Transcript Stat 281 Chapter 6

Chapter 6
Continuous Distributions
The Gaussian (Normal) Distribution
When Discrete Distributions Aren’t
Enough
• Discrete distributions are used in situations involving
counts. (Others are possible but this is the vast majority.)
• What happens when you want to measure things?
– Height
– Weight
– Miles per Gallon
• These aren’t counts. (Why not?)
• Measurements involve rounding and precision.
• When any level of precision is theoretically possible, we
call this a “continuous” variable.
• The values come from the set of Real Numbers, ie, the
number line.
Real Numbers
• -------------------------------------------------------
-∞ … -3 -2 -1 0 1 2
3 … ∞
• The Real Numbers include all possible values
between the pictured integers.
• That includes rational numbers like ½, 1/3,
237/573, etc.
• It also includes irrational numbers like π and √2.
• Real numbers have an infinite string of decimal
places.
• There are “uncountably many” real numbers
between any two specified real numbers.
Intervals
• An interval is a “piece” of the number line, or a
subset of the Real numbers.
• There are no “gaps.” For any two numbers in it,
all real numbers between them are included.
• Therefore an interval is described by its
endpoints—with a few special considerations.
• The endpoints may or may not be included.
Round brackets are used to exclude the
endpoints, square brackets to include them.
Write in order. Ex.: [0,1], (9, 100), [3,6), (0,7].
• If an interval goes on to infinity, the ∞ or -∞
symbol is used with a round bracket, since
infinity is not a number. Ex.: [0,∞), (-∞,-10).
Definition of Continuous R.V.
• A continuous random variable takes on values in
some Real Interval.
• -------------------------------------------------------
-∞ … -3 -2 -1 0 1 2
3 … ∞
• Suppose a r.v. X takes values in [0,1].
How many different values are there?
• Suppose you assign some tiny probability to
each Real Number in [0,1]. What is the total
probability?
• Suppose you divide [0,1] up into 10 subintervals.
• Can you assign probabilities to these so the total
is 1?
Definition of Continuous R.V.
• This illustrates the problem with assigning
probabilities to individual numbers, and the
contrasting ease of assigning probability to
intervals.
• Summary:
– Any continuous distribution has infinitely many values.
– No single point has a positive probability.
– Said another way: Every individual value of a
continuous random variable has probability zero, and
as such is an impossible event.
– Intervals can be assigned positive probability.
The Paradox
• Obviously, a r.v., X, must take on some value, and if it
does, that value is not impossible (it has P>0).
• We never actually “mean” a single value. Measurements
are given with a certain precision.
• Example: temperature is continuous, but measured to
the nearest degree, “70” really means the interval
[69.5,70.5).
• Intervals can have positive probability, and we can make
them as small as we like.
• The fact that a continuous r.v. cannot take a single value
agrees nicely with the fact that it is impossible to
measure anything to the exact real number value.
• Instead, we divide up our scale using equal-width
subintervals based on the precision of the measuring
device. These subintervals have positive probability.
Continuous Probabilities
• Probabilities for a continuous random variable,
X, are given by a probability function, P.
• P(X=k)=0 for any k.
• We might find positive probabilities for
expressions like
– P(X>k),
– P(X<k), or
– P(a<X<b).
Note: the interval is (k,∞)
Note: the interval is (-∞,k)
Note: the interval is (a,b)
• A formula that gives probabilities for X would
need to give probabilities for intervals, rather
than single values.
Has anything prepared us for this?
• Tables of probability for discrete r.v.’s?
Not if only individual values were given.
• Ungrouped histograms? No, same.
• Grouped histograms? Let’s see….
• Each bar represents
a frequency for an
interval, even though
this is a discrete
example.
What about relative histograms?
• Look at the histogram for the number of three’s
showing in a two-dice toss.
• Notice it shows the probabilities for 3 discrete values.
• Replace the discrete
values with intervals,
[0,1), [1,2), and [2,3).
• Then this histogram
looks like it belongs
to a continuous
distribution with
values in [0,3).
0.7
Relative Frequency
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
Three's Face Up
2
Making the Leap
• Change the horizontal axis to show that the bars
belong to each interval.
• Each bar is 1 unit wide and its height represents the
probability for that interval.
• Each bar is a rectangle, whose area is 1 x height.
• Since the heights add up
to 1, the total area of the
shaded region is 1.
• Make the transition to the
continuous case: Instead
of representing probability
by height, use area.
What did we leap over?
• This has been more of an analogy than an explanation.
Many details that require calculus are glossed over.
• The problem: can’t represent probabilities by height at a
point, because points all have probability zero.
• Solution: switch to areas, where the bottom boundary
(on the x axis) represents an interval for which we want
to determine probability. The area of the graph above
that interval represents its probability.
• In calculus, these areas are called “definite integrals.”
You don’t really need to know that, but you may come
across the following symbol, which means “the integral
from a to b.” b

