CJUS-K300 1nd Edition Lecture 8 Outline of Last Lecture I. ProbabilityOutline of Current Lecture II. Z scoreCurrent LectureProperties of a normal curve – as close to symmetrical as possible given that it would be close toimpossible for it to be perfectly symmetricalTwo curves can be normal with the same standard deviations but different means, or the same mean and different standard deviations. (See power point for examples)In cases of normal distribution :Once you find the mean of the curve, for normal curves it is in the middle of the curve, 50% of the data is on the left side and 50% of the data is on the right. The area between the mean and one standard deviation is 34% meaning 34% of the data points lie in between the mean/middle and one standard deviation away in either direction. The area between the mean and two standard deviations is 47.5%. The area between the mean and 3 standard deviations is 49.5%Example: Janes IQ score is 3 standard deviations above the mean. What do we know about her IQ ?- She is in the 99.5th percentile. = 50 + 49.5 (from 3 standard deviations) American IQ test : x|=100 s=10 Marys score was 110pts . She was above the average by exactly 1 standard deviation (10) so we know that she is in the 84th percentile. (50 + 34) Comparing 2 different testsThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.American IQ test x| = 100, s=10 Mary = 110European IQ test x|= 350, s=36, Pierre = 382Who has the higher IQ? Mary does. Pierre is not a full standard deviation above the mean whereas Mary is. Z = X – Mean / SPierre: z=382-350/36 = .89Mary: z = 110-100/10 = 1If it is not a full SD above the mean you have to take it to the z table. Look up the z score on the table and find the corresponding number and add it to the 50%Z=88-80/15 = .53 .53 corresponds to .2019, so you add 20% to 50% and which makes it the
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