The Normal Distribution Section 4.4   A distribution that is nearly symmetric with most of the data in the center resembling a bell-shaped curve is a normal distribution.   Discrete Variables: values that are distinct and separate, i.e. number of children is discrete.  Values that can be counted: {0, 1, 2, 3, . . .}   Continuous Variables: any value in an interval of the real number system, not distinct or separate values, i.e. height is continuous.   Characteristics of a Normal Distribution: Frequency of data points occur more in the middle and less when farther away from the center. The distribution is symmetric, one side mirrors the other side.   Because of #1 and #2, the mean, median and mode all occur at the center in a normal distribution. Also, in a normal distribution, more than 2/3 of data (68.26%) falls within one standard deviation of the mean.  And 95.44% of the data falls within two standard deviations of the mean.  And 99.74% of the data lie within three standard deviations of the mean. Notation for Sample Data (Subset): Variance = s2 Standard deviation = s Mean = x   For Population Data (Universal Set):   Variance =  S Standard deviation = s  (lowercase sigma) Mean = m     (lower case mu)   Recall relative frequency was a type of probability, i.e. If there are 14 English Majors in a group of 100 students, then          rf = 14/100 = .14 = 14%   The bell-shaped curve represents probability with the area under the curve = 1         P(S) = 1 = 100%   Most of the data (a large percentage) falls in the middle where the area is greatest (dense).  Since normal distributions are symmetric, 50% falls to the right of the center (mean m ) and 50% falls to the left.         P( x < m ) = .5                        P( x > m ) = .5                           _________________|_____|__________|___________                                                             a      m                     b   To find probability of           x < a    or         x > b    or         a < x < b Find the area under the curve corresponding to the endpoints.   In order to find the area under the curve, mathematicians have standardized the bell curve so that the mean m  = 0 and the curve covers three standard deviations.                                   __________________________________                         -3         -2         -1         0          1          2          3            A standardized normal distribution is called a z-distribution, using values of z to differentiate from any normal distribution using x.  The z-value is the standard deviation from m  = 0.  Appendix VI is a body table.  It gives the probability of an interval in the body (middle) of the bell curve.   Example 1:  Using the body table, find: a)     P(0 < z < .75) = .2734   b)     P(z > 1.2)             Subtract the tail from .5                 .5 – .3849 = .1151 That means ~ 11.5% lie to the right of 1.2   c) P( 15 < z < 1.35)               Take the area of  P(0< z < .15) from the area P(0 < z < 1.35)         P(0 < z < 1.35) – P(0< z < .15) = .4115 – .0596 = .3519   d) P(-1.56 < z < 2.11)           Add the Left and Right areas together.                                                   ___________________________________   Since the normal distribution is symmetric, then          P(-1.56 < z < 0) = P(0 < z < 1.56)           P (0 < z < 1.56) + P(0 < z < 2.11)    = .4406 + .4826 = .9232 So 92% of the z values fall in this interval. Converting to Standard Normal Distribution. Every value x in a normal distribution has a corresponding value z in a standard normal distribution.  In order to use the body table, x values must be converted to z values.           Z = x – m                    m  = mean                 s         s                      s  = standard deviation         Example 2:  Let          s = 3,    m = 65 Find         P(62 < x < 70)   When x = 62,                       z = 62 – 65  = - 3  = -1                    3                          3 When x = 70,                       z = 70 – 65  = 5  = 1.67                 3                          3 So we must find          P(-1 < z < 1.67) = P(0 < z < 1) + P(0 < z < 1.67)         = .3413 + .4525 = .7938   Application:  Given heights of men:         m = 68",        s = 4" a)     find the percentage of men over 6 feet tall, find P(x > 72) b)     find the percentage of men between 66 inches and 71 inches tall, find P(66 < x < 71)   a)     z  =  72 – 68  =  4  =  1              4            4 Find          P(z > 1) = .5 – P(0 < z < 1) = .5 –  .3413 = .1587 ~ 16%   b)     z = 66 – 68  = -2 = -1 = .5                       z = 71 – 68  =  3 = .75           4            4      2                                          4           4         Find          P(-.5 < z < .75)  = P(0 < z < .5) + P(0 < z < .75)          = .1915 + .2734 = .4646 ~ 46% of the men are between 5’6” and 5’11”.   How to find z values given their probabilities (percentages)? Going backwards.   Example 3:  Find P(0 < z < c ) = .4495 Look on body table for area = .4495, the z value that corresponds = 1.64  So          P(0 < z < 1.64) = .4495   Find P(z > c) = .6950 = .5 + .1950 Look on body table for area = .1950 gives us z = .51, but our value is to the left of m, so         Z = - .51          P(z > - .51) = .6950   How to find x values given their probabilities  (percentages)? How tall is a man if he is taller than 85% of the men?  So he is at the top 15% of the distribution. Find x when P(x > c) = .15 First look on body table to find z value when area is .5 - .15 = .35                  Z = 1.04          Use conversion equation to find x.         1.04  =  x – 68         4.16  =  x – 68                                  4              x = 72.16        A man 6 ft tall will be taller than 85% of the men.       Back to Statistics Main Page Back to the Survey of Math Ideas Home Page e-mail Questions and Suggestions