a
Uniform Distribution
• A uniform distribution is defined for an interval outside of
which there is no positive probability. (This is to prevent
the area from being infinite.)
• Inside that interval, it has the same probabilities for any
sub-interval of a given size (they are “always the same”).
• A uniform distribution on the interval [0,3] is shown here.
Note that the height is 1/3, because 3 x 1/3 = 1.
• However, we should not
say that 1/3 is the
probability of anything
in particular.
Uniform Examples
• Let X be a uniform r.v. on the interval [1,5].
• Find P(X>3), P(X<5), P(2<X<3), and P(0<X<3).
• Solution: The width of the distribution is 4, so
the height of the graph is ¼ between 1 and 5.
The area for any interval will be ¼ x the width of
the interval.
–
–
–
–
P(X>3)=(5-3)/4=1/2
P(X<5)=(5-1)/4=1
P(2<X<3)=(3-2)/4=1/4
Careful! P(0<X<3)=P(1<X<3)=(3-1)/4=1/2
Uniform Examples
• Let X be a uniform r.v. on the interval [1,5].
• Find P(X>3), P(X<5), P(2<X<3), and P(0<X<3).
• Solution: The width of the distribution is 4, so
the height of the graph is ¼ between 1 and 5.
The area for any interval will be ¼ x the width of
the interval.
–
–
–
–
P(X>3)=(5-3)/4=1/2
P(X<5)=(5-1)/4=1
P(2<X<3)=(3-2)/4=1/4
Careful! P(0<X<3)=P(1<X<3)=(3-1)/4=1/2
More Uniform Examples
• Let X be a uniform r.v. on the interval [0,8].
– Find P(X>3), P(X<5), and P(2<X<3 or 7<X<8).
– Find the median and the 90th percentile.
• Solution: The width of the distribution is 8, so
the height of the graph is 1/8.
–
–
–
–
P(X>3)=(8-3)/8=5/8.
P(X<5)=(5-0)/8=5/8.
P(2<X<3 or 7<X<8)=P(2<X<3)+P(7<X<8)=1/4.
The median must have half the probability above it
and half below. Therefore the median is 4.
– P90 is a number such that 90% of the probability is
below it, so we have (P90-0)/8=.9, so P90 =7.2.
Probability Density Function
• We have been dealing with the uniform distribution in
terms of graphs. Before moving on, we need to put
these ideas into the form of mathematical notation.
• We were focusing on the areas of portions of a graph
like the one below. But how do we define the region we
want the area for?
– The bottom boundary is the x axis
– The sides are vertical lines going through the x values we want
– The top of the region is a special “curve” (straight lines are
curves too).
• This curve is defined by a function, called the probability
density function, or pdf. For our graph, it is:
1/ 4 if 1  x  5
f ( x)  
 0 otherwise
Normal Probability Distributions
• The normal probability distribution
(Gaussian Distribution) is the most
important distribution in all of statistics.
• Many continuous random variables have
normal or approximately normal
distributions.
• A normal distribution is defined by its pdf.
The Normal pdf
1
f ( x) 
 2
•
•
•
•
•
e
1  x 
 

