Transcript z – Scores and Probability







A standardized value
A number of standard deviations a given
value, x, is above or below the mean
z = (score (x) – mean)/s (standard deviation)
A positive z-score means the value lies above
the mean
A negative z-score means the value lies below
the mean
Round to 2 decimals




The mean for IQ scores is 50 with a standard
deviation of 5 with a normal distribution
What is the probability that scores will be
between 45 and 55?
Calculate z-score first
Use the Normal Distribution (z) statistical table





Score (x) = 45
Mean = 50
s=5
z = (score (x) – mean)/s (standard deviation)
z = (45 - 50)/5 (standard deviation)
o
o
= -5/5, = -1
Wait! There’s a second value to consider.





Score (x) = 55
Mean = 50
s=5
z = (score (x) – mean)/s (standard deviation)
z = (55 - 50)/5 (standard deviation)
o
= 5/5, = 1


Using the normal distribution (z) statistical
table:
Determine the area from the mean:
1 s up (mean to z) = .3413
 1 s down (mean to z) = .3413
 Add the 2 values together

 .3413 + .3413 = .6826 * 100% = 68.26%

So, the probability that a score will be between
45 and 55 is 68.26%!




The mean for IQ scores is 50 with a standard
deviation of 5 with a normal distribution
What is the probability that an IQ score will be
between 55 and 60?
Calculate z-score first
Use the Normal Distribution (z) statistical table





Score (x) = 55
Mean = 50
s=5
z = (score (x) – mean)/s (standard deviation)
z = (55 - 50)/5 (standard deviation)
o
o
= 5/5, = 1
Wait! There’s a second value to consider.





Score (x) = 60
Mean = 50
s=5
z = (score (x) – mean)/s (standard deviation)
z = (60 - 50)/5 (standard deviation)
o
= 10/5, = 2


Using the normal distribution (z) statistical
table:
Determine the area from the mean:
1 s up (mean to z) = .3413
 2 s up (mean to z) = .4772
 Subtract the 2 values (because we only want the
distance from 1z to 2z, not the mean to 2z)

 .4772 - .3413 = .1359 * 100% = 13.59%

So, the probability that an IQ score will be
between 55 and 60 is 13.59%!




The mean for IQ scores is 100 with a standard
deviation of 15 with a normal distribution
What percentage of scores will lie below 100?
Calculate z-score first
Use the Normal Distribution (z) statistical table





Score (x) = 100
Mean = 100
s = 15
z = (score (x) – mean)/s (standard deviation)
z = (100 - 100)/15 (standard deviation)
o
= 0/15 = 0


Using the normal distribution (z) statistical
table:
Determine the area from the mean:


0 s down (larger portion) = .5000
So, the probability that a student’s IQ score will
be below 100 is 50%.




The mean for IQ scores is 100 with a standard
deviation of 15 with a normal distribution
What percentage of scores will lie below 115?
Calculate z-score first
Use the Normal Distribution (z) statistical table





Score (x) = 115
Mean = 100
s = 15
z = (score (x) – mean)/s (standard deviation)
z = (115 - 100)/15 (standard deviation)
o
= 15/15 = 1


Using the normal distribution (z) statistical
table:
Determine the area from the mean:


1 s up (larger portion) = .8413
So, the percentage of scores that will be below
115 is 84.13%




The mean for IQ scores is 100 with a standard
deviation of 15 with a normal distribution
What percentage of scores will lie above 115?
Calculate z-score first
Use the Normal Distribution (z) statistical table





Score (x) = 115
Mean = 100
s = 15
z = (score (x) – mean)/s (standard deviation)
z = (115 - 100)/15 (standard deviation)
o
= 15/15 = 1


Using the normal distribution (z) statistical
table:
Determine the area from the mean:



1 s up (smaller portion) = .1587
So, the probability that a student’s IQ score will
be above 115 is 15.87%
Note: this area of 15.87% plus the area of scores
below 115, 84.13%, equal 100%.