MM207 Statistics
Welcome to the Unit 6 Seminar
Wednesday, March 7, 2012
8 to 9 PM ET
The Normal Shape
• This is a histogram
for a distribution of
300 natural births.
The left vertical axis
shows the number of
births for each 4-day
bin*. The right
vertical axis shows
relative frequencies
* A bin is a group or
class.
The Normal Shape
The distribution of the birth data has a fairly
distinctive shape, which is easier to see if we
overlay the histogram
with a smooth curve.
Characteristics of the Normal Curve
• The distribution is single-peaked. Its mode, or most common
birth date, is the due date.
• The distribution is symmetric around its single peak; therefore,
its median and mean are the same as its mode. The median is
the due date because equal numbers of births occur before and
after this date. The mean is also the due date because, for
every birth before the due date, there is a birth the same
number of days after the due date.
• The distribution is spread out in a way that makes it resemble
the shape of a bell, so we call it a “bell-shaped” distribution.
• The total area under the curve is equal to 1.00
• The curve approaches the horizontal axis but never touches it
Variation in Distributions
Both distributions are normal and have the same mean of 75, but
the distribution on the left has a larger standard deviation.
When Can We Expect a Normal
Distribution?
A data set that satisfies the following four criteria is likely to
be normally distributed
1. Most data values are clustered near the mean, giving
the distribution a well-defined single peak.
2. Data values are spread evenly around the mean, making
the distribution symmetric.
3. Larger deviations from the mean become increasingly
rare, producing the tapering tails of the distribution.
4. Individual data values result from a combination of many
different factors, such as genetic and environmental
factors.
An Example of a Normal Distribution
Consider a Consumer Reports survey in which participants
were asked how long they owned their last TV set before
they replaced it. The variable of interest in this survey is
replacement time for television sets.
•Based on the survey, the distribution of replacement times has a mean
of about 8.2 years, which we denote as µ (the Greek letter mu).
•The standard deviation of the distribution is about 1.1 years, which we
denote as σ (the Greek letter sigma).
Television Replacement Distribution
Making the reasonable assumption that the distribution of TV
replacement times is approximately normal, we can picture it
as shown
“mu” = µ = 8.2
“sigma”= σ = 1.1
68-95-99.7 Rule or Empirical Rule
This rule gives guidelines for the percentage of data values
that will lie within 1, 2, and 3 standard deviations of the mean
for any normal distribution.
“mu” = µ = 8.2
“sigma”= σ = 1.1
That is from 7.1 years to 9.3 years
That is from 6 years to 10.4 years
That is from 4.9 years to 11.5 years
Finding a Percentile
On a visit to the doctor’s office, your fourth-grade daughter is told that her
height is 1 standard deviation above the mean for her age and sex. What is
her percentile for height? Assume that heights of fourth-grade girls are
normally distributed.
•
•
•
•
•
Recall that a data value lies in the nth percentile of a distribution if n% of the data
values are less than or equal to it (see Section 4.3).
According to the 68-95-99.7 rule, 68% of the heights are within 1 standard
deviation of the mean.
Therefore, 34% of the heights (half of 68%) are between 0 and 1 standard
deviation above the mean.
We also know that, because the distribution is symmetric, 50% of all heights are
below the mean.
Therefore, 50% + 34% = 84% of all heights are less than 1 standard deviation
above the mean (Figure 5.21). Your daughter is in the 84th percentile for heights
among fourth-grade girls
Finding a Percentile
Interpretation: Find the percentile for 1 standard deviation above
the mean for her age and sex. Assume that heights of fourth-grade
girls are normally distributed.
What is her percentile if she were 1
standard deviation BELOW the mean?
Introduction to Standard Scores
• Remember the Empirical
Rule!!!
• Sample Curve
• μ = 500
• σ = 100
• How many Standard
Deviations away from
the mean is:
•
•
•
•
•
300
800
250
500
650
200
300
400
500
600
700
800
-3
-2
-1
0
+1
+2
+3
Computing Standard Scores
• The number of standard deviations a data value
lies above or below the mean is called its
standard score (or z-score), defined by
data value – mean
standard deviation
z = standard score =
= (x – µ) / σ
• The standard score is positive for data values
above the mean and negative for data values
below the mean.
