Chapter 6 The Normal
Probability Distribution
Homework #6 (Week 9):
Chapter 6, Exercises 6, 8, 14, 16, 18, 20, 24 &
28.
Probability
Distributions
Definition: "A probability
distribution lists, in some form, all the possible outcomes of an "experiment"
and the probability associated with each one."
Continuous Random
Variables
A continuous random variable
is assumed to be able to take any value in an interval, e.g. height measured
with complete accuracy, time measured with complete accuracy,
temperature measured with complete accuracy,
...
Continuous Probability
Distributions
The total area under the
probability distribution curve bounded by the x-axis is 1 (=100%). The area
under the curve between the lines x = a and x = b gives the probability that x
lies between a and b.
Note: The height of the
curve over a point on the x-axis has no direct probability
meaning.
i.e. P(X = a) = P(X = b) =
0
̃ P(a<X<b) =
P(a£X£b) = P(a<X£b) = P(a£x<b)
The Normal (Gaussian)
Distribution
Intuition: The normal
distribution is given by a symmetrical, bell-shaped curve. The position of the
curve is determined by the expected value (m) and variance
(s2) of the
distribution.
X ~ N(m, s2)
Note: If we fix
m and let s2 vary we get a family of
curves with the same expected values but different variances (i.e. different
levels/degrees of "spreadoutness").
Note: If we fix
s2 and let m vary we get a family of
curves with the same shape but different locations along the horizontal
axis.
Examples: Men's heights,
Women's heights, Men's weights, Women's weights, Number of goals scored by
Premiership teams in a season, Introduction to Sociology grades, Mathematics
& Statistics grades (??), …
The Standard Normal
Distribution
Let Z be a standard normal
variable, that is, a normal variable for which m = 0 and s2 = 1. Statistical tables
allow us to calculate the area under any portion of the standard normal
curve.
Z ~ N(0, 1)
Examples:
(a) P(Z > 1.60) = 0.055
(interpretation?)
(b) P(1.60 < Z < 2.30)
= P(Z > 1.60) - P(Z > 2.30) = 0.055 - 0.011 = 0.044
(interpretation?)
(c) P(Z < 1.64) = 1 - P(Z
> 1.64) = 1 - 0.051 = 0.949 (» 0.95)
(interpretation?)
(d) P(-1.64 < Z <
-1.02) = P(1.02 < Z < 1.64) = P(Z > 1.02) - P(Z > 1.64) = 0.154 -
0.051 = 0.103 (interpretation?)
(e) P(0 < Z < 1.96) =
0.5 - P(Z > 1.96) = 0.5 - 0.025 = 0.475
(interpretation?)
(f) P(-1.96 < Z <
1.96) = 2xP(0 < Z < 1.96) = 0.95
(interpretation?)
(g) P(-1.50 < Z <
0.67) = 1 - P(Z > 0.67) - P(Z < -1.50) = 1 - P(Z > 0.67) - P(Z >
1.50) = 1 - 0.251 - 0.067 = 0.682 (interpretation?)
(h) P(Z < -2.50) = P(Z
> 2.50) = .006 (interpretation?)
The General Normal
Distribution
E(X) = m and Var (X) = s2 i.e. X ~ N(m, s2)
Let Z = (X - m)/s
̃ E(Z) = 0 and Var (Z) = 1
(see previous notes if unconvinced) i.e. Z ~ N(0, 1)
Examples:
E(X) = 25, Var (X) = 25
[Standard Deviation (X) = 5]
X ~ N(25,
25)
(a) P(X > 20) = P((X -
m)/s > (20 - 25)/5) = P(Z
> -1.0) = 1 - P(Z > 1.0) = 1 - 0.159 = 0.841
(interpretation?)
(b) P(X < 40) = P((X -
m)/s < (40 - 25)/5) = P(Z
< 3.0) = 1 - P(Z > 3.0) » 1.0
(interpretation?)
(c) P(21 < X < 30) =
P(-0.8 < Z < 1.0) = 1 - P(Z < -0.8) - P(Z > 1.0) = 1 - 0.212 - 0.159
= 0.629
(interpretation?)
(d) P(18 < X < 23) =
P(-1.4 < Z < -0.4) = P(0.4 < Z < 1.4) = P(Z > 0.4) - P(Z >
1.4) = 0.345 - 0.081 = 0.264
(interpretation?)
IQ (the intelligence
quotient) is Normally distributed with (population) mean 100 and (population)
standard deviation 16.
16
(a) What proportion of the population has an IQ above 120?
P(IQ > 120) = P(Z >
[120-100]/16) = P(Z > 1.25) = .1056
(b) What proportion of the
population has IQ between 90 and 110?
P(90 < IQ < 110) =
P([90–100]/16 < Z < [110–100]/16) = P(-.625 < Z < .625) = 1 – {2 x
P(Z > .625)} = 1 – {2 x .266} = .47
(c) In the past, about 10%
of the population went to university. Now the proportion is about 30%. What was
the IQ of the ‘marginal' student in the past? What is it
now?
Note: Offensive
assumption.
(i) P(IQ > ?) = .10
̃ P(Z > [?-100]/16) =
.10
̃ [?-100]/16 = 1.28
̃ ? =
120.5
(ii) P(IQ > ?) = .30
̃ P(Z > [?-100]/16) =
.30
̃ [?-100]/16 = .525
̃ ? =
108.4