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STA 291 Summer 2010

STA 291 Summer 2010. Lecture 9 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 Summer 2010

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  1. STA 291Summer 2010 Lecture 9 Dustin Lueker

  2. 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 Summer 2010 Lecture 9

  3. Examples • For a normally distributed random variable, find the following • P(Z>.82) = • P(-.2<Z<2.18) = STA 291 Summer 2010 Lecture 9

  4. 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 Summer 2010 Lecture 9

  5. 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 Summer 2010 Lecture 9

  6. 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 Summer 2010 Lecture 9

  7. 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 Summer 2010 Lecture 9

  8. 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 Summer 2010 Lecture 9

  9. 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 Summer 2010 Lecture 9

  10. 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 probability that PDI is less than 90 • Would you be surprised to observe a value of 90? STA 291 Summer 2010 Lecture 9

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