Answer:
The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the z-score that has a p-value of
.
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.
This means that 
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
Answer:

And we can find this probability like this:

And we can replace the values and we got:

Step-by-step explanation:
For this case we define the random variable Y as: the number of slices of cheesecake left on the platter after the first serving. And we have the following distribution for Y:
Y | 0 1 2 3 4
P(Y) | 0.005 0.114 0.429 0.381 0.071
And we want to find the following probability:

And we can find this probability like this:

And we can replace the values and we got:

63 is only 63% of the number 100.
Answer:
19.209 miles from each other.
Step-by-step explanation:

Using this we get:


