Standard Deviation A measure of variability Commonly used to calculate other statistical measures Shows the distance (difference) between the mean and the steepest part of a normal curve Indicates dispersion, or spread, of scores in a distribution

Standard Deviation The smaller the standard deviation, the more tightly clustered the scores The larger the standard deviation, the more spread out the scores

Standard Deviation On an average day in Los Angeles, the number of sudden deaths from cardiac causes is about 4.6. On the day of the Northridge earthquake, there were 25 such deaths, 24 of which occurred after 4:31 a.m., when the earthquake struck.

Standard Deviation Make your prediction: Will the average male or the average female who died from sudden cardiac causes on the day of the earthquake be older? A. the average female was older B. the average male was older

Predict whether the ages of the males or the ages of the females who died on the day of the earthquake were more variable. A. ages of males were more variable B. ages of females were more variable

The Data Ages of females who died: 66 92 75 84 83 80 Ages of males who died: 90 38 59 45 47 67 79 75 66 71 68 62 51 55 62 56 56 64

Calculate the Mean 66 92 75 480/6 = 80 84 83 80 480 90 38 59 45 47 67 79 75 66 71 68 62 51 55 62 56 56 64 1111 1111/18 = 61.7

Subtract the mean from each score X - X = s 66 – 80 = -14 92 – 80 = 12 75 – 80 = -5 84 – 80 = 4 83 – 80 = 3 80 – 80 = 0 s s2 -142 = 196 122 = 144 - 52 = 25 42 = 16 32 = 9 02 = 0 399 (sum of squares)

Find the variance; find the square root of the variance There it is! The standard deviation for the ages of the females who died that day is 8.93. Now you try the same formula for the ages of the males. s2/N (N-1) 399/5 = 79.8 sd = s2 79.8 = 8.93