4.3 NORMAL PROBABILITY
DISTRIBUTIONS
The Most Important Probability
Distribution in Statistics
Normal Distributions

A random variable X with mean m and
standard deviation s is normally distributed if
its probability density function is given by
 x m 
(1/ 2)

s


e
2
1
f ( x) 
  x  
s 2
where   3.14159 ... and e  2.71828 ...
The Shape of Normal
Distributions
Normal distributions are bell shaped, and
symmetrical around m.
90
m
Why symmetrical? Let m = 100. Suppose x = 110.
f (110) 
1
s 2
 110 100 
(1/ 2)

s 

e
2

1
s 2
 10 
(1/ 2) 
s
e
110
Now suppose x = 90
2
f (90) 
1
s 2
 90 100 
(1/ 2)

s 

e
2

1
s 2
 10 
(1/ 2)

s 

e
2
Normal Probability
Distributions
The expected value (also called the
mean) E(X) (or m) can be any number
 The standard deviation s can be any
nonnegative number
 The total area under every normal curve
is 1
 There are infinitely many normal
distributions

Total area =1; symmetric
around µ
The effects of m and s
How does the standard deviation affect the shape of f(x)?
s= 2
s =3
s =4
How does the expected value affect the location of f(x)?
m = 10 m = 11 m = 12
µ = 3 and s = 1
m
0
3
6
8
9
12
A family of bell-shaped curves that differ
only in their means and standard
deviations.
µ = the mean of the distribution
s = the standard deviation
X
µ = 3 and s = 1
m
0
3
6
3
12
µ = 6 and s = 1
m
0
9
X
6
9
12
X
m
0
3
µ = 6 and s = 2
6
8
3
12
µ = 6 and s = 1
m
0
9
X
6
8
9
12
X
µ = 6 and s = 2
P(6 < X < 8)
0
3
6
9
12
X
Probability = area under the density curve
P(6a < X < 8)
b = area under the density curve
between 6a and 8.
b
X
µ = 6 and s = 2
P(6 < X < 8)
0
3
6
8
9
12
X
Probability = area under the density curve
6
8
P(6a < X < 8)
b = area under the density curve
between 6a and 8.
b
6
8
X
Probability = area under the density curve
6
8
P(6a < X < 8)
b = area under the density curve
between 6a and 8.
b
6
8
X
Probabilities:
area under
graph of f(x)
P(a < X < b)
f(x)
a
b
X
P(a < X < b) = area under the density curve
b
between a and b.
P(a  X  b) =  f(x)dx
P(X=a) = 0
a
P(a < x < b) = P(a < x < b)
Standardizing
Suppose X~N(ms
 Form a new random variable by
subtracting the mean m from X and
dividing by the standard deviation s:
(Xms
 This process is called standardizing the
random variable X.

Standardizing (cont.)
(Xms is also a normal random
variable; we will denote it by Z:
Z = (Xms
  has mean 0 and standard deviation 1:
E(Z) = m = 0; SD(Z) = s 1.
1
 The probability distribution of Z is called
the standard normal distribution.

Standardizing (cont.)
If X has mean m and stand. dev. s,
standardizing a particular value of x tells how
many standard deviations x is above or below
the mean m.
 Exam 1: m=80, s=10; exam 1 score: 92
Exam 2: m=80, s=8;
exam 2 score: 90
Which score is better?

92  80 12
z1 

 1.2
10
10
90  80 10
z2 

 1.25
8
8
90 on exam 2 is better than 92 on exam 1
µ = 6 and s = 2
0
3
6
8
9
12
X
(X-6)/2
µ = 0 and s = 1
.5
-3
-2
-1
.5
0
1
2
3
Z
Pdf of a standard normal rv

A normal random variable x has the
following pdf:
f ( x) 
1
s 2
e

 12  ( x m2 )
s

2


, 
x
Z ~ N (0,1) substitute0 for m and 1 for s
pdf for the standard normal rv becomes
 ( z) 
1
2
e

1 2
z
2
,   z
Standard Normal Distribution
.5
-3
-2
-1
.5
0
1
2
3
Z = standard normal random variable
m = 0 and s = 1
Z
Important Properties of Z
#1. The standard normal curve is
symmetric around the mean 0
#2. The total area under the curve is 1;
so (from #1) the area to the left of 0 is
1/2, and the area to the right of 0 is 1/2
Finding Normal Percentiles by Hand
(cont.)


