Business Statistics: Quantitative Methods and Techniques - Lecture 7: Continuous Probability Distributions
Continuous Probability Distribution
Continuous Probability Distributions: This lecture focuses on continuous probability distributions, including exponential and normal distributions.
Agenda: The lecture covers continuous probability distributions, probability density functions, exponential distribution, normal distribution, and the use of the standard normal table.
Probability Distributions
Types of Probability Distributions:
Discrete Probability Distributions: Binomial, Poisson
Continuous Probability Distributions: Exponential, Normal
Continuous Random Variables
Discrete Random Variables:
Finite number of possible values (e.g., binomial variable: x = 0, 1, 2, …, n).
Countably infinite number of possible values (e.g., Poisson variable: x = 0, 1, …).
Continuous Random Variables:
Uncountably infinite number of possible values within an interval (e.g., x can take all values in the interval [a, b]).
Example: Amount of time (seconds) workers take to complete a task ➔ x can take all non-negative real numbers.
Properties of Random Variables
Discrete Random Variables:
All possible values of random variable X can be listed.
It is meaningful to consider probability P(X = x).
Continuous Random Variables:
All possible values cannot be listed; there is always another possible value between any two of its values.
The probability that a continuous random variable X will assume any particular value is zero: P(X = x) = 0.
The only meaningful events for a continuous random variable are intervals.
Probability Density Functions
Definition: To calculate probabilities, we define a probability density function f(x).
Probability Calculation: The probability that X falls between a and b is found by calculating the area under the graph of f(x) between a and b.
Conditions for f(x):
f(x) is non-negative.
The total area under the curve representing f(x) equals 1.
Important: f(x)
eq P(X = x).
Normal Distribution
Importance: Most important continuous distribution in statistics.
Modeling: Many random variables can be properly modeled as normally distributed.
Approximation: Many distributions can be approximated by a normal distribution.
Statistical Inference: The normal distribution is the cornerstone distribution in statistical inference; the (asymptotic) distribution of sample means is normal.
Normal Distribution - Definition
Notation:
Normal Distribution: A random variable X is said to have a normal distribution with parameters \mu and \sigma if its p.d.f. is:
f(x) = \frac{1}{\sigma \sqrt{2 \pi}} exp \left[ -\frac{1}{2} \left( \frac{x - \mu}{\sigma} \right)^2 \right]
where \pi = 3.14159… and e = 2.71828… and - \infty < x < \inftyE(X) = \mu
V(X) = \sigma^2
Also: X \,\tilde{}\ N(\mu, \sigma^2)
Normal Distribution - Shape
The p.d.f. of the normal distribution is bell-shaped and symmetrical around \mu.
Normal Distribution - Shape (cont.)
How the standard deviation affects the shape of f(x):
\sigma = 2
\sigma = 3
\sigma = 4
How the expected value affects the location of f(x):
\mu = 10
\mu = 11
\mu = 12
Standard Normal Distribution
Facts: Two facts help calculate normal probabilities:
The normal distribution is symmetrical.
Any normal distribution can be transformed into a specific normal distribution, called standard normal distribution, with mean 0 and variance 1.
Example: The time it takes to complete a standard entrance exam is assumed to be normally distributed, with a mean of 60 minutes and a standard deviation of 8 minutes. What is the probability that a student will complete the exam within 70 minutes?
Standard Normal Distribution – cont.
Solution:
If X denotes the time taken to complete the exam, we look for the probability P(X < 70) where \mu = 60 and \sigma = 8.
This probability can be calculated by creating a new normal variable, the standard normal variable (Z).
E(Z) = 0
V(Z) = 1
Z measures the number of standard deviation units \sigma that X deviates from its mean \mu.
Therefore, once probabilities for Z are calculated, probabilities of any normal variable can be found.
Every normal variable X with some \mu and \sigma can be transformed into this Z:
Z = \frac{X - \mu}{\sigma}
Standard Normal Distribution - cont.
To complete the calculation, we need to compute the standard normal probability P(Z < 1.25).