2  
2
The parameters are μ and σ.
The mean of the distribution is μ.
The standard deviation is σ.
The median and mode are also μ.
There is a normal distribution for every
combination of values of μ and σ
Basic Shape
• Here we see the
basic shape of a
normal distribution.
• The blue band is an
example of an “area
under the curve” that
we might want to
calculate.
• This particular distribution has μ=110 and σ=10.
• The “x” axis represents values of the r.v. X.
Effect of Changing μ
Changing μ just causes a horizontal shift,
centering the graph in a different place.
Effect of Changing σ
• Changing σ causes the graph to stretch out or squeeze
together around the mean.
What does this mean?
• The normal pdf is a complicated formula. It is not easy
to calculate probabilities from it, even if you know
calculus. So, we use tables (or computers).
• We can’t have a table for every possible normal
distribution.
• We have one table for
the “standard” normal
distribution, which has
μ=0 and σ=1. This r.v.
is called Z.
• It is easy to convert probability statements from
other normal distributions
to Z.
Table 3, Appendix B entries:
0
z
The table contains the area under the standard normal
curve between 0 and a specific value of z.
Example: Find the area under the standard normal
curve between z = 0 and z = 1.45.
0
1.45
A portion of Table 3:
z
0.00
0.01
1.4
P(0  z  1.45)  0.4265
0.02
0.03
0.04
0.05
0.4265
0.06
Example: Find the area under the normal curve to the
right of Z = 1.45; P(Z > 1.45).
Area asked for
0.4265
0
1.45
P( Z  1.45)  0.5000  0.4265  0.0735
Example: Find the area to the left of Z = 1.45; P(Z <
1.45).
0.5000
0.4265
0
1.45
P( Z  1.45)  0.5000  0.4265  0.9265
Example:
Find the area between Z = 1.26 and the mean (Z = 0).
Area from table
0.3962
Area asked for
1.26
0
P(1.26  Z  0)  0.3962
1.26
Example: Find the area to the left of .98; P(Z < .98).
Area from table
0.3365
Area asked for
Same as
area asked
for
.98
0
.98
P( Z  .98)  0.5000  0.3365  0.1635
Applications of Normal Distributions
• Apply the techniques learned for the Z distribution to
all normal distributions.
• Start with a probability question in terms of x-values.
• Convert, or transform, the question into an equivalent
probability statement involving z-values.
Standardization
Suppose X is a normal r.v. with mean  and standard
deviation .
X 
The r.v. Z 
has a standard normal distribution.


0
x
x

Example: A bottling machine is adjusted to fill bottles with a
mean of 32.0 oz of soda and standard deviation of 0.02.
Assume the amount of fill is normally distributed and a bottle
is selected at random.
1. Find the probability the bottle contains between 32 oz and
32.025 oz.
2. Find the probability the bottle contains more than 31.97 oz.
32   32  32
When x  32; z 

0

.02
32   32.025  32
When x  32.025; z 

 1.25

.02
 32  32 X  32 32.025  32 
P(32  X  32.025)  P 



.02
.02
 .02

 P(0  Z  1.25)  .3944
Other Normal Applications
Find a cutoff point: a value of X such that there is a
certain probability in a specified interval defined by x.
Example:
The waiting time X at a certain bank is approximately
normally distributed with a mean of 3.7 minutes and a
standard deviation of 1.4 minutes. The bank would
like to claim that 95% of all customers are waited on
by a teller within c minutes. Find the value of c that
makes this statement true.
Solution:
0.0500
0.5000 0.4500
3.7
0
P( X  c)  .95
 X  3.7 c  3.7 
P

  .95
1.4 
 1.4
c  3.7 

PZ 
  .95
1.4 

c
1645
.
x
z
c  3.7
 1645
.
14
.
c  (1645
. )(14
. )  3.7  6.003
c  6 minutes
Notation:
If X is a normal random variable with mean  and
standard deviation , this is often denoted: X ~ N(, 2).
Example: Suppose X is a normal random variable with 
= 35 and  = 6. A convenient notation to identify this
random variable is: x ~ N(35, 36).
z(a) and za are commonly used notations for the zscore (point on the z axis) such that there is a of the
area (probability) to the right of z(a) or za .
Illustrations:
z(0.10) represents the
value of Z such that the
area to the right under
the standard normal
curve is 0.10
010
.
0
z(0.80) represents the
value of Z such that the
area to the right under
the standard normal
curve is 0.80
z(010
. )
z
0.80
z(0.80) 0
z
Example: Find the numerical value of z(0.10).
Table shows this area (0.4000)
0.10 (area information
from notation)
0
z(010
. )
z
Use Table 3: look for an area as close as possible to
0.4000
z(0.10) = 1.28
Note:
The values of Z that will be used regularly come from
one of the following situations:
1. The z-score such that there is a specified area in one
tail of the normal distribution.
2. The z-scores that bound a specified middle
proportion of the normal distribution.
Example: Find the z-scores that bound the middle 0.99
of the normal distribution.
0.005
0.005
0.495
z(0.995)
or
 z(0.005)
0.495
0
z(0.005)
Use Table 3:
z(0.005)  2.575 and z(0.995)   z(0.005)  2.575