Sta 291 fall 2009
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STA 291 Fall 2009. Lecture 10 Dustin Lueker. Normal Distribution. Perfectly symmetric and bell-shaped Characterized by two parameters Mean = μ Standard Deviation = σ Standard Normal μ = 0 σ = 1 Solid Line. Examples. For a normally distributed random variable, find the following

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STA 291 Fall 2009

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Sta 291 fall 2009

STA 291Fall 2009

Lecture 10

Dustin Lueker


Normal distribution

Normal Distribution

  • Perfectly symmetric and bell-shaped

  • Characterized by two parameters

    • Mean = μ

    • Standard Deviation = σ

  • Standard Normal

    • μ = 0

    • σ = 1

      • Solid Line

STA 291 Fall 2009 Lecture 10


Examples

Examples

  • For a normally distributed random variable, find the following

    • P(Z>.82) =

    • P(-.2<Z<2.18) =

STA 291 Fall 2009 Lecture 10


Finding z values for percentiles

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

STA 291 Fall 2009 Lecture 10


Working backwards

Working backwards

  • We can also use the table to find z-values for given probabilities

  • Find the following

    • P(Z>a) = .7224

      • a =

    • P(Z<b) = .2090

      • b =

STA 291 Fall 2009 Lecture 10


Standard normal distribution

Standard Normal Distribution

  • When values from an arbitrary normal distribution are converted to z-scores, then they have a standard normal distribution

  • The conversion is done by subtracting the mean μ, and then dividing by the standard deviation σ

STA 291 Fall 2009 Lecture 10


Z scores

z-Scores

  • The z-score for a value x of a random variable is the number of standard deviations that x is above μ

    • If x is below μ, then the z-score is negative

  • The z-score is used to compare values from different normal distributions

  • Calculating

    • Need to know

      • x

      • μ

      • σ

STA 291 Fall 2009 Lecture 10


Example

Example

  • SAT Scores

    • μ=500

    • σ=100

      • SAT score 700 has a z-score of z=2

      • Probability that a score is above 700 is the tail probability of z=2

      • Table 3 provides a probability of 0.4772 between mean=500 and 700

        • z=2

      • Right-tail probability for a score of 700 equals 0.5-0.4772=0.0228

        • 2.28% of the SAT scores are above 700

    • Now find the probability of having a score below 450

STA 291 Fall 2009 Lecture 10


Z scores1

z-Scores

  • The z-score is used to compare values from different normal distributions

    • SAT

      • μ=500

      • σ=100

    • ACT

      • μ=18

      • σ=6

    • What is better, 650 on the SAT or 25 on the ACT?

      • Corresponding tail probabilities?

        • How many percent have worse SAT or ACT scores?

          • In other words, 650 and 25 correspond to what percentiles?

STA 291 Fall 2009 Lecture 10


Example1

Example

  • The scores on the Psychomotor Development Index (PDI) are approximately normally distributed with mean 100 and standard deviation 15. An infant is selected at random.

    • Find the probability that the infant’s PDI score is at least 100

      • P(X>100)

    • Find the probability that PDI is between 97 and 103

      • P(97<X<103)

    • Find the z-score for a PDI value of 90

      • Would you be surprised to observe a value of 90?

STA 291 Fall 2009 Lecture 10


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