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Slide 1
Lecture 6 Bell Shaped Curves
Slide 2
Thought Question 1: The heights of adult women in the United
States follow, at least approximately, a bell-shaped curve. What do
you think this means?
Slide 3
Thought Question 2: What does it mean to say that a mans weight
is in the 30 th percentile for all adult males?
Slide 4
Thought Question 3: A standardized score is simply the number
of standard deviations an individual falls above or below the mean
for the whole group. Male heights have a mean of 70 inches and a
standard deviation of 3 inches. Female heights have a mean of 65
inches and a standard deviation of 2 inches. Thus, a man who is 73
inches tall has a standardized score of 1. What is the standardized
score corresponding to your own height?
Slide 5
Thought Question 4: Data sets consisting of physical
measurements (heights, weights, lengths of bones, and so on) for
adults of the same species and sex tend to follow a similar
pattern. The pattern is that most individuals are clumped around
the average, with numbers decreasing the farther values are from
the average in either direction. Describe what shape a histogram of
such measurements would have.
Slide 6
8.1Populations, Frequency Curves, and Proportions Move from
pictures and shapes of a set of data to Pictures and shapes for
populations of measurements.
Slide 7
Note: Height of curve set so area under entire curve is 1.
Frequency Curves Smoothed-out histogram by connecting tops of
rectangles with smooth curve. Frequency curve for population of
British male heights. The measurements follow a normal distribution
(or a bell-shaped or Gaussian curve).
Slide 8
Frequency Curves Not all frequency curves are bell-shaped!
Frequency curve for population of dollar amounts of car insurance
damage claims. The measurements follow a right skewed distribution.
Majority of claims were below $5,000, but there were occasionally a
few extremely high claims.
Slide 9
Proportions Recall: Total area under frequency curve = 1 for
100% Mean British Height is 68.25 inches. Area to the right of the
mean is 0.50. So about half of all British men are 68.25 inches or
taller. Key: Proportion of population of measurements falling in a
certain range = area under curve over that range. Tables will
provide other areas under normal curves.
Slide 10
8.2The Pervasiveness of Normal Curves Many populations of
measurements follow approximately a normal curve: Physical
measurements within a homogeneous population heights of male
adults. Standard academic tests given to a large group SAT
scores.
Slide 11
Normal Distribution Probability Probability is area under
curve!
Slide 12
The height of a normal density curve at any point x is given by
is the mean is the standard deviation Normal Distribution
Slide 13
Importance of Normal Distribution 1.Describes Many Random
Processes or Continuous Phenomena 2.Basis for Classical Statistical
Inference
Slide 14
Examples with approximate Normal distributions Height Weight IQ
scores Standardized test scores Body temperature Repeated
measurement of same quantity These distributions which are like
generalised relative frequency histograms can take many different
shapes, some symmetrical some skewed. There is one shape however
that crops up all through the natural world and that is
Slide 15
The Normal Distribution is Symmetric. There are many different
Normal curves, some are fat some are thin. Some are centred at 0
some at 1 some at 5 etc. Each normal curve can be uniquely
identified by two parameters. The Mean and the Standard Deviation
Once you know the mean and the S.Deviation for a Normal curve then
it is possible to draw the curve. Normal curves are centred at the
Mean. And the Standard Deviation describes how spread out they
are.
Slide 16
A Normal Frequency Curve for the Population of SAT scores
Slide 17
The area under a Normal curve to the left of the mean is.5.
This indicates that the probability that something which is
normally distributed is less than its mean is.5. The area under the
curve to the left of any point A on the X axis represents the
probability that a Normal variable is less than A.
Slide 18
Slide 19
8.3Percentiles and Standardized Scores Your percentile = the
percentage of the population that falls below you. Finding
percentiles for normal curves requires: Your own value. The mean
for the population of values. The standard deviation for the
population. Then any bell curve can be standardized so one table
can be used to find percentiles.
Slide 20
Percentiles Example: Have you ever wondered what percentage of
the population (of your gender) is taller than you are? Your
percentile in a population represents the position of your
measurement in comparison with everyone elses. It gives the
percentage of the population that fall below you. For example, if
you are in the 98 th percentile, it means that 98% of the
population falls below you and only 2% is above you. Your
percentile value is easy to find if the population of values has an
approximate bell shape. Although there are an unlimited number of
potential bell-shaped curves, each one can be completely determined
once you know the mean and standard deviation of the population. In
addition, each curve can be standardized in a way such that the
same table can be used to find percentiles for any of them.