Table Z is the standard Normal table. We have to convert
our data to z-scores before using the table.
The figure shows us how to find the area to the left when
we have a z-score of 1.80:
Areas Under the Z Curve:
Using the Table
P(0 < Z < 1) = .8413 - .5
= .3413
.50
.3413 .1587
0
1
Z
Standard normal probabilities have been
calculated and are provided in table Z.
P(- <Z<z0)
The tabulated probabilities correspond
to the area between Z= - and some z0
Z = z0
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.5000
0.5040
0.5080
0.5120
0.5160
0.5199
0.5239
0.5279
0.5319
0.5359
0.1
0.5398
0.5438
0.5478
0.5517
0.5557
0.5596
0.5636
0.5675
0.5714
0.5753
0.2
0.5793
0.5832
0.5871
0.5910
0.5948
0.5987
0.6026
0.6064
0.6103
0.6141
0.8413
0.8438
0.8461
0.8485
0.8508
0.8531
0.8554
0.8577
0.8599
0.8621
0.8849
0.8869
0.8888
0.8907
0.8925
0.8944
0.8962
0.8980
0.8997
0.9015
…
1
…
…
…
…
1.2
…
0.01
…
0.00
…
z
…
…
…
…

Example – continued X~N(60, 8)
 X  60 70  60 
P ( X  70)  P 


8 
 8
 P ( z  1.25)
0.8944
0.8944
0.8944
0.8944
0.8944
0.8944
P(z < 1.25) = 0.8944
In this example z0 = 1.25
z
0.0
0.1
0.2
0.8438
0.8461
0.8849
0.8869
0.8888
0.05
0.5199
0.5596
0.5987
0.8485
0.8508
0.8531
0.8907
0.8925
0.8944
0.06
0.5239
0.5636
0.6026
0.07
0.5279
0.5675
0.6064
0.08
0.5319
0.5714
0.6103
0.09
0.5359
0.5753
0.6141
0.8554
0.8577
0.8599
0.8621
0.8962
0.8980
0.8997
0.9015
…
…
…
1.2
0.04
0.5160
0.5557
0.5948
…
0.8413
…
1
0.03
0.5120
0.5517
0.5910
…
0.02
0.5080
0.5478
0.5871
…
0.01
0.5040
0.5438
0.5832
…
0.00
0.5000
0.5398
0.5793
…
…
…
…
Examples
Area=.3980
0

P(0  z  1.27) =
1.27
.8980-.5=.3980
z
A2
0
.55
P(Z  .55) = A1
= 1 - A2
= 1 - .7088
= .2912
Examples
Area=.4875
Area=.0125 -2.24

0
P(-2.24  z  0) = .5 - .0125 = .4875
z
P(z  -1.85)
= .0322
Examples (cont.)
.9968
A1
A1
.1190



A2
A
-1.18
0
P(-1.18 z 2.73)
2.73
z
= A - A1
= .9968 - .1190
= .8778
vi) P(-1≤ Z ≤ 1)
.6826
.1587
.8413
P(-1 ≤ Z ≤ 1) = .8413 - .1587 =.6826
6. P(z < k) = .2514
.2514 .5
.5
-.67
Is k positive or negative?
Direction of inequality; magnitude of probability
Look up .2514 in body of table; corresponding entry is -.67
Examples (cont.) viii)
.7190
.2810
250  275
P( X  250)  P( Z 
)
43
25
P( Z 
)  P ( Z  .58)  1  .2810  .7190
43
Examples (cont.) ix)
.8671
.1230
.9901
ix) P (225  x  375)
 P  22543 275 
x  275
43

375  275
43

 P(1.16  z  2.33)  .9901  .1230  .8671
X~N(275, 43) find k so that P(x<k)=.9846

.9846 P ( x  k )  P x  275  k  275
43
43

P z  k  275
43


 k  275 2.16 ( from standard normal table)
43
 k  2.16 ( 43 )275 367 .88
P( Z < 2.16) = .9846
.9846
Area=.5
.4846
.1587
0
2.16
Z
Example
Regulate blue dye for mixing paint; machine
can be set to discharge an average of m
ml./can of paint.
 Amount discharged: N(m, .4 ml). If more than
6 ml. discharged into paint can, shade of
blue is unacceptable.
 Determine the setting m so that only 1% of
the cans of paint will be unacceptable

Solution
X =amount of dye discharged into can
X ~N(m , .4); determine m so that
P ( X  6)  .01
Solution (cont.)
X =amount of dye discharged into can
X ~N(m , .4); determine m so that
P ( X  6)  .01
.01  P ( x  6)  P

xm
.4

6 m
.4


P z
6 m
.4

m
 6.4
 2.33(from standard normal table)
 m = 6-2.33(.4) = 5.068