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

there would be 365 different 4 digit numbers

step-by-step explanation:

i belive that it is 365 because i asked google

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph