#### Transcript Chapter 3

```NORMAL
DISTRIBUTION
Chapter 3
DENSITY CURVES
Example: here is a histogram
of vocabulary scores of 947
The smooth curve drawn
over the histogram is a
mathematical model for
the distribution.
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DENSITY CURVES
Example: the areas of the
histogram represent the
proportion of scores in the
observed data that are less
than or equal to 6.0. This
proportion is equal to
0.303.
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DENSITY CURVES
Example: now the area under
the smooth curve to the left of
6.0 is shaded. If the scale is
under the curve is exactly 1,
then this curve is called a
density curve. The proportion
of the area to the left of 6.0 is
now equal to 0.293.
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DENSITY CURVE
A density curve is a curve that
is always on or above the horizontal axis, and
has area exactly 1 underneath it.
A density curve describes the overall pattern of a distribution. The area
under the curve and above any range of values is the proportion of all
observations that fall in that range.
A normal distribution is described by a density curve that is a
continuous, symmetric, bell-shaped distribution of a variable.
A normal distribution curve is the theoretical counterpart to a relative
frequency histogram for a large number of data values with a very small class width.
EXAMPLE: COLLEGE ENROLLMENT OF
FEMALE STUDENTS
PROPERTIES OF A NORMAL DISTRIBUTION
A normal distribution curve is bell-shaped
The mean, median, and mode are equal and are located at the center of the
distribution
A normal distribution curve is unimodal (i.e., it has only one mode)
The curve is symmetric about the mean, which is equivalent to saying that its shape is
the same on both sides of a vertical line passing through the center
The curve is continuous, that is, there are no gaps or holes. For each value of X, there
is a corresponding value of Y
The curve never touches the x axis. Theoretically, no matter how far in either
direction the curve extends, it never meets the x axis – but it gets increasingly
closer
The total area under a normal distribution curve is equal to 1.00, or 100%
The Empirical Rule applies
THE EMPIRICAL RULE
68-95-99.7 RULE
The area under the part of a normal curve that lies within 1 standard deviation of
the mean is approximately 0.68, or 68%; within 2 standard deviations, about 0.95, or
95%; and within 3 standard deviations, about 0.997, or 99.7%.
WHAT SHAPES THE CURVE?
The standard deviation σ, is the natural measure of spread for the Normal
Distribution. Between µ – σ, and µ + σ, (in the center of the curve), the graph
curves downward. The graph curves upward to the left of µ – σ, and to the
right of µ + σ. The points at which the curve changes from curving upward
to curving downward are called the inflection points.
STANDARDIZING THE NORMAL
DISTRIBUTION
A z-score tells us how many standard deviations the original observation falls away
from the mean, and in which direction. Observations larger than the mean are
positive when standardized, and observations smaller than the mean are negative.
If x is an observation from a distribution that has mean μ and standard deviation σ, that is N(μ, σ),
the standardized value of x is
A standardized value is often called a z-score.
The standard normal distribution of the new value is the normal distribution N(0,1)
with mean 0 and standard deviation 1.
AREA UNDER THE NORMAL CURVE
The proportion of observations on a standard normal variable, z can be taken from
the area under the normal curve.
The cumulative proportion for a value x in a distribution is the proportion of
observations in the distribution that are less than or equal to x.
The area for the
cumulative proportion is
usually found from a table
of areas. The entry for
each value x is the area
under the curve to the left
of x.
Example 1 – Find the Area
What proportion of observations on a standard
Normal variable z take values less than 1.47?
Solution: To find the area to the left of 1.47, locate
1.4 in the left-hand column of Table A, then locate
the remaining digit 7 as .07 in the top row. The entry
opposite 1.4 and under .07 is 0.9292.
Table A – Cumulative Probability
Example 2 – Who qualifies for college
sports?
Scores of high school seniors on the SAT follow the Normal distribution with mean μ
= 1026 and standard deviation σ = 209. These seniors qualify for NCAA sports with a
score of 820. What percent of seniors scored at least 820?
Step 1. Draw a picture.
Step 2. Standardize. Call the SAT score x. Subtract the mean, then divide by the standard deviation, to
transform the problem about x into a problem about a standard Normal z:
Step 3. Use the table. The picture shows that we need the cumulative proportion for x = 820. Step 2 says this
is the same as the cumulative proportion for z = − 0.99. The Table A entry for z = − 0.99 says that this
cumulative proportion is 0.1611.The area to the right of − 0.99 is therefore 1 − 0.1611 = 0.8389. So 84% of
high school seniors scored at least 820.
Example 3 – Who qualifies for college
sports?
The NCAA considers a student a “partial qualifier” for Division II athletics if high school
grades are satisfactory and the combined SAT score is at least 720. What proportion of
all students who take the SAT would be partial qualifiers?
Step 1. Draw a picture. Call the SAT score x. The variable x has the N(1026, 209) distribution. What proportion of SAT
scores fall between 720 and 820? Here is the picture:
Step 2. Standardize. Subtract the mean, then divide by the standard deviation, to turn x into a standard Normal z:
Step 3. Use the table.
area between −1.46 and −0.99 = (area left of −0.99) − (area left of −1.46) = 0.1611 − 0.0721 = 0.0890
About 9% of high school seniors would be partial qualifiers.
EXAMPLE 4 - HEALTH AND
NUTRITION EXAMINATION STUDY
OF 1976-1980
How tall must a man be to place in the lower 10% for men aged 18 to
24?
.10
? 70
CHAPTER 3
(height values)
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TABLE A:
STANDARD NORMAL PROBABILITIES
Look up the closest probability
(to .10 here) in the table.
Find the corresponding
standardized score.
The value you seek is that
many standard deviations from
the mean.
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z
.07
.08
.09
1.3
.0853 .0838 .0823
1.2
.1020
.1003
.0985
1.1
.1210
.1190
.1170
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EXAMPLE 4 - HEALTH AND NUTRITION
EXAMINATION STUDY OF 1976-1980
 How tall must a man be to place in the lower 10% for men aged 18
to 24?
.10
? 70
-1.28 = z
CHAPTER 3
(height values)
(standardized values)
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OBSERVED VALUE FOR A
STANDARDIZED SCORE
Need to “unstandardize” the z-score to find the observed value (x) :
z
x
x    z

observed value =
mean plus [(standardized score)  (std dev)]
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OBSERVED VALUE FOR A STANDARDIZED
SCORE
observed value =
mean plus [(standardized score)  (std dev)]
= 70 + [(1.28 )  (2.8)]
= 70 + (3.58) = 66.42
 A man would have to be approximately 66.42 inches tall or less to
place in the lower 10% of all men in the population.
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