Download What`s a Z-Score

What’s a Z-Score?
Z-scores are standardized values based on the random variable’s value, population mean and standard
deviation X ~ N(, ). We are assuming the random variable has a normal distribution so the
distribution of the z-scores will also be normally distributed with a population mean of zero and
standard deviation of 1, Z ~ N(0, 1).
The formula for relating the four unknowns, z, x, , and  is
z
x

.
Assuming X ~ N(66, 2), use the formula to calculate the corresponding z-scores for the x-values of 60, 62,
64, 66, 68, 70, and 72.
If we don’t know the values of x, but we know X ~ N(40, 4), then we can calculate the missing
corresponding x-values when the z-score is –3, -2, -1, 0, 1, 2, and 3. Remember Z ~N(0, 1).
For the following problems, first sketch and label a normal distribution with all the information provided.
x
Then write the formula z 
. Substitute values of z, x, , and  that are given and solve for the one

value not given.
1. Suppose the average height of freshmen at OHS is 60 inches with a standard deviation of 1.5 inches.
What is the z-score for a freshman who has a height of
a) 58 inches?
b) 60.15 inches?
c) 56.25 inches?
d) 70 inches?
2. Suppose the average height of sophomores at OHS is 62 inches with a standard deviation of 2
inches. What is the height of the sophomore (x-value) that corresponds to a
a) z-score = 0?
b) z-score = -2.44?
c) z-score = 1.76?
d) z-score = 3.1?
3. Suppose the average height of juniors at OHS is unknown but the standard deviation is 2.5 inches.
What is the population mean height of juniors if
a) a junior 66 inches tall corresponds to a z-score of -.75?
b) a junior 71 inches tall corresponds to a z-score of 1.55? (This resulting population mean
should be different from the answer to a.)
4. Suppose the height of seniors at OHS is 67 inches but the standard deviation is unknown. What is
the standard deviation knowing
a) a senior 68.5 inches tall has a corresponding z-score of .87?
b) a senior 63 inches tall has a corresponding z-score of –2.43? (This resulting population
standard deviation should be different from the answer to a.)
Remember that there are four unknowns (z, x, , and ) in the formula z 
x

. You will be given three
of the four values to calculate the last unknown.
Why do we calculate z-scores?
The z-score “marks” a spot to shade to the left or right under the normal curve Z ~ N(0, 1). The shaded
area represents the likelihood of a sample statistic x to have been randomly drawn from a population
distribution X~N(, ).
Example A: Suppose teachers at OHS have an age distribution X ~ N(40, 8). What is the likelihood that a
randomly selected teacher from this population would have an age of 25 or younger?
z
x

z 
25  40
 z  1.875
8
P(z  -1.875) = 0.03036
So how likely is this?


 





Example B: Assuming the same distribution exists for age of teachers at OHS, how likely is it for a
randomly selected teacher from OHS to be older than 50 years of age?
z
x

z 
50  40
 z  1.25
8
P(z > 1.25) = 0.10562
Again, how likely is this?


 





Example C: Again using the same teacher age distribution at OHS, what is the probability that a randomly
selected teacher’s age would fall somewhere between 30 and 50 years of age?
We know the z-score of 1.25 corresponds to teacher’s age of 50 as calculated in example B.
x
30  40
z 
z = -1.25
z

8


 





P(-1.25 < z < 1.25) = 0.78870


 




 

 





P(z < 1.25) – P(z < -1.25) = 0.89432 – 0.10562 = 0.78870
Let’s try a few problems.
Suppose the average teenage romantic relationship is normally distributed with a mean number of 100 days
with a standard deviation of 30 days.
5. What is the likelihood that a randomly selected couple’s relationship has lasted for 90 days or
fewer?
6. What is the probability that a randomly selected teenage bliss goes on less than 14 days?
7. How likely is it to randomly select a couple to find their admiration (relationship) has lasted 180
days or more?
8. What is the probability that a randomly selected love relationship has lasted more than one year
(hint 365 days)?
9. What is the likelihood that a randomly selected couple’s relationship survived between 90 and 150
days?
10. How likely is it for teenage bliss to endure between 30 and 60 days?
Finding a z-score when we know probability
Example D: What is the z-score if the area shaded to the left under the normal curve with population mean
of 0 and standard deviation of 1 is 0.25?
As we can see the z-score falls between –1 and 0, we can use TI-83+ to
calculate the z-score for us. The function key we need is the invNorm
command. This can be found by using the catalog or distribution
commands. To use the invNorm enter the area shaded to the left in the
normal curve, then a comma, the population mean (in this case, 0), a
comma, and the standard deviation (in this case, 1).


 





Example E: What is the z-score if the area under the curve to the right is 84%, knowing the normal
distribution has a mean of 0 and standard deviation of 1?
If the shaded area to the right represents 84%, then the non-shaded area to
the left must represent the remaining
16% of the normal curve. The z-score
separating the two sections can be found
easier by using the area to the left with
the technology of the TI-83+ calculator.
Remember, the first number entered
must represent the area to the left
    



 endpoint of the distribution.
Finding an x-value when we know probability
Example F: What is the value of a randomly selected value from a normal distribution with a population
mean of 50 and standard deviation of 5, if the area shaded to the left represents 40% of the curve?
As we can view in the graph, the xvalue falls closer to 50 than 45. Using
the TI-83+, the invNorm command
calculates the x-value from the amount
of area shaded to the left, the
population mean, and the standard
deviation. Therefore the x-value is
approximately 48.733.
Example G: What is the age at which 75% of more of the teachers at OHS are as old or older knowing the
normal distribution of teacher ages has a mean of 40 years and standard deviation of 8 years.
The area shaded in the normal
distribution represents the 75% of
the teachers we could randomly
select from so that their age would
be as old or older than the cut-off xvalue. That x-value falls between
32 and 40 as seen in the graph.
Therefore, approximately 75% of
the teachers are 34.6 years or older at OHS.