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Lecture 8
Dustin Lueker
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Can not list all possible values with probabilities
◦ Probabilities are assigned to intervals of numbers
 Probability of an individual number is 0
 Probabilities have to be between 0 and 1
◦ Probability of the interval containing all possible values
equals 1
◦ Mathematically, a continuous probability distribution
corresponds to a (density) function whose integral equals 1
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Let X=Weekly use of gasoline by adults in
North America (in gallons)
P(16<X<21)=0.34
 The probability that a randomly chosen adult in North
America uses between 16 and 21 gallons of gas per
week is 0.34
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Discrete Variables:
◦ Histogram
◦ Height of the bar represents the probability
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Continuous Variables:
◦ Smooth, continuous curve
◦ Area under the curve for an interval represents the
probability of that interval
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Gaussian Distribution
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Perfectly symmetric and bell-shaped
◦ Carl Friedrich Gauss (1777-1855)
◦ Empirical rule applies
 Probability concentrated within 1 standard deviation
of the mean is always 0.68
 Probability concentrated within 2 standard
deviations of the mean is always 0.95
 Probability concentrated within 3 standard
deviations of the mean is always 0.997
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Characterized by two parameters
◦ Mean = μ
◦ Standard Deviation = σ
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Assume that adult female height has a
normal distribution with mean μ=165 cm
and standard deviation σ=9 cm
◦ With probability 0.68, a randomly selected adult
female has height between μ - σ = 156 cm and
μ + σ = 174 cm
 This means that on the normal distribution graph of
adult female heights the area under the curve between
156 and 174 is .68
◦ With probability 0.95, a randomly selected adult
female has height between μ - 2σ = 147 cm and
μ + 2σ = 183 cm
 This means that on the normal distribution graph of
adult female heights the area under the curve
between 147and 183 is .95
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So far, we have looked at the probabilities
within one, two, or three standard deviations
from the mean using the Empirical Rule
(μ + σ, μ + 2σ, μ + 3σ)
How much probability is concentrated within
1.43 standard deviations of the mean?
More general, how much probability is
concentrated within any number (say z) of
standard deviations of the mean?
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Our table shows for different values of z the
probability between 0 and μ + zσ
◦ Probability that a normal random variable takes any
value between the mean and z standard deviations
above the mean
◦ Example
 z =1.43, the tabulated value is .4236
 That is, the probability between 0 and 1.43 of the standard
normal distribution equals .4236
 Symmetry
 z = -1.43, the tabulated value is .4236
 That is, the probability between -1.43 and 0 of the
standard normal distribution equals .4236
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So, within 1.43 standard deviations of the
mean is how much probability?
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P(-1<Z<1) should be about 68%
◦ P(-1<Z<1) = P(-1<Z<0) + P(0<Z<1) = 2*P(0<Z<1)
=?
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P(-2<Z<2) should be about 95%
◦ P(-2<Z<2) = P(-2<Z<0) + P(0<Z<2) = 2*P(0<Z<2)
=?
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P(-3<Z<3) should be about 99.7%
◦ P(-3<Z<3) = P(-3<Z<0) + P(0<Z<3) = 2*P(0<Z<3)
=?
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We can also use the table to find z-values for
given probabilities
Find the z-value corresponding to a righthand tail probability of 0.025
This corresponds to a probability of 0.475
between 0 and z standard deviations
Table: z = 1.96
◦ P(Z>1.96) = .025
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For a normal distribution, how many standard
deviations from the mean is the 90th percentile?
◦ What is the value of z such that 0.90 probability is less
than z?
 P(Z<z) = .90
◦ If 0.9 probability is less than z, then there is 0.4
probability between 0 and z
 Because there is 0.5 probability less than 0
 This is because the entire curve has an area under it of 1,
thus the area under half the curve is 0.5
 z=1.28
 The 90th percentile of a normal distribution is 1.28 standard
deviations above the mean
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P( -.25 < Z < 1.61) =
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P(Z > 1.13) =
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P(Z < -3.54) =
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P(Z < m) = .44, m =
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