1. Probability Density Functions
Recall that a random variable X is continuous if
1). possible values of X comprise either a single interval on the number line (for some A < B, any number x between A and B is a possible value) or a union of disjoint intervals;
2). P(X = c) = 0 for any number c that is a possible value of X .
Examples:
X = the temperature in one day. X can be any value between L and H, where L represents the lowest temperature and H represents the highest temperature.
Example 4.3: X = the amount of time a randomly selected customer spends waiting for a haircut before his/her haircut commences.
Is X really a continuous rv? No.
The point is that there are customers lucky enough to have no wait whatsoever before climbing into the barber’s chair, which means
P(X = 0) > 0. Only conditioned on no chairs being empty, the waiting time will be continuous.

2. Probability Density Functions
Let’s consider the temperature example again. We want to know the probability that the temperature is in any given interval. For example, what’s the probability for the temperature between 70◦ and 80◦? Ultimately, we want to know the probability distribution for X .
One way to do that is to record the temperature from time to time and then plot the histogram.
However, when you plot the histogram, it’s up to you to choose the bin size.
But if we make the bin size finer and finer (meanwhile we need more and more data), the histogram will become a smooth curve which will represent the probability distribution for X .

3. Probability Density Functions
bin size 2
bin size 0.8
limit case
(•)
(b)
(c)

4. Probability Density Functions
Definition
Let X be a continuous rv. Then a probability distribution or probability density function (pdf) of X is a function f (x ) such that for any two numbers a and b with a ≤ b, b
P(a ≤ X ≤ b) = f (x )dx a
That is, the probability that X takes on a value in the interval [a, b] is the area above this interval and under the graph of the density function. The graph of f (x ) is often referred to as the density curve.
Liang Zhang (UofU) I

5. Probability Density Functions
CD 0
"q' q N
-I- --+
60 70 50 80 90
Figure : P(60 :':'.: X :':'.: 70)

6. Probability Density Functions
Remark:
For f (x ) to be a legitimate pdf, it must satisfy the following two conditions:
1. f (x ) ≥ 0 for all x ;
2. r ∞ f (x )dx = area under the entire graph of f (x ) = 1. −∞
Liang Zhang (fU)

7. Probability Density Functions
Example:
A clock stops at random at any time during the day. Let X be the time (hours plus fractions of hours) at which the clock stops. The pdf for X is known as
( 1
24 0 ≤ x ≤ 24
0 otherwise
f (x ) =
The density curve for X is showed below:
Liang Zhang (UofU)

8. Probability Density Functions
Example: (continued)
A clock stops at random at any time during the day. Let X be the time (hours plus fractions of hours) at which the clock stops. The pdf for X is known as
( 1
24 0 ≤ x ≤ 24
0 otherwise
f (x ) =
If we want to know the probability that the clock will stop between 2:00pm and 2:45pm, then r 1
P(14 ≤ X ≤ 14.75) =
dx = |14 = 14
Liang Zhang (UofU)

9. Probability Density Functions
Definition
A continuous rv X is said to have a uniform distribution on the interval [A, B], if the pdf of X is ( 1
A−B
A ≤ x ≤ B
otherwise
f (x ; A, B) =
The graph of any uniform pdf looks like the graph in the previous example:
Liang Zhang (UofU)

10. Probability Density Functions
Comparisons between continuous rv and discrete rv:
For discrete rv Y , each possible value is assigned positive probability;
For continuous rv X , the probability for any single possible value is 0!
c+E c−E c
P(X = c) = f (x )dx = lim
f (x )dx = 0 c
E→0
Since P(X = c) = 0 for continuous rv X and P(Y = ct) > 0, we have
P(a ≤ X ≤ b) = P(a < X < b) = P(a < X ≤ b) = P(a ≤ X < b)
while P(at ≤ Y ≤ bt), P(at < Y < bt), P(at < Y ≤ bt) and
P(at ≤ Y < bt) are different.
Liang Zhang (UofU)

11. Normal Distributions
One particularly important class of density curves are the Normal curves, which describe Normal distributions.
All Normal curves are symmetric, single-peaked, and bell-
shaped
A Specific Normal curve is described by giving its mean µ and standard deviation σ.

12. The Rule
The Rule
In the Normal distribution with mean µ and standard deviation σ:
Approximately 68% of the observations fall within σ of µ.
Approximately 95% of the observations fall within 2σ of µ.
Approximately 99.7% of the observations fall within 3σ of µ.

13. The Standard Normal Distribution
All Normal distributions are the same if we measure in units of size σ from the mean µ as center.
The standard Normal distribution is the Normal distribution with mean 0 and standard deviation 1.
If a variable x has any Normal distribution N(µ,σ) with mean µ and
standard deviation σ, then the standardized variable
z  x - μ 
has the standard Normal distribution, N(0,1).
Stems = 10’s Leaves = 1’s
Key
20|3 means 203 pounds

14. The Standard Normal Table
Because all Normal distributions are the same when we standardize, we can find areas under any Normal curve from a single table.
The Standard Normal Table
Table A.3 is a table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the left of z.
Suppose we want to find the proportion of observations from the standard Normal distribution that are less than 0.81.
We can use Table A.3:
P(z < 0.81) = Z
.00
.01
.02
0.7
.7580
.7611
.7642
0.8
.7881
.7910
.7939
0.9
.8159
.8186
.8212

15. Normal Calculations How tall must a man be in the lower 10% for
According to the Health and Nutrition Examination Study of , the heights (in inches) of adult men aged are N(70, 2.8).
How tall must a man be in the lower 10% for
men aged 18 to 24?
N(70, 2.8)
.10
? 70

16. Normal Calculations Z = -1.28
How tall must a man be in the lower 10% for men aged 18 to 24?
N(70, 2.8)
.10
Look up the closest
? 70
probability (closest to 0.10) in z
.07
.08
.09
the table.
Find the corresponding
1.3
.0853
.0838
.0823
standardized score.
-1.2
.1020
.1003
.0985
The value you seek is that many standard deviations
1.1
.1210
.1190
.1170
from the mean.
Z = -1.28

17. Normal Calculations x   z  Z = -1.28 x   z  .10
How tall must a man be in the lower 10% for men aged 18 to 24?
N(70, 2.8)
Z = -1.28
.10
? 70
We need to “unstandardize” the z-score to find the observed value (x):
x  
x   z
x = 70 + z(2.8)
= 70 + [(1.28 )  (2.8)]
= 70 + (3.58) = 66.42
z  
A man would have to be approximately inches tall or less to
place in the lower 10% of all men in the population.