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# STA 291 Fall 2009 - PowerPoint PPT Presentation

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 291Fall 2009

Lecture 10

Dustin Lueker

• 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

• For a normally distributed random variable, find the following

• P(Z>.82) =

• P(-.2<Z<2.18) =

STA 291 Fall 2009 Lecture 10

• 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

• 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

• 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

• 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

• 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

• 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

• 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