Download Continuous Probability Distributions

MGMT 201: Statistics
Continuous Probability Distributions (ASW Chapter 6)
 What do we mean when we say continuous random variable? Roughly speaking, a continuous
random variable is one for which we cannot devise some method for counting the possible outcomes.
Note that this specifically requires having an infinite number of outcomes.
 Conditions
 Let f(x) be the probability density associated with sample point x. We can loosely interpret
“density” as a measure of the probability of getting something near x.
 Then,
 f(x)  0
 f(x) = 1. Said differently, the area under the curve f(x) must be one.
 These are necessary conditions for a continuous probability distribution.
 Interpreting a Continuous Probability Density Function
 The probability of getting any point x is zero!
 Fundamental concept: The probability of getting some number between two values x1 and x2 is
equal to the area under the curve f(x) between x1and x2. Graphically, we have the following.
f(x)
P(x1<x<x2)
x1
x
x2
 Types of Continuous Distributions
 Uniform Density Function (Probability Distribution): the probability density function is across
for all values in a specified range.



 1
if a  x  b

f x    b  a
 0 otherwise
 = (a+b)/2
2 = (b-a)2/12
f(x)
1/(a-b)
a



b
x
The probability of getting a number between x1 and x2 is then (x2-x1)/(b-a) (assuming that x2
and x1 are between a and b and x2>x1).
example: Suppose a=6 and b=10. P(x<7) = (7-6)/(10-6) = 0.25.
The uniform distribution has few natural applications, but is often used to generate random
samples. More on this later.

Normal Density Function (Probability Distribution) (i.e., bell curve)

f x  


1
2 
  x   2
e
2 2
  3.14159
e  2.71828
f(x)

x



The two parameters of the distribution are  and .
  can be any number.  is at the peak of the distribution and is equal to both the
median and mode for the distribution.
  can be any positive number.
The distribution is always symmetric.
The greater  is, the more “spread out” the curve is.
f(x)
“small” 
“large” 




x
Excel function: NORMDIST
We know from the Empirical Rule that….
 observations will fall within one standard deviation of the mean 68.26% of the time.
 observations will fall within two standard deviations of the mean 95.44% of the time.
 observations will fall within three standard deviation of the mean 99.72% of the time.
Standard Normal Probability Distribution
 Excel function: NORMSDIST
 We often want to use a normal distribution with =0 and =1. Intuitively, this is
desirable because we can always convert a normal random variable to its standard normal
equivalent by simply replacing each observation with its z-score!
 standard normal tables are easily located (Table 1 of Appendix B) and make calculations
easy.
 example: A random variable has mean 8 and standard deviation 2. If the variables is
normally distributed, what is the probability of getting a number less than 5? …more
than 9? …between 7.44 and 10.68?


less than 5:
 z = (5-8)/2 = -1.5
 The table gives the area between the z and z=0 (the mean). It only lists positive
z-scores because the distribution is symmetric. In other words, the entry for
z=-1.5 is the same as the entry for z=1.5.
 At 1.5, the table reads 0.4332. This means that the area under the curve between
5 and 8 is 0.4332. The area below 8 is, of course, 0.5 so the area below z=-1.5 is
0.5-0.4332 = 0.0668. This is the probability of getting a number less than 5.
 more than 9:
 z=(9-8)/2 = 0.5
 The table gives 0.1915, which is the probability of getting a number between 8
and 9. The probability of getting a number greater than 9 is then 0.5-0.1915 =
0.3085.
 between 7.44 and 10.68
 z7 = (7.44-8)/2 = -0.28
 The table gives 0.1103, which is the probability of getting a number between
7.44 and 8.
 z10 = (10.68-8)/2 = 1.34
 The table gives 0.4099, which is the probability of getting a number between
8 and 10.68.
 The probability of getting a number between 7.44 and 10.68 is then
0.1103+0.4099 = 0.5202.
 example: Suppose a normal r.v. (random variable) has =10 and =2.5. At what value
x0 is P(x<x0) = 0.8?
 We want to find an entry in the table that is close to 0.3. Why? We must add 0.5 to
what is in the table to get P(x<x0). From the normal table, we see that at z=0.84,
N=0.2995.
 Using z=0.84, we see that 0.84 = (x0-10)/2.5. Solving for x0 gives x0 = 12.1.
The Normal Approximation to the Binomial
 When n is “large”, np is not small, and n(1-p) is not small, the normal distribution is a
reasonable approximation of the binomial.
 Rule of Thumb
 n  20; np  5; n(1-p)  5  normal is a reasonable approximation of the binomial.
 To apply the approximation…
…we set  = np and   np1  p  .
…we calculate the probability between x-0.5 and x+0.5. This is necessary to account
for the conceptual difference between the discrete distribution and the continuous
distribution.
example: Suppose we flip a fair coin 100 times.
 What is the probability of getting 54 heads?
 np = 50
 np(1-p) = 25, so  = 5.
 z53.5 = (53.5-50)/5 = 0.7
 z54.5 = (54.5-50)/5 = 0.9
 From the table, z=0.7 corresponds to 0.2580 in probability.
 From the table, z=0.9 corresponds to 0.3159 in probability.
 P(x = 54) = 0.3159-0.2580 = 0.0579
 Binomial model: 0.05796



f(x)
50
54.5
53.5
x


What is the probability of getting between 48 and 60 heads (inclusive)?
 z47.5 = (47.5-50)/5 = -0.5
 z60.5 = (60.5-50)/5 = 2.1
 N(-0.5) = 0.5-0.1915 = 0.3085
 N(2.1) = 0.5+0.4821 = 0.9821
 P(48  x  60) = 0.9821–0.3085 = 0.6736
 Binomial model: 0.6738
The Exponential Density Function (Probability Distribution): If occurrences are Poissondistributed, what is the distribution of the time until the next occurrence?

f x  

1


e
x

Here x  0 and  is the average time between occurrences.
f(x)
1/
x





Excel function: EXPONDIST
The exponential distribution is important for reasons similar to those offered for the Poisson.
Instead of asking the question “how many sales might we get today?”, we can ask “how long
can I expect to wait before the next sale?”
One nice thing about the exponential distribution is that it can be integrated to get the
cumulative probabilities.
P x  x 0   1  e

x0

x0
and Px  x0   e 
example: On average a car dealer sells a car every 3 hours. What is the probability that the
dealer will sell no cars today (assuming an 8 hour day)?
 =3




8
 Px  8  1  Px  8  e 3  0.0695
Note: We must be careful when using the exponential and Poisson distributions because  is
used in both formulas, yet has a different meaning.
example: A customer service line receives, on average, 48 calls in an eight hour day.
 What is the probability that no calls will be received over the next hour?
 Poisson Analysis:
  = 48/8 = 6 calls.



f 0 
6 x e 6
 0.00248
0!
Exponential Analysis:
  = 8/48 = 1/6
 P(x > 1) = e-1/(1/6) = 0.00248
What is the probability that the next call will occur within 15 minutes?
  = 10 minutes (480 minutes / 48 calls).

Px  15  1  e

15
10
 0.7769