Slide 21
Infinite Number of Tables Normal distributions differ by mean
& standard deviation. Each distribution would require its own
table. Thats an infinite number!
Slide 22
Standardize the Normal Distribution One table! Normal
Distribution Standardized Normal Distribution
Slide 23
The standardized score is often called the z-score. Once you
know the z-score for an observed value, you can easily find the
percentile corresponding to the observed value by using the table
that gives the percentiles for a normal distribution with mean 0
and standard deviation 1. A normal curve with a mean of 0 and a
standard deviation of 1 is called a standard normal curve. It is
the curve that results when any normal curve is converted to
standardized scores.
Slide 24
Standardized Scores Standardized Score (standard score or
z-score): observed value mean standard deviation IQ scores have a
normal distribution with a mean of 100 and a standard deviation of
16. Suppose your IQ score was 116. Standardized score = (116
100)/16 = +1 Your IQ is 1 standard deviation above the mean.
Suppose your IQ score was 84. Standardized score = (84 100)/16 = 1
Your IQ is 1 standard deviation below the mean. A normal curve with
mean = 0 and standard deviation = 1 is called a standard normal
curve.
Slide 25
Table 8.1: Proportions and Percentiles for Standard Normal
Scores
Slide 26
Finding a Percentile from an observed value: 1.Find the
standardized score = (observed value mean)/s.d., where s.d. =
standard deviation. Dont forget to keep the plus or minus sign.
2.Look up the percentile in Table 8.1. Suppose your IQ score was
116. Standardized score = (116 100)/16 = +1 Your IQ is 1 standard
deviation above the mean. From Table 8.1 you would be at the 84 th
percentile. Your IQ would be higher than that of 84% of the
population.
Slide 27
Finding an Observed Value from a Percentile: 1.Look up the
percentile in Table 8.1 and find the corresponding standardized
score. 2.Compute observed value = mean +(standardized score)(s.d.),
where s.d. = standard deviation. Jury urges mercy for mother who
killed baby. The mother had an IQ lower than 98 percent of the
population. (Scotsman, March 8, 1994,p. 2) Mother was in the 2 nd
percentile. Table 8.1 gives her standardized score = 2.05, or 2.05
standard deviations below the mean of 100. Her IQ = 100 +
(2.05)(16) = 100 32.8 = 67.2 or about 67. Example 1: Tragically Low
IQ
Slide 28
The Standard Normal Table: 8.1 Table 8.1 is a table of areas
under the standard normal density curve. The table entry for each
value z is the area under the curve to the left of z.
Slide 29
The Standard Normal Table: Table A Table 8.1 can be used to
find the proportion of observations of a variable which fall to the
left of a specific value z if the variable follows a normal
distribution.
Slide 30
Example 2: Calibrating Your GRE Score GRE Exams between 10/1/89
and 9/30/92 had mean verbal score of 497 and a standard deviation
of 115. (ETS, 1993) Suppose your score was 650 and scores were
bell-shaped. Standardized score = (650 497)/115 = +1.33. Table 8.1,
z = 1.33 is between the 90 th and 91 st percentile. Your score was
higher than about 90% of the population.
Slide 31
Example 3: Removing Moles Company Molegon: remove unwanted
moles from gardens. Standardized score = (68 150)/56 = 1.46, and
Standardized score = (211 150)/56 = +1.09. Table 8.1: 86% weigh 211
or less; 7% weigh 68 or less. About 86% 7% = 79% are within the
legal limits. Weights of moles are approximately normal with a mean
of 150 grams and a standard deviation of 56 grams. Only moles
between 68 and 211 grams can be legally caught.
Slide 32
Standardizing Example Normal Distribution Standardized Normal
Distribution
Slide 33
Some Examples Suppose it is know that verbal SAT scores are
normally distributed with a mean of 500 and a standard deviation of
100. First we need to find the standardized score:
Z-score=(observed value-mean)/(standard deviation) =(600-500)/100 =
+1 Find the proportion of the population of SAT scores are less
than or equal to 600. From Table 8.1 we see that a z-score of +1 is
the 84 th percentile and the proportion of population SAT scores
that are less than or equal to 600 is 0.84.
