Download The Standard Normal Distribution - MATH 130, Elements of Statistics I

The Standard Normal Distribution
MATH 130, Elements of Statistics I
J. Robert Buchanan
Department of Mathematics
Fall 2015
Objectives
During this lesson we will learn to:
I
use the uniform probability distribution,
I
graph a normal curve,
I
state the properties of the normal curve,
I
explain the role of area in the normal density function.
Properties of the Standard Normal Curve
1. It is symmetric about its mean µ = 0 and has standard
deviation σ = 1.
2. The mean = mode = median, and thus the peak of the
curve occurs at µ = 0.
3. It has inflection points at µ − σ = −1 and µ + σ = 1.
4. The area under the curve is 1.
5. The area under the curve to the left of µ = 0 is 1/2 which
is also the area under the curve to the right of µ = 0.
6. The PDF approaches 0 but never reaches 0.
7. The Empirical Rule applies.
Empirical Rule
0.34
PDF
0.34
0.0235
-3
-2
0.135
0.135
-1
0
Z
1
0.0235
2
3
Finding the Area to the Left of Z
We will use Table V (pages A–11 and A–12) to look up areas
associated with different values of the standard normal random
variable Z .
0.9505
1.65
Z
On the “Formulas and Tables” card, Table V is only one page.
Examples
Find the areas under the standard normal curve associated
with the following Z -scores.
1. To the left of Z = 1.43.
2. To the left of Z = −1.34.
3. To the right of Z = 1.17.
4. To the right of Z = −0.66.
5. Between Z = −0.49 and Z = 1.08.
6. Between Z = 1.13 and Z = 2.05.
Left of Z = 1.43
1.43
Z
Left of Z = 1.43
1.43
Area = 0.9236
Z
Left of Z = −1.34
Z
-1.34
Left of Z = −1.34
Z
-1.34
Area = 0.0901
Right of Z = 1.17
1.17
Z
Right of Z = 1.17
1.17
Area = 1 − 0.8790 = 0.1210
Z
Right of Z = −0.66
Z
-0.66
Right of Z = −0.66
Z
-0.66
Area = 1 − 0.2546 = 0.7454
−0.49 < Z < 1.08
-0.49
1.08
Z
−0.49 < Z < 1.08
-0.49
Area = 0.8599 − 0.3121 = 0.5478
1.08
Z
1.13 < Z < 2.05
1.13
2.05
Z
1.13 < Z < 2.05
1.13
Area = 0.9798 − 0.8708 = 0.1090
2.05
Z
Finding Z -scores Given the Area
If we are given the area under the standard normal curve, we
can search for the closest area found in Table V and look up the
Z -score corresponding to this area.
Finding Z -scores Given the Area
If we are given the area under the standard normal curve, we
can search for the closest area found in Table V and look up the
Z -score corresponding to this area.
Example
Find the Z -scores associated with the following areas under the
standard normal curve.
1. Area to the left is 0.68.
2. Area to the left is 0.25
3. Area to the right is 0.72.
4. Area to the right is 0.85.
Left Area 0.68
z
Z
Left Area 0.68
z
z = 0.47
Z
Left Area 0.25
z
Z
Left Area 0.25
z
z = −0.67
Z
Right Area 0.72
z
Z
Right Area 0.72
z
z = −0.58
Z
Right Area 0.85
z
Z
Right Area 0.85
z
z = −1.04
Z
Area in the Middle
Example
Find the Z -scores associated with the following areas under the
standard normal curve.
1. Middle 80%.
2. Middle 90%.
3. Middle 95%.
4. Middle 99%.
Middle 80%
-z
z
Z
Middle 80%
-z
z = ±1.28
z
Z
Middle 90%
-z
z
Z
Middle 90%
-z
z = ±1.645
z
Z
Middle 95%
-z
z
Z
Middle 95%
-z
z = ±1.96
z
Z
Middle 99%
-z
z
Z
Middle 99%
-z
z = ±2.575
z
Z
zα
Definition
The notation zα is the Z -score such that the area under the
standard normal curve to the right of zα is α.
Α
zΑ
Z
Examples
Find the Z -scores associated with the following zα ’s.
1. z0.20
2. z0.15
3. z0.10
4. z0.05
z0.20
z0.20
Z
z0.20
z0.20
z0.20 = 0.841
Z
z0.15
z0.15
Z
z0.15
z0.15
z0.15 = 1.036
Z
z0.10
z0.10
Z
z0.10
z0.10
z0.10 = 1.282
Z
z0.05
z0.05
Z
z0.05
z0.05
z0.05 = 1.645
Z
Area and Probability
The area under the standard normal curve is the probability
that Z lies in a particular interval.
P(a < Z < b) represents the probability that a standard normal
random variable is between a and b.
P(Z > a) represents the probability that a standard normal
random variable is greater than a.
P(Z < a) represents the probability that a standard normal
random variable is less than a.
We will not distinguish between strict (<, >) and non-strict (≤,
≥) inequality.
Examples
Find the following probabilities associated with the standard
normal random variable Z .
1. P(Z < 1.39)
2. P(Z < −1.21)
3. P(Z ≥ 1.42)
4. P(Z > −1.84)
5. P(−2.03 ≤ Z < 1.09)
6. P(−1.30 ≤ Z < 3.03)
P(Z < 1.39)
1.39
Z
P(Z < 1.39)
1.39
P(Z < 1.39) = 0.9177
Z
P(Z < −1.21)
Z
-1.21
P(Z < −1.21)
Z
-1.21
P(Z < −1.21) = 0.1131
P(Z ≥ 1.42)
1.42
Z
P(Z ≥ 1.42)
1.42
Z
P(Z ≥ 1.42) = 1 − P(Z < 1.42) = 1 − 0.9222 = 0.0778
P(Z > −1.84)
Z
-1.84
P(Z > −1.84)
Z
-1.84
P(Z > −1.84) = 1 − P(Z ≤ −1.84) = 1 − 0.0329 = 0.9671
P(−2.03 ≤ Z < 1.09)
-2.03
1.09
Z
P(−2.03 ≤ Z < 1.09)
-2.03
1.09
P(−2.03 ≤ Z < 1.09) = 0.8621 − 0.0212 = 0.8409
Z
P(−1.30 ≤ Z < 3.03)
-1.30
3.03
Z
P(−1.30 ≤ Z < 3.03)
-1.30
3.03
P(−1.30 ≤ Z < 3.03) = 0.9988 − 0.0968 = 0.9020
Z