Unit 4: Continuous Distributions
QBA 201 – Summer 2013
Instructor: Michael Malcolm
4.1: Characteristics of continuous distributions
4.2: The uniform distribution
4.1: Characteristics of continuous distributions
Recall from the previous unit that a discrete random variable takes on a countable number of possible values, and the associated probability distribution function gives the probability with which the random variable takes on various values.
By contrast, a continuous random variable takes on an uncountable number of values over an interval. While a discrete random variable might be number of children or the number of accidents at an intersection, a continuous random variable might be height or the time between customer arrivals.
The key difference between a continuous and a discrete random variable is that, for a continuous random variable, each individual point has zero probability. Since the values of the variable are defined over some uncountable set, it only makes sense for continuous random variables to talk about probabilities over intervals. To repeat, probabilities for continuous distributions only make sense over intervals, not for single points.
While discrete random variables are characterized by a probability distribution function, continuous random variables are characterized by a probability density function. The total area under a probability density function is equal to 1, and the area under the probability density function between two points represents the probability that the random variable will take a value that falls between these two points.
EXERCISES
1. Consider a continuous random variable 𝑋 that can take on any value in the interval [0,2] and whose probability density function is 𝑓(𝑥) =3
8𝑥
2. For this particular distribution, the
probability that 𝑥 is less than or equal to 𝑘 is 𝑃(𝑥 ≤ 𝑘) =18𝑘3.
4.2: The uniform distribution
The simplest example of a continuous distribution is the uniform distribution. When the random variable 𝑋 obeys a uniform distribution over the interval [𝑎, 𝑏], what this means is that any sub-interval of the same size on [𝑎, 𝑏] is equally likely.
For example, if the travel time for a flight from Chicago to Atlanta is uniformly distributed over the interval [120,140], what this means is that any one-minute span over this interval is equally likely. For example, the plane is just as likely to take between 122 and 123 minutes as it is to take between 135 and 136 minutes. Again, because the distribution is continuous, the probability that the flight will take exactly 135 minutes is zero. Probabilities are only positive over intervals.
Formally, when 𝑋 obeys a uniform distribution over the interval [𝑎, 𝑏], its probability density function is as follows:
𝑓(𝑥) =𝑏−𝑎1 for 𝑎 ≤ 𝑥 ≤ 𝑏 and equal to 0 otherwise.
It is easy to compute probabilities for the uniform distribution simply by drawing a graph of the
probability distribution. It is a rectangle with height 𝑏−𝑎1 over the interval [𝑎, 𝑏]. You can then compute the probability of any interval using the area under this graph.
There are simple expressions for the expected value and the variance of a uniform distribution. For any random variable 𝑋 that is uniformly distributed over [𝑎, 𝑏], the expected value and the variance are:
𝐸(𝑋) =𝑎+𝑏2
𝑉𝑎𝑟(𝑋) =(𝑏−𝑎)12 2
For example if 𝑋 is travel time from Chicago to Atlanta, then 𝑋 is distributed uniformly over the interval [120,140]. The probability distribution is a rectangle over the interval [120,140] with a height of 140−1201 = 201. Notice that the area under the entire interval [120,140] is 1, as required.
What is the probability that the trip will take between 125 and 130 minutes?
What is the probability that the trip will take more than 130 minutes?
Note that the rectangle on the diagram extends only to 𝑋 = 140. Thus, the probability is 1
20(130 − 140) = 1
20⋅ 10 = 1 2.
What are the mean travel time and the variance in travel times?
Using the formulas for the expected value and variance of a uniform distribution:
𝐸(𝑋) =120+1402 = 130
EXERCISES
1. You are bidding on a piece of land against one other competitor. You both submit a private bid, and whichever one of you submits a higher bid wins the piece of land. You do not know how much your competitor will bid, but you know that his bids are uniformly distributed between $10,000 and $15,000.
a. If you bid $12,000, what is the probability that you will win the auction? b. If you bid $14,000, what is the probability that you will win the auction? c. What should you bid to be sure that you get the property?