Getting More Precise
Standard Scores and Percentiles
Once we know the standard score of a data value, the properties of
the normal distribution allow us to find its percentile in the
distribution. This is usually done with a standard score table. (In
eText see “chapter BM” for Back matter to get Appendix A on
pages 446-447)
Example 1
Example 2
A college admissions test is scaled so
that scores have a mean of 500 and a standard
deviation of 100.(You will use StatCrunch, but you
must understand theory.)
Finding Z Scores from Percentiles i.e. (working
backwards)
Example: Given the mean cholesterol level of 178 and the standard
deviation of 41, What cholesterol level corresponds to the 90th
percentile?
The 90th percentile would be on the POSITIVE Z table since it is larger than the 50th
percentile. Right? Go to that table and SCAN the body looking for the value closest to
.9000 (the 90th percentile). Move your fingers back to the left to get the x.y part of the
Z xcore. Move you finger up to see the .0w part of the score. Now add these values to
make the score x.yz. All z scores have 2 digits to the right of the decimal.
So moving to left we get 1.2; moving up we see .08; add these gives
1.2 + .08 = 1.28 as our z score.
Finding Z Scores from Percentiles i.e. (working
backwards)
Example: Given the mean cholesterol level of 178 and the standard
deviation of 41, What cholesterol level corresponds to the 90th
percentile?
Now z = 1.28. Thus, the 90th percentile is about 1.28 standard deviations
above the mean.
Finally, give this z score in terms of the problem application or the x value.
Use the formula z = (x – µ) / σ and solve for x. You can do the algebra or
just trust me that is x = µ + (z)* σ For our problem,
178 + (1.28 * 41) = 230.48
Therefore, A cholesterol level of about 230.48 or 230 corresponds
to the 90th percentile.
The Central Limit Theorem
Suppose we take many random samples of size n for a
variable with any distribution (not necessarily a normal
distribution) and record the distribution of the means of each
sample. Then,
1. The distribution of means will be approximately a normal
distribution for large sample sizes. n>30 is magic number
2. The mean of the distribution of means approaches the
population mean, µ, for large sample sizes.
3. The standard deviation of the distribution of means approaches
σ/√n for large sample sizes, where σ is the standard deviation of
the population.
The Interpretation of the Central
Limit Theorem
If you have a group of size n, instead of one individual selection (like the
problems we did earlier) the only difference in working the problem is
how you COMPUTE the Z Score.
Use the formula z = (given sample mean – µ) / [σ/√n ]
Also, see Example 1 of Section 5.3 called Predicting Test Score. Be sure
to notice the difference in part a and part b. In part a, you have ONE
person and in part b you have a group of 100 people.
CLT Demonstrated
(Figure 5.26)
The Value of the Central Limit Theorem
• The Central Limit Theorem allows us to say something about
the mean of a group if we know the mean, µ, and the standard
deviation, σ, of the entire population. This can be useful, but it
turns out that the opposite application is far more important.
• Two major activities of statistics are making estimates of
population means and testing claims about population means.
Is it possible to make a good estimate of the population mean
knowing only the mean of a much smaller sample?
• As you can probably guess, being able to answer this type of
question lies at the heart of statistical sampling, especially in
polls and surveys. The Central Limit Theorem provides the key
to answering such questions.
Computing Probabilities in MSL EASY!
Example 1
Note: the icon here is
not data for the problem
but a standard scores
for the specific distribution
given in this problem.
Example 1-Part a: find the percentage of
scores greater than 1866.
Choose Calculator -> Normal then put in the mean, st dev and value in
question, 1866. Be sure to choose => for “greater than or equal”. Click
Compute to get the graph and answer shown in the second picture below.
The answer is .15865 which as a percentage is 15.865 and rounds to
15.87% with two decimals as asked for in the question.
Example 1 - Part c: find the percentage
of scores between 1389 and 2184.
There are 3 steps to this one. To get area between you must subtract the
Area of the LEFTMOST value FROM the area of the RIGHTMOST value.
Compute percentage less than 2184; compute percentage of 1389; then
Subtract. From below you see .97724986 - .30853754 = .66871232
Which is 66.871% and rounded to 2 places is 66.87%.
Example 2 - Part c: find the probability
the mean blood pressure is less than 111
for a sample of 280 women.
Since n > 1, this is Central Limit
Theorem. Be sure to compute new
Standard deviation as sigma / sqrt(n)
before plugging into StatCrunch.
Standard deviation = 13.22 /
sqrt(280) = 13.22 / 16.73 = .79 ~.8
Questions?