So if Emily had y baseball cards and distributes them evenly among 4 cousins and herself, then you would start off with y / 5 (don't forget herself) and then she gives 6 of them to her brother the final expression would be
(y / 5) - 6
a. Find the probability that an individual distance is
greater than 214.30 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (214.30 – 205) / 8.3
z = 1.12
Since we are looking for x > 214.30 cm, we use the
right tailed test to find for P at z = 1.12 from the tables:
P = 0.1314
b. Find the probability that the mean for 20 randomly
selected distances is greater than 202.80 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (202.80 – 205) / 8.3
z = -0.265
Since we are looking for x > 202.80 cm, we use the
right tailed test to find for P at z = -0.265 from the tables:
P = 0.6045
c. Why can the normal distribution be used in part (b),
even though the sample size does not exceed 30?
I believe this is because we are given the population
standard deviation sigma rather than the sample standard deviation. So we can
use the z test.
To answer the problems correctly, you have to follow and pay close attention to whether or not it says "Area" or "Perimeter". For example, for the first problem, area of 63 square units, you would have to create a shape thats sides multiply out to equal 63 units squared. He could draw a square with 9 units by 7 units. This would work because 9x7=63. For perimeter, all four sides must add up to equal 38 units. Feel free to get creative with the shape on this one! If you have any other questions, just ask. Hope I helped some. :)