Solution (cont.):
P(X < 70) = P(X - \mu < 70 - 60) = P\left( \frac{X - \mu}{\sigma} < \frac{70 - 60}{8} \right) = P(Z < 1.25)
Standard Normal Table
Solution (cont.):
Standard normal probabilities have been calculated and are provided in a table.
The tabulated probabilities correspond to the area between Z = -\infty and some Z = z_0
Standard Normal Table
Solution (cont.):
In this example, z_0 = 1.25
P(X < 70) = P(Z < 1.25) = 0.8944
Calculating Normal Probabilities
Probabilities of the type P(Z > z_0) can easily be derived from Table 3 by applying the complement rule.
P(Z > 1.25) = 1 - P(Z < 1.25) = 1 - 0.8944 = 0.1056
From the symmetry of the normal distribution, it follows that P(Z < -1.25) = P(Z > 1.25) = 0.1056
Hence, P(Z < -z) + P(Z < z) = 1 for any z
P(Z < -z) = P(Z > +z)
Calculating Normal Prob. – examples
(a) Find the probability P(Z > 1.47)
P(Z > 1.47) = 1 - P(Z < 1.47) = 1 - 0.9292 = 0.0708
Tip: Always first sketch the distribution and the relevant area!
Calculating Normal Prob. – examples
(b) Find the probability P(-2.25 < Z < 1.85)
P(Z < -2.25) = 0.0122
P(Z < 1.85) = 0.9678
P(-2.25 < Z < 1.85) = P(Z < 1.85) - P(Z < -2.25) = 0.9678 - 0.0122 = 0.9556
Normal Distribution - Applications
Example:
The rate of return (X) on an investment is normally distributed with a mean of 30% and a standard deviation of 10%.
What is the probability that the return will exceed 55%?
P(X > 55) = P\left(Z > \frac{55 - 30}{10} \right) = P(Z > 2.5) = 1 - P(Z < 2.5) = 1 - 0.9938 = 0.0062
Normal Distribution - Applications
Example (cont.):
What is the probability that the return will be less than 22%?
P(X < 22) = P\left(Z < \frac{22 - 30}{10} \right) = P(Z < -0.8) = 0.2119
Normal Distribution - Applications
Example:
If Z is a standard normal variable, determine the value z for which P(Z < z) = 0.6331
P(Z < 0.34) = 0.6331 ➔ z = 0.34
Normal Distribution - Applications
Example: Determine z_{0.025} (critical value).
Solution: The value zA is defined as the z-value for which the area to the right(!) of zA under the standard normal curve is A:P(Z > zA) = A (right-tail probability)
P(Z < 1.96) = 0.975 ➔ z_{0.025} = 1.96
-1.96 = -z_{0.025}
Normal Distribution - properties and
If X \sim N(\muX, \sigmaX^2) and Y \sim N(\muY, \sigmaY^2) and X and Y are independent random variables, then:
aX + bY \sim N(a\muX + b\muY, a^2\sigmaX^2 + b^2\sigmaY^2)
aX - bY \sim N(a\muX - b\muY, a^2\sigmaX^2 + b^2\sigmaY^2)
Exercise: Sum of Independent Normal Random Variables
Let Xi denote the weight of a randomly selected prepackaged one-pound bag of carrots. Xi is normally distributed with a mean of 1.18 pounds and a standard deviation of 0.07 pound.
Let W denote the weight of randomly selected prepackaged three-pound bag of carrots. W is normally distributed with a mean of 3.22 pounds and a standard deviation of 0.09 pound.
Selecting bags at random, what is the probability that the sum of three one-pound bags exceeds the weight of one three-pound bag?
