Lecture 6
THE NORMAL
DISTRIBUTION
Probability
Density Function of a Continious Random variable:
Consider a Histogram where the heights of the bars equal the relative
frequencies for each class interval. Let the number of observations in the
sample grow indefinitely. Then the number of class intervals can be increased
indefinitely, and the width of the class intervals can be made increasingly
narrower. In the limit the histogram will be approximated by a smooth curve.
This is called the probability density function.
The total area
under the curve must equal 1. The probability that a random variable will
assume a value between any two points, a and b, equals the area
under the random variable’s probability density function over the interval from
a and b.
The Normal
Distribution:
The most popular and widely use continuous probability density is Normal
density given by the formula:
F(x) = 1/{(2p )1/2 s } exp( -1/2[(x-m )/s ]2,
where m is the expected value
(mean), s the standard deviation, e @ 2.718 - the base of the natural logarithms,
and p @ 3.1416.
Normal curves vary
in shapes because differences in mean and variance, but all have the following
characteristics:
1. Symmetrical and bell-shaped. So its
mean=median=mode.
2. Probability that a value will lie within k standard deviations of the mean
is always given by the following rule:
within one s.d. of the mean = 68.3%
within two s.d. of the mean = 95.4%
within three s.d. of the mean= 99.7%
See Diagram 6.4 (page 193).
3. Location and shape determined entirely by m and s . So the distribution X
with mean m and s.d. s 2 is denoted as
N (m , s 2).
Normal distribution
has been found to approximate a wide variety of real life distributions, but
not always.
STANDARD NORMAL
CURVE
For any normal distribution X which is N(m , s 2),
Z = (X-m )/s
has
a standard normal density. So if X (weight of an adult male) is distributed as
N(170, 25) then (X-170)/5 is a standard normal variable or variate.
How to use the
table of the standard normal distribution: See Table A.2 (Page A8 in Text).
The executives at
Liz Claiborne want to know the probability that the height of an adult male lies
between 65 and 68 inches.
We know that the
height distribution is characterized by N(66, 4). So the two points on N(0, 1)
are (65-66)/2 = -0.5 and (68-66)/2 = 1.0.
Now let us find the
area under the standard normal curve between –.5 and 1.0.
From the table A.2,
The probability from 0 to 1.0 is .3413;
The probability from 0 to .5 is .1915;
But by symmetry,
Prob (0 to -.5) = Prob (0 to .5) = .1915.
Hence, the total probability
from -.5 to 1.0 is .3413 + .1915 = .5328.