2. A parachutist is equally likely to land anywhere between markers A and B. What is the probability that the parachutist will land closer to A than to B?
3. The time for hauling concrete to a highway construction site is uniformly distributed between 50 and 70 minutes.
a. What is the probability that the delivery will take more than 65 minutes?
b. If 55 minutes have already elapsed, what is the probability that the delivery will take more than 65 minutes?
4.3: The exponential distribution
While the uniform distribution assigns equal probabilities over an entire interval, the exponential distribution is defined for positive values 0 ≤ 𝑥 ≤ ∞, with the probabilities over various intervals consistently falling as 𝑥 increases.
The exponential distribution is frequently encountered in business applications to model the time between successive arrivals. For example, the time before your next telephone call comes and the time until default on a loan are modeled using an exponential distribution. Geologists use it to model the amount of time before an earthquake occurs in a region.
If 𝑋 follows an exponential distribution with mean 𝛽, then its probability density function is as follows:
𝑓(𝑥) =𝛽1𝑒−𝑥/𝛽 for 𝑥 ≥ 0 and equal to 0 otherwise.
Again, this probability density function does not give the probability that 𝑋 is equal to a particular value 𝑥, since this is equal to 0. However, areas under the probability density function between two points can be used to calculate the probability that 𝑋 will fall between these two points. To actually compute this, a useful fact is that, for an exponential distribution with mean
𝛽, the probability that 𝑥 will be less than or equal to any positive constant 𝑘 is given by:
𝑃(𝑥 ≤ 𝑘) = 1 − 𝑒−𝑘/𝛽
The expected value and variance for any random variable 𝑋 that follows an exponential distribution with mean 𝛽 are given as follows:
𝐸(𝑋) = 𝛽
𝑉𝑎𝑟(𝑋) = 𝛽2
To illustrate how to use the exponential distribution, suppose that the time to download all of the content on a webpage follows an exponential distribution with a mean of 15 seconds.
What is the probability that the webpage will be loaded in 10 seconds or fewer?
Applying the formula given earlier for an exponential distribution with mean 𝛽 = 15 gives:
What is the probability that the webpage will take between 15 and 20 seconds to load?
To find 𝑃(15 ≤ 𝑥 ≤ 20), we can simply take 𝑃(𝑥 ≤ 20) − 𝑃(𝑥 ≤ 15). Applying the formula given earlier:
𝑃(15 ≤ 𝑥 ≤ 20) = (1 − 𝑒−2015) − (1 − 𝑒−1515) = 0.1043
What is the probability that the webpage will take more than 2 minutes to load?
Note that 𝑃(𝑥 ≥ 120) can be calculated as 1 − 𝑃(𝑥 ≤ 120). Applying the formula given earlier:
𝑃(𝑥 ≥ 120) = 1 − (1 − 𝑒−12015) = 0.0003
What are the mean and variance of the webpage’s loading time?
Using the known results for an exponential distribution with 𝛽 = 15 gives:
𝐸(𝑋) = 15
𝑉𝑎𝑟(𝑋) = 152 = 225
One final point is that, if we are applying the exponential distribution to model the time between successive arrivals and 𝛽 is the mean time between arrivals, then 1/𝛽 is interpreted as the arrival rate. For example, if customers arrive on average 𝛽 = 1/4 of an hour apart (i.e 15 minutes apart), then 𝛽1 = 4 is the arrival rate of 4 customers per hour.
EXERCISES
1. Carbon monoxide samples from a city follow an exponential distribution with a mean of 3.6 parts per million.
a. What is the probability that the carbon monoxide concentration exceeds 9 parts per million for a particular reading?
b. Answer (a) again if pollution control can reduce the mean reading to 2.5 parts per million.
2. The magnitude of earthquakes striking North America follows an exponential distribution with a mean of 2.4, as measured on the Richter scale.
a. What is the probability that a single earthquake will exceed 5 on the Richter scale?
b. Of the next 10 earthquakes, what is the probability that at least one will exceed 5 on the Richter scale?
