Download Normal distribution

NORMAL DISTRIBUTION
Normal distribution- continuous
distribution.
Normal density:

bell shaped,

unimodal- single peak at the
center, symmetric.

Completely described by its
center of symmetry - mean μ
and spread - standard
deviation σ.
μ
Random variable with normal
distribution – normal random
variable with mean μ and st. dev.
σ: X~N(μ, σ)
Standard normal random variable:
mean 0 and st. dev. 1: Z~N(0, 1)
NORMAL DISTRIBUTION-CHANGING LOCATION AND SCALE
CHANGING SCALE σ
0.0
0.0
0.1
0.1
0.2
0.2
0.3
0.3
0.4
0.4
CHANGING LOCATION μ
-4
-2
μ1 μ2
0
0=μ1 < μ2 =1
2
4
-4
“peaky”
density
Changes in mean/location cause
shifts in the density curve along
the x-axis.
-2
0
1=σ1 < σ2 = 2
2
4
“flatter”
density
Changes in spread/standard
deviation cause changes in
the shape of the density
curve.
Why Bother with Normal Distributions?

Normal distributions are great descriptions-modelsapproximations for many data sets such as weights,
heights, exam scores, experimental errors, etc.

Great descriptions of results-outcomes of many chance
driven experiments.

Statistical inference based on normal distribution works
well for many (approximately) symmetric distributions.

HOWEVER, remember that not everything or everybody
is normal!
THE EMPIRICAL RULE - THE 68-95-99.7 RULE
Areas under the normal curve are probabilities (as under any density curve).
Special areas under the normal density curve:
 Approximately 68% of the observations fall within 1 standard
deviation of the mean
 Approximately 95% of the observations fall within 2 standard
deviations of the mean
 Approximately 99.7% of the observations fall within 3 standard
deviations of the mean
THE EMPIRICAL RULE - contd
The range "within one/two/three standard deviation(s) of the mean" is
highlighted in green.
The area under the curve over this range is the rel. frequency of observations in
the range.
That is, 0.68/95/99.7 = 68%/95%/99.7% of the observations fall within
one/two/three standard deviation(s) of the mean, or, 68%/95%/99.7% of the
observations are between (μ – 1/2/3σ) and (μ + 1/2/3σ).
AREAS UNDER THE NORMAL CURVE
Normal probabilities = areas under the normal curve are tabulated for
the standard normal distribution (front cover of your text book).
In looking for probabilities keep in mind:
Symmetry of the normal curve and P(Z=a)=0 for any a.
FIND:
P(Z < 0.01) = 0.504
P(Z ≤ -0.01 ) = 0.496
P(Z < 0) = 0.5
P( Z < 2.92)= 0.9982
P(Z>2.92)=1-0.9982=0.0018 or, by symmetry =P(Z< - 2.92)=0.0018
P(-1.32< Z <1.2)=0.8849 – 0.0934=0.7915
SUMMARY OF RULES we used above: P(Z>a)=1-P(Z< -a)
P(a < Z < b) = P(Z < b)- P( Z < a)
GENERAL NORMAL DISTRIBUTION
X 
IF X~N(μ, σ) then Z =
~N(0, 1) standard normal.

standardization
Example. Suppose that the weight of people in NV follows normal
distribution with mean 150 and standard deviation 20 lb. Find the
probability that a randomly selected Nevadan weighs
at most 160 lb; b) over 160 lb.
Solution. Let X= weight of a randomly selected Nevadan. X~N(150, 20).
a)
 X  150 160  150 
P

P(X ≤ 160) = 
  P( Z  0.5)  0.6515.
20
 20

b)
P(X>160)= 1 - P(X ≤ 160) =1 – 0.6915 = 0.3085.
NORMAL PERCENTILES
Given that P(Z < p)=0.95 find p. Here p is called 95th percentile of Z.
Inside the table I looked for 0.95.
Found 0.9495 and 0.9505.
Used z-value corresponding to
the midpoint (0.95) between the
two available probabilities 1.645.
p=1.645
If an available probability is closer
to the one we need, use the
z-value corresponding to
that probability.
0.95
p=?
NORMAL PERCENTILES, CONTD.
EXAMPLE. Suppose scores X on a test follow a normal distribution with mean
430 and standard deviation 100. Find 90th percentile of the scores, that is find
score x such that P(X ≤ x)=0.9.
Solution. Since we start with a normal but NOT STANDARD normal distribution,
we have to standardize at some point:


x  430
 X  430 x  430 
P

 P( Z 
).

100 
100
 100

Z

z
0.9 = P(X ≤ x) =
get equation:
0.90
z =1.28
x  430
 1.28
100
x - 430 =128
x = 558
90% of students scored 558 or less.
EXAMPLE
Height of women follows normal distribution with mean 64.5 and
standard deviation of 2.5 inches. Find
a) The probability that a woman is shorter than 70 in.
b) The probability that a woman is between 60 and 70 in tall.
c) What is the height 10% of women are shorter than, i.e. what is the 10th
percentile of women heights?
SOLUTION. X= women height; X~N(64.5, 2.5).
a) P(X <70)=P(Z< (70-64.5)/2.5)=P(Z<2.2)=0.9861
b) P( 60 < X < 70) = P( (60-64.5)/2.5) < Z < (70-64.5)/2.5)=P(-1.8< Z <
2.2)= P( Z <-2.2) – P( Z < -1.8) = 0.9861 – 0.0359 = 0.9502.
c) 10th percentile of X =?
0.1=P( X< x) = P( Z< (x-65.5)/2.5), so -2.33=(x-65.5)/2.5; x=59.675.
10% of women are shorter than 59.675 in.