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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
0
µ1 µ2
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 −µ
IF X~N(µ, σ) then Z =
~N(0, 1) standard normal.
σ
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.