For each of the following z-score locations in a normal distribution, determine whether the tail is on the left side or the right side of the distribution and find/give the proportion that is located in the tail (There will be two answers for each question and drawn a distrubtion.

a. z = +1.75
b. z = +0.80
c. z = –0.70

a) For this case we have , and this value is higher than 0, so then would be on the right tail. And we can find the probability in the tail like this:

b) For this case we have , and this value is higher than 0, so then would be on the right tail. And we can find the probability in the tail like this:

c) For this case we have , and this value is lower than 0, so then would be on the left tail. And we can find the probability in the tail like this:

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

For this case we have , and this value is higher than 0, so then would be on the right tail. And we can find the probability in the tail like this:

Part b

For this case we have , and this value is higher than 0, so then would be on the right tail. And we can find the probability in the tail like this:

Part c

For this case we have , and this value is lower than 0, so then would be on the left tail. And we can find the probability in the tail like this:

The results are on the figure attached for this case.

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt