# Normal distribution, Central limit theory, and z scores

Normal distribution, Central limit theory, and z scores PA330 Review - Problem 1.3 - frequency distribution n u m b e r o f a p p lic a n ts Civil Service Exam Score 60 50 40 30 20

10 0 61-65 66-70 71-75 76-80 81-85 86-90 91-9596-100 Exam score Problem 1.3 - frequency polygon n u m b e r o f a p p lic a n ts Civil Service Exam Score 60 50 40

30 20 10 0 58 63 68 73 78 83 88 93 98 103 Exam score Mean, problem 2.7 class interval # of pe rsons (F) 0-4 213 5-9

215 10-14 242 15-19 194 20-24 168 25-29 162 30-34 111 35-39 82 40-44 74

45-49 50 50-54 53 55-59 46 60-64 24 65-85 86 Total Mean midpoint (M)

2 7 12 17 22 27 32 37 42 47 52 57 62 75

F X M 426 1505 2904 3298 3696 4374 3552 3034 3108 2350 2756 2622 1488 6450

41,563 / 1720 = 41563 24.16 1720 Problem 2.7 Group median (lecture method) There are a total of 1720 people. Therefore, the median person is #860. This person is in the interval 15-19. The midpoint of this interval is 17.

Therefore, the median age is 17. Problem 2.7 Group median (textbook method) There are a total of 1720 people. Therefore, the median person is #860. This person is in the interval 15-19. The 190th person in that interval. The width of the interval is 5. The median is 15 + 190/194 of the 4th class. Or, 15 + 190/194(5) = 15 + 4.89 = 19.89 Problem 3.7

Weapon Mean A B C 22.4 18.7 24.6 St dev 15.9 36.5 19.7

Which weapon, on average, comes closest to the target? Which weapon is most reliable? Which should be selected and why? Problem 3.14 X X 17

21 42 32 16 24 31 15 22 26 Count Sum Mean St Dev 10

246 24.6 X -7.6 -3.6 17.4 7.4 -8.6 -0.6 6.4 -9.6 -2.6

1.4 X X 2 57.76 12.96 302.76

54.76 73.96 0.36 40.96 92.16 6.76 1.96 644.4 64.44 8.03 Frequency Distributions Shape

Modality The number of peaks in the curve Skewness An asymmetry in a distribution where values are shifted to one extreme or the other. Kurtosis The degree of peakedness in the curve Frequency Distributions Modality Unimodal Bimodal Multi-modal

Frequency Distributions Skewness Right Skewed (Positively Skewed) Left Skewed (Negatively Skewed) Right (or positively) skewed Note - long tail to right Left (or negatively) skewed Note: long tail to left Kurtosis Platykurtic

Fewer values near the mean, more in the tails. Appears flatter. Leptokurtic More values near the mean, fewer in the tails. Appears taller. Mesokurtic Evenly distributed values. The normal distribution is mesokurtic. The normal distribution The normal distribution, or normal curve, is a family of distributions that have the same general shape. This group of distributions differs in

how spread out their values occur but each share some common characteristics. Overhead/handout Characteristics of the normal distribution Curve is perfectly symmetrical The mean divides the curve exactly in half. Each half is a mirror image of the other. Scores above and below mean are equally like to occur. So, half of the area under the curve is above (or right of) the mean and half is below (or left of) the mean. In other words, half of the cases

fall above and half below the mean. The mean, median, and mode are equal and occur at the highest point of the curve. Characteristics of the normal distribution Most values fall close to the mean. As values fall further from the mean, towards the tails of the distribution, their probability of occurrence decreases. The curve is asymptotic. The tails of the curve get closer and closer to the X axis but never touch it.

A fixed proportion of the values lie between the mean and any other point. Thus, we can compute the probability of any particular value occurring. Differences in normal curves Two parameters affect the shape and location on the X axis of a normal curve. mean standard deviation Differences in normal curves 2 curves with same standard deviation but

different means. Differences in normal curves 2 curves with same mean but different standard deviations. Differences in normal curves 2 curves with same mean but different standard deviations. Standard normal curve The standard normal curve is a normal distribution whose mean = 0 standard deviation = 1

See handout/overhead Portion of values falling under normal curve For all normal distributions, the distance between the mean and one unit of standard deviation includes approximately 34% of the observations. This is true independent of the mean and the standard deviation. Approximately 14% of cases fall between 1 st. dev and 2 st. dev. Approximately 2% of cases fall between

2 st. dev and 3 st. dev. Using the Normal Distribution We can use this information in the following fashion The P(0.0 Xi 1.0) = .3413 Thus 68% of the Xis will fall between 1 Thus 96% of the Xis will fall between 2 Example You make a 95 on a methods exam. How do you feel? What if the highest possible score was 200? What if the highest grade in the class

was 101? What if the average grade was 98? What if the lowest grade was 95? Raw numbers by themselves dont tell us enough. You have to know the distribution. Z scores One of the ways we can take a raw number and interpret it (relative to the mean and standard deviation of the distribution) is to convert it to a standard normal variable or z score. A z score simply converts a raw score

into the number of standard deviation units from the mean that it represents. Z scores can only be computed for data that is normally distributed. Formula case m ean z s ta n d a rd d e v ia tio n x z

Z scores A z score represents a specific point on the x axis of a normal curve. If the z score is positive, that point is above (to right of) the mean. If the z score is negative, that point is below (to left of) the mean. Using a normal distribution table, given any z score, we can determine the percentage of cases falling above/below that score AND the probability of that score occurring. Example - you and your father have the

same IQ score of 132. He took the test in 1968; you in 1998. Did you score the same? Are you equally intelligent? In 1968: In 1998 X 132 X 132 X 100 X 110 16

16 To compare, compute a z score 1968 132 100 z = 2 16 So, dad was smarter than 97.72% of the population in 1968. (50% + 47.72%)

1998 z = 132 110 1 .1 20 So, you are smarter than 86.43% of the population in 1998. (50% + 36.43%)

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