Download Chapter 4 Probability Distributions

Chapter 7 Continuous Probability Distributions
Homework #8 Part 1 (Week 2 Hilary Term): Chapter
7 Exercises 6, 16, 22 & 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 xaxis 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 () and
variance (2) of the distribution.
X ~ N(, 2)
Note: If we fix  and let 2 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 2 and let  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  = 0 and 2 = 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) = 1 - .4452 = .0548 (interpretation?)
(b) P(1.60 < Z < 2.30) = .4893 - .4452 = 0.0441
(interpretation?)
(c) P(Z < 1.64) = .5 + .4495 = .9495 ( 0.95)
(interpretation?)
(d) P(-1.64 < Z < -1.02) = P(1.02 < Z < 1.64) = .4495 .3461 = .1034 (interpretation?)
(e) P(0 < Z < 1.96) = .4750 (interpretation?)
(f) P(-1.96 < Z
(interpretation?)
<
1.96)
=
2x.4750
=
.95
(g) P(-1.50 < Z < 0.67) = .4332 + .2486 = .6818
(interpretation?)
(h) P(Z < -2.50) = P(Z > 2.50) = .5 - .4938 = .0062
(interpretation?)
The General Normal Distribution
E(X) =  and Var (X) = 2 i.e. X ~ N(, 2)
Let Z = (X - )/
 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 - )/ > (20 - 25)/5) = P(Z > -1.0)
= .5 + .3413 = 0.8413 (interpretation?)
(b) P(X < 40) = P((X - )/ < (40 - 25)/5) = P(Z < 3.0)
= .5 + .4987 = .9987  1.0 (interpretation?)
(c) P(21 < X < 30) = P(-0.8 < Z < 1.0) = .2881 + .3413 =
0.6294 (interpretation?)
(d) P(18 < X < 23) = P(-1.4 < Z < -0.4) = P(0.4 < Z <
1.4) = .4192 - .1554 = .2638 = 0.264 (interpretation?)
IQ Example
IQ (the intelligence quotient) is Normally distributed
with (population) mean 100 and (population)
standard deviation 16.
IQ ~ N(100, 162)  IQ – 100 ~ N(0, 1)
16
(a) What proportion of the population has an IQ
above 120?
P(IQ > 120) = P(Z > [120-100]/16) = P(Z > 1.25) = .5 .3944 = .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) = 2 x.2340 = .4680
(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