Download The Normal Distribution

The Normal Distribution
April 4, 2009
The Bell Curve
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The normal distribution is the most important of all the
distributions.
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It is widely used and even more widely abused.
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Its graph is bell-shaped
The Bell Curve
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The location of the curve on the x axis is determined by the
mean µ.
The Bell Curve
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The location of the curve on the x axis is determined by the
mean µ.
This is the x value where the peak is located.
The Bell Curve
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The location of the curve on the x axis is determined by the
mean µ.
This is the x value where the peak is located.
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The width (and height) of the curve is determined by the
standard deviation σ.
The Bell Curve
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The location of the curve on the x axis is determined by the
mean µ.
This is the x value where the peak is located.
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The width (and height) of the curve is determined by the
standard deviation σ.
The larger σ is the wider (and lower) the curve is.
Used and Abused
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Why is it so widely abused?
Used and Abused
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Why is it so widely abused?
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Good question.
Examples of Good Use (Theoretically)
Figure: Class Heights (in inches) and Sleep Times (in hours)
Examples of Poor Use
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Random numbers generated by a calculator
Examples of Poor Use
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Random numbers generated by a calculator (uniform
distribution)
Examples of Poor Use
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Random numbers generated by a calculator (uniform
distribution)
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Earnings on the stock market
Examples of Poor Use
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Random numbers generated by a calculator (uniform
distribution)
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Earnings on the stock market (outliers dominate)
Examples of Poor Use
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Random numbers generated by a calculator (uniform
distribution)
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Earnings on the stock market (outliers dominate)
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Number of friends a person has
Examples of Poor Use
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Random numbers generated by a calculator (uniform
distribution)
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Earnings on the stock market (outliers dominate)
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Number of friends a person has (outliers dominate)
Examples of Poor Use
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Random numbers generated by a calculator (uniform
distribution)
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Earnings on the stock market (outliers dominate)
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Number of friends a person has (outliers dominate)
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Number of published works
Examples of Poor Use
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Random numbers generated by a calculator (uniform
distribution)
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Earnings on the stock market (outliers dominate)
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Number of friends a person has (outliers dominate)
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Number of published works (outliers dominate)
Questionable Use
Figure: Class Homework Scores and Class Test Scores
So, When do We Use it?
So, When do We Use it?
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For now, use it only when told that the given random variable
is normally distributed.
How do We Use it?
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Conceptually, we find the area under the bell curve and over
the event of interest.
How do We Use it?
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Conceptually, we find the area under the bell curve and over
the event of interest.
Figure: P(X < 2) is equal to the area of the shaded region
How do We Use it?
Let X be normally distributed with mean µ and standard deviation
σ.
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We use our calculator
How do We Use it?
Let X be normally distributed with mean µ and standard deviation
σ.
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We use our calculator
P(xL < X < xR ) = normalcdf(xL , xR , µ, σ).
How do We Use it?
Let X be normally distributed with mean µ and standard deviation
σ.
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We use our calculator
P(xL < X < xR ) = normalcdf(xL , xR , µ, σ).
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We can also reference tables or use computer software.
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
P(0 ≤ X ≤ 80.6) = .339
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
P(0 ≤ X ≤ 80.6) = .339
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P(X ≤ 0)
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
P(0 ≤ X ≤ 80.6) = .339
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P(X ≤ 0) = 0
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
P(0 ≤ X ≤ 80.6) = .339
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P(X ≤ 0) = 0
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P(90 ≤ X ≤ 100)
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
P(0 ≤ X ≤ 80.6) = .339
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P(X ≤ 0) = 0
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P(90 ≤ X ≤ 100) = .242
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
P(0 ≤ X ≤ 80.6) = .339
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P(X ≤ 0) = 0
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P(90 ≤ X ≤ 100) = .242
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P(X > 100)
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
P(0 ≤ X ≤ 80.6) = .339
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P(X ≤ 0) = 0
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P(90 ≤ X ≤ 100) = .242
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P(X > 100) = .097
Example 1
Let X ∼ N(85.3, 11.3).
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Find the probability that 0 ≤ X ≤ 80.6.
P(0 ≤ X ≤ 80.6) = .339
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P(X ≤ 0) = 0
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P(90 ≤ X ≤ 100) = .242
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P(X > 100) = .097
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P(X ≤ 80.6) = P(X ≥ 90) = .339 because the normal
distribution is symmetric around its mean and both scores are
the same distance from the mean.
Example 1
Let X ∼ N(85.3, 11.3).
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The mean and standard deviation for X are taken from the
class test.
Example 1
Let X ∼ N(85.3, 11.3).
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The mean and standard deviation for X are taken from the
class test.
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Do you think the probabilities we found accurately reflect the
class performance?
Example 1
Let X ∼ N(85.3, 11.3).
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Problem 1: Theoretically, scores could be any real number.
But, we know scores are only in [0, 100].
Example 1
Let X ∼ N(85.3, 11.3).
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Problem 1: Theoretically, scores could be any real number.
But, we know scores are only in [0, 100].
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Problem 2: The distribution was clearly skewed.
Figure: Class Homework Scores and Class Test Scores
Example 1
Let X ∼ N(85.3, 11.3).
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Problem 1: Theoretically, scores could be any real number.
But, we know scores are only in [0, 100].
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Problem 2: The distribution was clearly skewed.