Exercise Solution
Because the bags are selected at random, we can assume that X1, X2, X3, and W are mutually independent. The theorem helps us determine the distribution of Y, the sum of three one-pound bags: Y = (X1 + X2 + X3) \sim N(1.18+1.18+1.18, 0.07^2 +0.07^2 + 0.07^2) = N(3.54, 0.0147)
That is, Y is normally distributed with a mean of 3.54 pounds and a variance of 0.0147. Now, Y - W, the difference in the weight of three one-pound bags and one three-pound bag is normally distributed with a mean of 0.32 and a variance of 0.0228, as the following calculation suggests:
(Y – W) \sim N(3.54 – 3.22, (1)^2 (0.0147) + (-1)^2 (0.09^2)) = N(0.32, 0.0228)Therefore, finding the probability that Y is greater than W reduces to a normal probability calculation:
P(Y > W) = P(Y – W > 0) = P\left(Z > \frac{0 - 0.32}{\sqrt{0.0228}} \right) = P(Z > -2.12) = P(Z < 2.12) = 0.9830That is, the probability that the sum of three one-pound bags exceeds the weight of one three-pound bag is 0.9830.
Exponential Distribution
The exponential distribution is frequently applied when interested in elapsed time intervals.
Examples:
The length of time between telephone calls.
The length of time between arrivals at a service station.
The lifetime of electronic components.
When the number of occurrences of an event has a Poisson distribution with average \mu per unit of time, the time between the occurrences has an exponential distribution with average 1/\mu time unit.
Exponential Distribution - Definition
Notation: A random variable X is said to be exponentially distributed with parameter \lambda if its p.d.f. is:
f(x) = \lambda e^{-\lambda x}, x \geq 0 where \lambda > 0The expected value and the variance of X are:
E(X) = \frac{1}{\lambda}
V(X) = \frac{1}{\lambda^2}
X \sim Exp(\lambda)
Exponential Distribution - Definition
Also, the probability that an exponentially distributed r.v. X will take a value greater than a specified nonnegative number a can be easily computed using:
P(X > a) = P(t \leq X < a) = \intt^a \lambda e^{-\lambda t} dt = [-e^{-\lambda t}]0^a = 1-e^{-\lambda a} = e^{-\lambda t}
Exponential Distribution - Shape
Exponential distribution for \lambda = 0.5, 1, 2
P(a < X < b) = P(X > a) - P(X > b) = e^{-\lambda a} - e^{-\lambda b} ial Distribution - Application
Example:
Cars arrive randomly and independently at a tollbooth at an average of 360 cars per hour.
Let Y = number of arrivals per minute, then we know already that Y \sim Poisson(\mu = 6) [average arrival rate is 360/60 = 6 cars per minute].
(a) Use the exponential distribution to find the probability that the next car will not arrive within half a minute.
(b) Use the Poisson distribution to find the probability that no car will arrive within the next half minute.
Exponential Distribution - Application
Solution (a):
Let X = time (in minutes) that elapses before the next car arrives, then the desired probability is P(X > 0.5), i.e., it takes more than 0.5 minute before the next car arrives.
If cars arrive at an average rate of 6 cars per minute, then the average inter-arrival time is 1/6 minute [= E(X) = 1/\lambda], hence X \sim Exp(\lambda = 6).
It follows that P(X > 0.5) = e^{-6 \times 0.5} = e^{-3} = 0.0498
Note that the parameter of the Poisson variable Y (\mu = 6) always equals the parameter of the corresponding exponential variable X (\lambda = 6) when same units of time (here: minutes) are used.
Exponential Distribution - Application
Solution (b):
Now we use Y = number of arrivals per minute.
We know that Y \sim Poisson(\mu = 6 per minute), or Y \sim Poisson(\mu = 3 per half minute).
Desired probability P(Y = 0 (per half minute)):
P(Y = 0) = \frac{e^{-3} 3^0}{0!} = e^{-3} = 0.0498
Comment: If the first car will not arrive within the next half minute, then no car will arrive within the next half minute. Therefore, not surprisingly, the probability found here is the exactly same probability found in the previous question.
Exponential Distribution - Application
Example:
The lifetime of a transistor is exponentially distributed, with a mean of 1,000 hours.
What is the probability that the transistor will last between 1,000 and 1,500 hours?
Solution:
Let X denote the lifetime of a transistor (in hours).
E(X) = 1000 = 1/\lambda, so \lambda = 1/1000 = 0.001
P(1000 < X < 1500) = e^{-0.001 \times 1000} - e^{-0.001 \times 1500} = 0.3679 - 0.2231 = 0.1448