerns about the validity of the test used, and considering the sample techniques only, how many students should she sample in order to be 95% confident that her estimate of freshman emotional maturity will be within 6 units of the true mean? (The test publishers indicate that the population variance is 100 units.)

So the answer for this case would be n=11 rounded up to the nearest integer

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".  

Solution to the problem

Teh population variance is given by and the deviation is

The margin of error is given by this formula:

   (a)

And on this case we have that ME =6 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got , replacing into formula (5) we got:

So the answer for this case would be n=11 rounded up to the nearest integer

answer: need picture of problem

step-by-step explanation:

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option