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The Normal Distribution The Normal Distribution Distribution – any collection of scores, from either a sample or population Can be displayed in any form, but is usually represented as a histogram Normal Distribution – specific type of distribution that assumes a characteristic bell shape and is perfectly symmetrical The Normal Distribution Can provide us with information on likelihood of obtaining a given score 60 people scored a 6 – 6/350 = .17 = 17% 9 people scored a 1 – 3% The Normal Distribution Why is the Normal Distribution so important? Almost all of the statistical tests that we will be covering (ZTests, T-Tests, ANOVA, etc.) throughout the course assume that the population distribution, that our sample is drawn from (but for the variable we are looking at), is normally distributed Also, many variables that psychologists and health professionals look at are normally distributed Why this is requires a detailed examination of the derivation of our statistics, that involves way more detail than you need to use the statistic. The Normal Distribution Ordinate Density – what is measured on the ordinate (more on this in Ch. 7) Abscissa The Normal Distribution Mathematically defined as: 1 X 2 / 2 2 f (X ) (e) 2 Since and e are constants, we only have to determine μ (the population mean) and σ (the population standard deviation) to graph the mathematical function of any variable we are interested in Don’t worry, understanding this is not necessary to understanding the normal distribution, only a helpful aside for the mathematically inclined The Normal Distribution Using this formula, mathematicians have determined the probabilities of obtaining every score on a “standard normal distribution” (see Table E.10 in your book) To determine these probabilities for the variable you’re interested in we must plug in your variable to the formula Note: This assumes that your variable fits a normal distribution, if not, your results will be inaccurate The Normal Distribution However, this table refers to a Standard Normal Distribution Μ = 0; σ = 1 How do you get your variable to fit? z X The Normal Distribution Z-Scores Range from +∞ to -∞ Represent the number of standard deviations your score is from the mean i.e. z = +1 is a score that is 1 standard deviation above the mean and z = -3 is a score 3 standard deviations below the mean Now we can begin to use the table to determine the probability that our z score will occur using table E.10 The Normal Distribution z Mean to z Larger Portion Smaller Portion 0.000 .0000 .5000 .5000 0.100 .0398 .5398 .4602 0.200 .0793 .5793 .4207 1.000 .3413 .8413 .1587 1.500 .4332 .9332 .0668 1.645 .4500 .9500 .0500 1.960 .4750 .9750 .0250 The Normal Distribution Mean to Z Normal Distribution Cutoff at +1.645 1200 1000 800 600 400 200 0 50 3. 00 3. 50 2. 00 2. 50 1. 00 1. 0 .5 00 0. 0 -.5 0 .0 -1 0 .5 -1 0 .0 -2 0 .5 -2 0 .0 -3 0 .5 -3 0 .0 -4 z The Normal Distribution The Normal Distribution Larger Portion Normal Distribution Cutoff at +1.645 1200 1000 800 600 400 200 0 50 3. 00 3. 50 2. 00 2. 50 1. 00 1. 0 .5 00 0. 0 -.5 0 .0 -1 0 .5 -1 0 .0 -2 0 .5 -2 0 .0 -3 0 .5 -3 0 .0 -4 z The Normal Distribution Smaller Portion Normal Distribution Cutoff at +1.645 1200 1000 800 600 400 200 0 50 3. 00 3. 50 2. 00 2. 50 1. 00 1. 0 .5 00 0. 0 -.5 0 .0 -1 0 .5 -1 0 .0 -2 0 .5 -2 0 .0 -3 0 .5 -3 0 .0 -4 z Normal Distribution Cutoff at +1.645 1200 1000 800 600 400 200 0 50 3. 00 3. 50 2. 00 2. 50 1. 00 1. 0 .5 00 0. 0 -.5 0 .0 -1 0 .5 -1 0 .0 -2 0 .5 -2 0 .0 -3 0 .5 -3 0 .0 -4 z Reminder: Z-Scores represent # of standard deviations from the mean For this distribution, if μ = 50 and σ = 10, what score does z = -3 represent? z = +2.5? z = -.1 z = 1.645 (z = -.1, “Mean to Z”) + (z = 1.645, “Mean to Z”) .0398 + .4500 = .4898 = 49% z = -1.00, “Smaller Portion” = Red + Blue z = -1.645, “Smaller Portion” = Blue (Red + Blue) - Blue = Red z = -1.645 z = -1.00 (z = -1.00, “Smaller Portion”) – (z = -1.645, “Smaller Portion”) .1587 - .0500 = .1087 = 11% The Normal Distribution What are the scores that lie in the middle 50% of a distribution of scores with μ = 50 and σ = 10? Look for “Smaller Portion” = .2500 on Table E.10 z = .67 Solve for X using z-score formula Scores = 56.7 and 43.3 The Normal Distribution z X X 50 .67 10 X 50 .67(10) X 50 6.7 X 6.7 50 X 56.7 and 43.3 The Normal Distribution Other uses for z-scores: 1. Converting two variable to a standard metric You took two exams, you got an 80 in Statistics and a 50 in Biology – you cannot say which one you did better in without knowing about the variability in scores in each If the class average in Stats was a 90 and the s.d. 15, what would we conclude about your score now? How is it different than just using the score itself? If the mean in Bio was a 30 and the s.d. was a 5, you did 4 s.d’s above the mean (a z-score of +4) or much better than everyone else The Normal Distribution Other uses for z-scores: 1. Converting variables to a standard metric This also allows us to compare two scores on different metrics 2. i.e. two tests scored out of 100 = same metric one test out of 50 vs. one out of 100 = two different metrics Is 20/50 better than 40/100? Is it better when compared to the class average? Allows for quick comparisons between a score and the rest of the distribution it is a part of The Normal Distribution Standard Scores – scores with a predetermined mean and standard deviation, i.e. a z-score Why convert to standard scores? You can compare performance on two different tests with two different metrics You can easily compute Percentile ranks but they are population-relative! Percentile – the point below which a certain percent of scores fall i.e. If you are at the 75th%ile (percentile), then 75% of the scores are at or below your score The Normal Distribution How do you compute %ile? Convert your raw score into a z-score Look at Table E.10, and find the “Smaller Portion” if your zscore is negative and the “Larger Portion” if it is positive Normal Distribution Multiply by 100 Cutoff at +1.645 1200 1000 800 600 400 200 0 50 3. 00 3. 50 2. 00 2. 50 1. 00 1. 0 .5 00 0. 0 -.5 0 .0 -1 0 .5 -1 0 .0 -2 0 .5 -2 0 .0 -3 0 .5 -3 0 .0 -4 z The Normal Distribution New Score = New s.d. (z) + New Mean New IQ Score = 15 (2) + 100 = 130 T-Score – commonly used standardized normal distribution w/ mean = 50 and s.d. = 10 T-Score = 10 (2) + 50 = 70