lecture 9 dustin lueker. can not list all possible values with probabilities ◦ probabilities are...
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STA 291Fall 2009
Lecture 9Dustin 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
Continuous Probability Distribution
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Example of a Continuous Probability Distribution 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
STA 291 Fall 2009 Lecture 9
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Graphs for Probability Distributions Discrete Variables:
◦ Histogram◦ Height of the bar represents the probability
Continuous Variables: ◦ Smooth, continuous curve◦ Area under the curve for an interval represents
the probability of that interval
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Continuous Distributions
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The Normal Probability Distribution
Gaussian Distribution◦ Carl Friedrich Gauss (1777-1855)
Perfectly symmetric and bell-shaped◦ 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
Characterized by two parameters◦ Mean = μ◦ Standard Deviation = σ
STA 291 Fall 2009 Lecture 9
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Different Normal Distributions
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Normal Distribution and Empirical Rule 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
STA 291 Fall 2009 Lecture 9
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Normal Distribution 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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Normal Distribution Table Table 3 (page B-8) 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
So, within 1.43 standard deviations of the mean is how much probability?
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Verifying the Empirical Rule 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) = ?
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) = ?
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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Working backwards We can also use the table to find z-values
for given probabilities Find the z-value corresponding to a right-
hand 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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Finding z-Values for Percentiles 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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Finding z-Values for Two-Tail Probabilities
What is the z-value such that the probability is 0.1 that a normally distributed random variable falls more than z standard deviations above or below the mean?
Symmetry◦ We need to find the z-value such that the right-tail probability
is 0.05 (more than z standard deviations above the mean)◦ z=1.65◦ 10% probability for a normally distributed random variable is
outside 1.65 standard deviations from the mean, and 90% is within 1.65 standard deviations from the mean
Find the z-value such that the probability is 0.5 that a normally distributed random variable falls more than z standard deviations above or below the mean
STA 291 Fall 2009 Lecture 9