A study conducted by a commuter train transportation authority involved surveying a random sample of 200 passengers. The results show that a customer had to wait on the average 9.3 minutes with a standard deviation of 6.2 minutes to buy his or her ticket. Find the upper confidence limit (UCL) for a 95% confidence interval for population mean , the true mean waiting time. Round your answer to two decimal places.

The UCL of the 95% confidence interval for population mean is 10.16 minutes

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The upper end of the interval is the sample mean added to M. So it is 9.3 + 0.86 = 10.16 minutes.

The UCL of the 95% confidence interval for population mean is 10.16 minutes

da freeck kinda math is this just drop out

step-by-step explanation:

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation