15 September 2003 15 September 2003 1.2 Describing Distributions with Numbers Five-number summaries and boxplots Changing the unit of measurement 1.3 The Normal Distributions Density curves The normal distribution Standardized scores Areas under a standard normal curve (Table A) Normal quantile plots FIVE-NUMBER SUMMARIES and BOXPLOTS 50 ◄ Maximum 45 40 35 30 ◄ 75th percentile (Q3) 25 20 ◄ Median (Q2) 15 ◄ 25th percentile (Q1) 10 5 ◄ Minimum Calculating the variance and the standard deviation Mean Variance Standa rd Deviation 2 6 8 5 9 6 Deviations from Mean -4 0 +2 -1 +3 Squared Deviations 16 0 4 1 9 7.5 2.74 What if we add 10 to each number? 12 16 18 15 19 Mean 16 Variance Standa rd Deviation Deviations from Mean -4 0 +2 -1 +3 Squared Deviations 16 0 4 1 9 7.5 2.74 What if we add 10 to each number? 12 16 18 15 19 Mean 16 Variance Standa rd Deviation Deviations from Mean -4 0 +2 -1 +3 Squared Deviations 16 0 4 1 9 7.5 2.74 What if we multiply each number by 10? 20 60 80 50 90 Mean 60 Variance Standa rd Deviation Deviations from Mean -40 0 +20 -10 +30 Squared Deviations 1600 0 400 100 900 750 27.4 What if we multiply each number by 10? 20 60 80 50 90 Mean 60 Variance Standa rd Deviation Deviations from Mean -40 0 +20 -10 +30 Squared Deviations 1600 0 400 100 900 750 27.4 What if we multiply each number by 10? 20 60 80 50 90 Mean 60 Variance Standa rd Deviation Deviations from Mean -40 0 +20 -10 +30 Squared Deviations 1600 0 400 100 900 750 27.4 CHANGING THE UNIT OF MEASUREMENT If you add or subtract the same amount from each value in a distribution, then the mean is increased or decreased by that amount the spread is not changed If you multiply or divide each value in a distribution by the same amount, then the mean is multiplied or divided by that amount the variance is multiplied or divided by the square of that amount the standard deviation is multiplied or divided by that amount STANDARDIZED SCORES A standardized score (or z score) tells us far above or below the mean a given number falls, in standard deviation units. z deviation standard deviation x x s STANDARDIZED SCORES For example, suppose the grades on a quiz have mean of 85 and standard deviation of 5 points. If your grade is 95, your z score is +2.00, because you are two standard deviations above the mean. Your friend whose grade is 83 has what z score? Another friend tells you that his z score is -1.00. What’s his grade? 65 70 75 80 85 90 95 100 ——+—————+—————+—————+—————+—————+—————+—————+—— -4 -3 -2 -1 0 +1 +2 +3 Standardized scores 2 6 8 5 9 Mean 6 Standa rd Deviation 2.74 -4 / 2.74 0 / 2.74 2 / 2.74 -1 / 2.74 3 / 2.74 z score = -1.46 = 0.00 = 0.73 = -0.36 = 1.09 What if we add 10 to each number? 12 16 18 15 19 Mean 16 Standa rd Deviation 2.74 -4 / 2.74 0 / 2.74 2 / 2.74 -1 / 2.74 3 / 2.74 z score = -1.46 = 0.00 = 0.73 = -0.36 = 1.09 What if we multiply each number by 10? 20 60 80 50 90 Mean 60 Standa rd Deviation 27.4 -40 / 27.4 0 / 27.4 20 / 27.4 -10 / 27.4 30 / 27.4 z score = -1.46 = 0.00 = 0.73 = -0.36 = 1.09 STANDARDIZED SCORES If you take a list of numbers and add the same amount to each number, subtract the same amount from each number, multiply each number by the same amount, or divide each number by the same amount, the z scores do not change. THE NORMAL CURVE THE NORMAL CURVE The normal curve for women’s heights How many women are at least 67 inches tall? How many women are at least 67 inches tall? z = (67 - 64.5) / 2.5 = +1.00 The relative frequency of women at least 67 inches tall is approximately equal to the fraction of a standard normal curve which lies to the right of +1.00. Table A says that 84% of the area lies to the left of +1.00, so 16% must lie to the right. So about 16% of women are at least 67” tall. How many women are at least 68 inches tall? z = (68 - 64.5) / 2.5 = +1.40 The relative frequency of women at least 68 inches tall is approximately equal to the fraction of a standard normal curve which lies to the right of +1.40. Table A says that 91.92% of the area lies to the left of +1.40, so 8.08% must lie to the right. So about 8% of women are at least 68” tall. How many women are at least 68 inches tall? What SAT-verbal score is at the 90th percentile? What SAT-verbal score is at the 90th percentile? Table A says that about 90% of a normal histogram lies left of +1.28 ( X - 505 ) / 110 = 1.28 X = 110 (1.28) + 505 X = 645.8 So approximately 90% of SAT-v scores are less than 645.8 What fraction of SAT-v scores are less than 645.8? z = ( 645.8 - 505 ) / 110 = ( 140.8 ) / 110 = 1.28 According to Table A, 89.97% of a normal histogram lies to the left of +1.28, so about 90% of the SAT scores will be less than 645.8. 1.2 Describing Distributions with Numbers Five-number summaries and boxplots Changing the unit of measurement 1.3 The Normal Distributions Density curves The normal distribution Standardized scores Areas under a standard normal curve (Table A) Normal quantile plots Sections to skip 1.3 Normal Quantile Plots (pp 78-83) 2.1 Categorical Explanatory Variables (pp 113-114)