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The Normal Distribution April 4, 2009 The Bell Curve I The normal distribution is the most important of all the distributions. I It is widely used and even more widely abused. I Its graph is bell-shaped The Bell Curve I The location of the curve on the x axis is determined by the mean µ. The Bell Curve I The location of the curve on the x axis is determined by the mean µ. This is the x value where the peak is located. The Bell Curve I The location of the curve on the x axis is determined by the mean µ. This is the x value where the peak is located. I The width (and height) of the curve is determined by the standard deviation σ. The Bell Curve I The location of the curve on the x axis is determined by the mean µ. This is the x value where the peak is located. I The width (and height) of the curve is determined by the standard deviation σ. The larger σ is the wider (and lower) the curve is. Used and Abused I Why is it so widely abused? Used and Abused I Why is it so widely abused? I Good question. Examples of Good Use (Theoretically) Figure: Class Heights (in inches) and Sleep Times (in hours) Examples of Poor Use I Random numbers generated by a calculator Examples of Poor Use I Random numbers generated by a calculator (uniform distribution) Examples of Poor Use I Random numbers generated by a calculator (uniform distribution) I Earnings on the stock market Examples of Poor Use I Random numbers generated by a calculator (uniform distribution) I Earnings on the stock market (outliers dominate) Examples of Poor Use I Random numbers generated by a calculator (uniform distribution) I Earnings on the stock market (outliers dominate) I Number of friends a person has Examples of Poor Use I Random numbers generated by a calculator (uniform distribution) I Earnings on the stock market (outliers dominate) I Number of friends a person has (outliers dominate) Examples of Poor Use I Random numbers generated by a calculator (uniform distribution) I Earnings on the stock market (outliers dominate) I Number of friends a person has (outliers dominate) I Number of published works Examples of Poor Use I Random numbers generated by a calculator (uniform distribution) I Earnings on the stock market (outliers dominate) I Number of friends a person has (outliers dominate) I Number of published works (outliers dominate) Questionable Use Figure: Class Homework Scores and Class Test Scores So, When do We Use it? So, When do We Use it? I For now, use it only when told that the given random variable is normally distributed. How do We Use it? I Conceptually, we find the area under the bell curve and over the event of interest. How do We Use it? I Conceptually, we find the area under the bell curve and over the event of interest. Figure: P(X < 2) is equal to the area of the shaded region How do We Use it? Let X be normally distributed with mean µ and standard deviation σ. I We use our calculator How do We Use it? Let X be normally distributed with mean µ and standard deviation σ. I We use our calculator P(xL < X < xR ) = normalcdf(xL , xR , µ, σ). How do We Use it? Let X be normally distributed with mean µ and standard deviation σ. I We use our calculator P(xL < X < xR ) = normalcdf(xL , xR , µ, σ). I We can also reference tables or use computer software. Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. P(0 ≤ X ≤ 80.6) = .339 Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. P(0 ≤ X ≤ 80.6) = .339 I P(X ≤ 0) Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. P(0 ≤ X ≤ 80.6) = .339 I P(X ≤ 0) = 0 Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. P(0 ≤ X ≤ 80.6) = .339 I P(X ≤ 0) = 0 I P(90 ≤ X ≤ 100) Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. P(0 ≤ X ≤ 80.6) = .339 I P(X ≤ 0) = 0 I P(90 ≤ X ≤ 100) = .242 Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. P(0 ≤ X ≤ 80.6) = .339 I P(X ≤ 0) = 0 I P(90 ≤ X ≤ 100) = .242 I P(X > 100) Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. P(0 ≤ X ≤ 80.6) = .339 I P(X ≤ 0) = 0 I P(90 ≤ X ≤ 100) = .242 I P(X > 100) = .097 Example 1 Let X ∼ N(85.3, 11.3). I Find the probability that 0 ≤ X ≤ 80.6. P(0 ≤ X ≤ 80.6) = .339 I P(X ≤ 0) = 0 I P(90 ≤ X ≤ 100) = .242 I P(X > 100) = .097 I P(X ≤ 80.6) = P(X ≥ 90) = .339 because the normal distribution is symmetric around its mean and both scores are the same distance from the mean. Example 1 Let X ∼ N(85.3, 11.3). I The mean and standard deviation for X are taken from the class test. Example 1 Let X ∼ N(85.3, 11.3). I The mean and standard deviation for X are taken from the class test. I Do you think the probabilities we found accurately reflect the class performance? Example 1 Let X ∼ N(85.3, 11.3). I Problem 1: Theoretically, scores could be any real number. But, we know scores are only in [0, 100]. Example 1 Let X ∼ N(85.3, 11.3). I Problem 1: Theoretically, scores could be any real number. But, we know scores are only in [0, 100]. I Problem 2: The distribution was clearly skewed. Figure: Class Homework Scores and Class Test Scores Example 1 Let X ∼ N(85.3, 11.3). I Problem 1: Theoretically, scores could be any real number. But, we know scores are only in [0, 100]. I Problem 2: The distribution was clearly skewed. Figure: Class Homework Scores and Class Test Scores I Problem 3: The numbers just don’t match up. For example P(90 ≤ X ≤ 100) = .24 