The Normal Distribution 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:
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 = standard deviation
Example 2: Let s = 3, m = 65 Find P(62 < x < 70) When x = 62,
z = 62 – 65 = -
3 = -1 When x = 70,
z = 70 – 65 = 5
= 1.67 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 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 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 x = 72.16 A man 6 ft tall will be taller than 85% of the men.
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