Figure: Class Homework Scores and Class Test Scores
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Problem 3: The numbers just don’t match up. For example
P(90 ≤ X ≤ 100) = .24 6= .54 = actual proportion ≥ 90
z-scores
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A z-score is the number of standard deviations a
(non-standard) normal random variable X is from the mean.
x = µ + zσ
z-scores
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A z-score is the number of standard deviations a
(non-standard) normal random variable X is from the mean.
x = µ + zσ
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e.g. z = −4 means x is 4 standard deviations to the left of
the mean.
If µ = 11 and σ = 2, then the x = 11 + (−4)(2) = 3
If µ = 11 and σ = .3, then the x = 11 + (−4)(.3) = 9.8
z-score Example
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Let X is a normal random variable with µ = 2 and σ = 3.
z-score Example
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Let X is a normal random variable with µ = 2 and σ = 3.
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Find the z-scores of x = −10, 2, 5, 21.
z-score Example
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Let X is a normal random variable with µ = 2 and σ = 3.
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Find the z-scores of x = −10, 2, 5, 21.
Hint: First, we solve the equation
x = µ + σz
i.e.,
x = 2 + 3z
x −µ
σ
i.e.,
z=
to get
z=
x −2
3
z-score Example
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Let X is a normal random variable with µ = 2 and σ = 3.
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Find the z-scores of x = −10, 2, 5, 21.
Hint: First, we solve the equation
x = µ + σz
to get
i.e.,
x = 2 + 3z
x −µ
x −2
i.e., z =
σ
3
Then we plug in the x values to get z(−10) = −4, z(2) = 0,
z(5) = 1 and z(21) = 6.33.
z=
The Standard Normal Distribution & Standard Normal
Random Variables
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The standard normal distribution is a normal distribution of
standardized values called z-scores.
The Standard Normal Distribution & Standard Normal
Random Variables
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The standard normal distribution is a normal distribution of
standardized values called z-scores.
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We use the symbol Z for the random variable of z-scores and
say Z is a standard normal random variable.
The Standard Normal Distribution & Standard Normal
Random Variables
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The standard normal distribution is a normal distribution of
standardized values called z-scores.
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We use the symbol Z for the random variable of z-scores and
say Z is a standard normal random variable.
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The mean of a Z is 0 and the standard deviation is 1.
µZ = 0
and
σZ = 1
The Point of z-scores and the Standard Normal
Distribution
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Let X be a normal random variable with mean µX and
standard deviation σX . Then
z(a) =
a−µ
b−µ
and z(b) =
σ
σ
are the z-scores of a and b respectively.
The Point of z-scores and the Standard Normal
Distribution
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Let X be a normal random variable with mean µX and
standard deviation σX . Then
z(a) =
a−µ
b−µ
and z(b) =
σ
σ
are the z-scores of a and b respectively.
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Using the equality below can use z-scores to find the
probability of an X event.
P(a < X < b) = P(z(a) < Z < z(b))
The Point of z-scores and the Standard Normal
Distribution
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Let X be a normal random variable with mean µX and
standard deviation σX . Then
z(a) =
a−µ
b−µ
and z(b) =
σ
σ
are the z-scores of a and b respectively.
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Using the equality below can use z-scores to find the
probability of an X event.
P(a < X < b) = P(z(a) < Z < z(b))
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This is because the area beneath the PDF of X between
x = a and x = b is the same as the area beneath the standard
normal curve between z(a) and z(b).
Example: Using z-Scores and a TI Calculator
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Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15)
Example: Using z-Scores and a TI Calculator
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Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15)
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The normalcdf distribution in the TI calculator defaults to a
standard normal random variable if no mean or standard
deviation are entered.
Example: Using z-Scores and a TI Calculator
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Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15)
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The normalcdf distribution in the TI calculator defaults to a
standard normal random variable if no mean or standard
deviation are entered.
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Thus, we have
1.0 − 1.2
1.3 − 1.2
P(1.1 < X < 1.3) = P
<Z <
0.15
0.15
= P(−1.33 < Z < 0.67)
= normalcdf(−1.33, .67)
= 0.657
Example: Using z-Scores and a Table
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Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15)
Example: Using z-Scores and a Table
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Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15)
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We can find the same probability as found in the last slide by
using our table of Areas of the Standard Normal Distribution.
Example: Using z-Scores and a Table
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Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15)
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We can find the same probability as found in the last slide by
using our table of Areas of the Standard Normal Distribution.
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First we recognize that the event −1.33 < Z < 0.67 can be
split into two mutually exclusive events
−1.33 < Z < 0 and 0 < Z < 0.67
Example: Using z-Scores and a Table
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Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15)
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We can find the same probability as found in the last slide by
using our table of Areas of the Standard Normal Distribution.
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First we recognize that the event −1.33 < Z < 0.67 can be
split into two mutually exclusive events
−1.33 < Z < 0 and 0 < Z < 0.67
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Then, from the symmetry of the bell curve we have that
P(−1.33 < Z < 0) = P(< Z < 1.33).
Example: Using z-Scores and a Table
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Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15)
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We can find the same probability as found in the last slide by
using our table of Areas of the Standard Normal Distribution.
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First we recognize that the event −1.33 < Z < 0.67 can be
split into two mutually exclusive events
−1.33 < Z < 0 and 0 < Z < 0.67
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Then, from the symmetry of the bell curve we have that
P(−1.33 < Z < 0) = P(< Z < 1.33).
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Thus, we simply find the entry corresponding to 0.67 and add
it to the value corresponding to 1.33, i.e.,
0.2486 + 0.4082 ≈ 0.657