Transcript The Normal Curve and Z Scores

Z Scores and The Standard
Normal Curve
Properties of the Normal
Distribution:
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Theoretical construction
Also called Bell Curve or Gaussian Curve
Perfectly symmetrical normal distribution
The mean (µ) of a distribution is the midpoint of the
curve
The tails of the curve are infinite
Mean of the curve = median = mode
The “area under the curve” is measured in standard
deviations (σ) from the mean (also called Z).
Total area under the curve is an area of 1.00
Properties (cont.)
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Has a mean µ = 0 and standard deviation σ = 1.
General relationships: ±1 σ = about 68.26%
±2 σ = about 95.44%
±3 σ = about 99.72%*
*Also, when z=±1 then p=.68, when z=±2, p=.95, and when z=±3, p=.997
68.26%
95.44%
99.72%
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0
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Z-Scores
 Are a way of determining the position of a single score under the
normal curve.
 Measured in standard deviations relative to the mean of the curve.
 The Z-score can be used to determine an area under the curve
which is known as a probability.
 Formula:
 The formula changes a “raw” score to a standardized score (Z).
Table A can then be used to determine the area (i.e. the
proportion) beyond z, the area between the mean and z, or the
area below z.
Using Table A to Find the Area
Beyond Z
 Table A (back of the book) can be used to find
the area beyond the part of the curve cut off by
the z-score.
 Column C is the probability associated with the
area of the curve beyond z (more extreme).
 Column B is the proportion of the curve
between the z-score and the mean.
 The probability of falling within ±z standard
deviations of the mean is found by doubling
Column B.
Example: Proportion Below the Mean
4.
1.
Z
X 
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15.87 % of
customers
X  20,   30, and  10
2.
20  30
Z
 1
10
3. Use Column C
Going the other way
 Example: Say a graduate school only
admits students with GRE math scores
in the top 10% (mean = 500, standard
deviation = 100). What score does one
need to be admitted?
 Like last time but in reverse: from the
proportion (10%, or .1000), look up Z in
the table (column C): Z = 1.28.
 Solve for x! Z=1.28=(x-500)/100. So, x = 628.
Using Table A (cont.)
 Example: Say you score 130 on an IQ test with mean = 100
and standard deviation = 15 in the general population. What
percentage of the population scores higher than you?
 Z = (130-100)/15 = +2.00. According to the table, the area
beyond (Z = +2.00) is .0228 (column C), and thus the
proportion of the population with scores exceeding this
score, would be .0228. So, the answer is 2.28%.
Using Table A (cont.)
 Example: Say your friend scores 115 on an IQ test with mean =
100 and standard deviation = 15 in the general population.
What percentage of the population scores below him?
 This is answered by Column B: if Z = (115-100)/15 = 1.00,
then the proportion scoring between the mean of 100 and
the score of 115 is .3413. We must add .5 to this to account
for the half of the population below the mean, arriving at
.8413 or 84.13%
Using Table A (cont.)
 Example: Say you score 130 on an IQ test with mean = 100
and standard deviation = 15 in the general population.Say your
friend scores 115.
 What proportion of the population is smarter than him but not
as smart as you?
 Be careful here: you can’t add/subtract Z scores! Instead,
find the difference in proportions: Your score of 130
(Z=2.00) exceeds all but .0228 of the population; his (115, Z
= 1.00), exceeds all but .1587 (use either Column B or C,
but use the same for both values). So: .1587-.0228 = .1359,
or 13.59% of the population.
Research: Rejection Region
 α=.05
 zcritical=+1.96 and zcritical=-1.96
Homework ch. 5
Say the # of candy bars eaten in a month by each student at
this school is normally distributed, with a mean of 15 and a
standard deviation of 3.
1. What percentage of students ate:
a. more than 21?
b. more than 12?
c. between 10 and 20?
d. either less than 10 or more than 20?
2. How many candy bars would one have to eat to be in:
a. The top 5%?
b. The bottom 10%?
c. The middle 50%?
d. At least 2 standard deviations away from the mean?