We intend to estimate the average driving time of Chicago commuters. From a previous study, we believe that the average time is 42 minutes with a standard deviation of 6 minutes. We want our 99 percent confidence interval to have a margin of error of no more than plus or minus 2 minutes. What is the smallest sample sizethat we should consider?

So the answer for this case would be n=60 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".

represent the sample mean

population mean (variable of interest)

represent the estimated population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

   (a)

And on this case we have that ME =2 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 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=60 rounded up to the nearest integer

answer: i think it's a

step-by-step explanation:

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3