Slide 34
SAT SCORES
Slide 35
Standardized Scores (Z-Scores)
Slide 36
Estimate the proportion of population SAT scores that are
greater than 400. First, we need to find the standardized score:
z-score=(400-500)/100 = -1 From Table 8.1 we see that 16% of
population values have a z-score less than or equal to -1 (or
equivalently, 16% of population values have an observed score less
than 400. However, we are interested in the proportion of the
population with scores GREATER than 400. proportion ABOVE 400 = 1 -
proportion BELOW 400 = 1 0.16 = 0.84
Slide 37
Slide 38
Estimate the proportion of population SAT scores that are
between 400 and 600. An observed value of 400 has a z-score of -1
and represents the 16 th percentile (proportion below z = -1 is
0.16). An observed value of 600 has a z-score of +1 and represents
the 84 th percentile (proportion below z = +1 is 0.84). Lets draw a
picture.
Slide 39
So the proportion with scores between 400 and 600 =Proportion
below 600 Proportion below 400 = 0.84 - 0.16 = 0.68
Slide 40
Find an SAT score such that 70% of the population had SAT
scores less than or equal to this number (i.e., estimate the 70th
percentile of the population). First we need to find the z-score
that corresponds to the 70 th percentile. From Table 8.1 we see
that this z-score is +0.52. Next we need to find the observed value
(from the z- score): Observed value = mean + (z-score)*(standard
deviation) = 500 + 0.52*100 = 552
Slide 41
8.4z-Scores and Familiar Intervals Empirical Rule For any
normal curve, approximately 68% of the values fall within 1
standard deviation of the mean in either direction 95% of the
values fall within 2 standard deviations of the mean in either
direction 99.7% of the values fall within 3 standard deviations of
the mean in either direction A measurement would be an extreme
outlier if it fell more than 3 s.d. above or below the mean.
Slide 42
The 68-95-99.7 Rule
Slide 43
The Empirical Rule Applet
http://www.stat.sc.edu/~west/applets/empiri
calrule.htmlhttp://www.stat.sc.edu/~west/applets/empiri
calrule.html
Slide 44
Heights of Adult Women 68% of adult women are between 62.5 and
67.5 inches, 95% of adult women are between 60 and 70 inches, 99.7%
of adult women are between 57.5 and 72.5 inches. Since adult women
in U.S. have a mean height of 65 inches with a s.d. of 2.5 inches
and heights are bell-shaped, approximately
Slide 45
For Those Who Like Formulas
Slide 46
Example In Tombstone, Arizona Territory people used Colt.45
revolvers. However people used different ammunition. Wyatt Earp
knew that his brothers and Doc Holliday were the only ones in the
territory who used Colt.45s with Winchester ammunition. The Earp
brothers conducted tests on many different combinations of weapons
and ammunition.They found that dataset of observations produced by
the combination of Colt.45 with Winchester shells showed a Mean
velocity of 936 feet/second and a Standard Deviation of 10
feet/second.
Slide 47
The measurements were taken at a distance of 15 feet from the
gun. When Wyatt examined the body of a cowboy shot in the back in
cold blood he concluded that he was shot at a distance of 15 feet
and that the velocity of the bullet at impact was 1,000
feet/second. The dastardly Ike Clanton claimed that this cowboy was
shot by the Earp brothers or Doc Holliday. Was Wyatt able to clear
his good name using the Empirical Rule?
Slide 48
The distribution of this bullet velocity data should be
approximately bell-shaped. This implies that the empirical rule
should give a good estimation of the percentages of the data within
each interval.
Slide 49
This table quite clearly demonstrates that since the bullet
velocity in the shooting was 1000 ft/sec and since this lies more
than 6 Standard Deviations away from the mean the probability is
extremely high that the Earps were not responsible for this
shooting. This is especially evident from looking at the column
showing percentages from the empirical rule. Practically 100% of
bullet velocities should be between 896 and 976 ft/sec.
Slide 50
Example P(3.8 X 5) Normal Distribution.0478 Standardized Normal
Distribution Shaded area exaggerated
Slide 51
Example P(2.9 X 7.1) Normal Distribution.1664.1664.0832.0832
Standardized Normal Distribution Shaded area exaggerated
Slide 52
Example P(X 8) Normal Distribution Standardized Normal
Distribution.1179.1179.5000.3821.3821 Shaded area exaggerated
Slide 53
Example P(7.1 X 8) Normal Distribution.0832.1179.0347.0347
Standardized Normal Distribution Shaded area exaggerated
Slide 54
Solution* P(2000 X 2400) Normal Distribution.4772.4772
Standardized Normal Distribution