6= .54 = actual proportion ≥ 90 z-scores I A z-score is the number of standard deviations a (non-standard) normal random variable X is from the mean. x = µ + zσ z-scores I A z-score is the number of standard deviations a (non-standard) normal random variable X is from the mean. x = µ + zσ I e.g. z = −4 means x is 4 standard deviations to the left of the mean. If µ = 11 and σ = 2, then the x = 11 + (−4)(2) = 3 If µ = 11 and σ = .3, then the x = 11 + (−4)(.3) = 9.8 z-score Example I Let X is a normal random variable with µ = 2 and σ = 3. z-score Example I Let X is a normal random variable with µ = 2 and σ = 3. I Find the z-scores of x = −10, 2, 5, 21. z-score Example I Let X is a normal random variable with µ = 2 and σ = 3. I Find the z-scores of x = −10, 2, 5, 21. Hint: First, we solve the equation x = µ + σz i.e., x = 2 + 3z x −µ σ i.e., z= to get z= x −2 3 z-score Example I Let X is a normal random variable with µ = 2 and σ = 3. I Find the z-scores of x = −10, 2, 5, 21. Hint: First, we solve the equation x = µ + σz to get i.e., x = 2 + 3z x −µ x −2 i.e., z = σ 3 Then we plug in the x values to get z(−10) = −4, z(2) = 0, z(5) = 1 and z(21) = 6.33. z= The Standard Normal Distribution & Standard Normal Random Variables I The standard normal distribution is a normal distribution of standardized values called z-scores. The Standard Normal Distribution & Standard Normal Random Variables I The standard normal distribution is a normal distribution of standardized values called z-scores. I We use the symbol Z for the random variable of z-scores and say Z is a standard normal random variable. The Standard Normal Distribution & Standard Normal Random Variables I The standard normal distribution is a normal distribution of standardized values called z-scores. I We use the symbol Z for the random variable of z-scores and say Z is a standard normal random variable. I The mean of a Z is 0 and the standard deviation is 1. µZ = 0 and σZ = 1 The Point of z-scores and the Standard Normal Distribution I Let X be a normal random variable with mean µX and standard deviation σX . Then z(a) = a−µ b−µ and z(b) = σ σ are the z-scores of a and b respectively. The Point of z-scores and the Standard Normal Distribution I Let X be a normal random variable with mean µX and standard deviation σX . Then z(a) = a−µ b−µ and z(b) = σ σ are the z-scores of a and b respectively. I Using the equality below can use z-scores to find the probability of an X event. P(a < X < b) = P(z(a) < Z < z(b)) The Point of z-scores and the Standard Normal Distribution I Let X be a normal random variable with mean µX and standard deviation σX . Then z(a) = a−µ b−µ and z(b) = σ σ are the z-scores of a and b respectively. I Using the equality below can use z-scores to find the probability of an X event. P(a < X < b) = P(z(a) < Z < z(b)) I This is because the area beneath the PDF of X between x = a and x = b is the same as the area beneath the standard normal curve between z(a) and z(b). Example: Using z-Scores and a TI Calculator I Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15) Example: Using z-Scores and a TI Calculator I Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15) I The normalcdf distribution in the TI calculator defaults to a standard normal random variable if no mean or standard deviation are entered. Example: Using z-Scores and a TI Calculator I Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15) I The normalcdf distribution in the TI calculator defaults to a standard normal random variable if no mean or standard deviation are entered. I Thus, we have 1.0 − 1.2 1.3 − 1.2 P(1.1 < X < 1.3) = P <Z < 0.15 0.15 = P(−1.33 < Z < 0.67) = normalcdf(−1.33, .67) = 0.657 Example: Using z-Scores and a Table I Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15) Example: Using z-Scores and a Table I Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15) I We can find the same probability as found in the last slide by using our table of Areas of the Standard Normal Distribution. Example: Using z-Scores and a Table I Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15) I We can find the same probability as found in the last slide by using our table of Areas of the Standard Normal Distribution. I First we recognize that the event −1.33 < Z < 0.67 can be split into two mutually exclusive events −1.33 < Z < 0 and 0 < Z < 0.67 Example: Using z-Scores and a Table I Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15) I We can find the same probability as found in the last slide by using our table of Areas of the Standard Normal Distribution. I First we recognize that the event −1.33 < Z < 0.67 can be split into two mutually exclusive events −1.33 < Z < 0 and 0 < Z < 0.67 I Then, from the symmetry of the bell curve we have that P(−1.33 < Z < 0) = P(< Z < 1.33). Example: Using z-Scores and a Table I Find P(1.0 < X < 1.3) if X ∼ N(1.2, 0.15) I We can find the same probability as found in the last slide by using our table of Areas of the Standard Normal Distribution. I First we recognize that the event −1.33 < Z < 0.67 can be split into two mutually exclusive events −1.33 < Z < 0 and 0 < Z < 0.67 I Then, from the symmetry of the bell curve we have that P(−1.33 < Z < 0) = P(< Z < 1.33). I Thus, we simply find the entry corresponding to 0.67 and add it to the value corresponding to 1.33, i.e., 0.2486 + 0.4082 ≈ 0.657