3. The time before a flashlight bulb fails follows an exponential distribution with a mean of 200 hours.
a. What is the probability that the bulb will last for 100 hours or more?
b. The bulb has already lasted for 300 hours. What is now the probability that it will last for 400 hours or more? (i.e. for an additional 100 hours)
c. The bulb has already lasted for 600 hours. What is now the probability that it will last for 700 hours or more? (i.e. for an additional 100 hours)
d. It is sometimes said that exponential distribution displays the “memoryless” property. Why does this name make sense?
4.4: The normal distribution
The normal distribution is, by far, the most important and frequently encountered distribution in probability and statistics. It is a continuous distribution that is bell-shaped. Specifically, a normally distributed random variable 𝑋 with mean 𝜇 and standard deviation 𝜎 has the following properties:
The distribution of 𝑋 is symmetric about its mean 𝜇.
The distribution of 𝑋 is mound-shaped, with the maximum value of the probability density function occurring at 𝑥 = 𝜇 and then falling in either direction.
The distribution of 𝑋 has positive probability over the entire real line from −∞ to ∞, although the probability that 𝑥 will take on values more than three standard deviations away from the mean is almost 0.
The median of 𝑋 is 𝜇.
When the random variable 𝑋 is normally distributed with mean 𝜇 and standard deviation 𝜎, a short way of writing this is 𝑋~𝑁(𝜇, 𝜎).
The probability density function for a normal distribution 𝑋 with mean 𝜇 and standard deviation
𝜎 is as follows:
𝑓(𝑥) = 1
𝜎√2𝜋𝑒 −( 1
2𝜎2)(𝑥−𝜇)2
Areas under this curve for particular intervals give the probability that 𝑋 will lie in this interval. Unfortunately, unlike the uniform distribution and the exponential distribution, there is no simple formula that can be used to evaluate these areas. Thus, calculating probabilities for a normal distribution requires either the use of tables or computer software.
A special case of the normal distribution is the standard normal distribution, which is the normal distribution with mean 𝜇 = 0 and standard deviation 𝜎 = 1. That is, if 𝑍 follows a standard normal distribution, then 𝑍~𝑁(0,1). Notice that the standard normal distribution is centered about 0 in the sense that probability 0.5 lies above 0 and probability 0.5 lies below 0.
Tables that tabulate standard normal probabilities are widely available. As usual, these tables give the probability 𝑃(𝑧 ≤ 𝑘) for particular values of 𝑘 when the random variable 𝑍 follows a standard normal distribution. Here are some examples of how to use the table.
𝑃(𝑧 ≤ 1.02) = 0.8461
𝑃(𝑧 ≥ 0.83) = 1 − 𝑃(𝑧 ≤ 0.83) = 1 − 0.7967 = 0.2033
𝑃(−0.10 ≤ 𝑧 ≤ 1.32) = 𝑃(𝑧 ≤ 1.32) − 𝑃(𝑍 ≤ −0.10) = 0.9066 − 0.4602 = 0.4464
Calculating probabilities is easy for variables that follow a standard normal distribution, but what about variables that are normally distributed, but not with a mean of 0 and a standard deviation of 1? For example, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.
An extremely useful property of the normal distribution is that any normal random variable 𝑋 with any mean 𝜇 and any standard deviation 𝜎 can be transformed to a standard normal random variable 𝑋 using the following transformation:
𝑍 = 𝑋−𝜇𝜎
In other words, we can use the standard normal table to calculate probabilities associated with
any normal distribution simply by implementing this transformation: subtracting the mean and dividing by the standard deviation.
For example, suppose that 𝑋 is a random variable measuring IQ scores and that 𝑋 follows a normal distribution with mean 𝜇 = 100 and standard deviation 𝜎 = 15. Using notation,
𝑋~𝑁(100,15).
What is the probability that a randomly selected person’s IQ exceeds 130
𝑃(𝑥 ≥ 130) = 𝑃 (𝑧 ≥130−10015 ) = 𝑃(𝑧 ≥ 2) = 0.0228
What is the probability that a randomly selected person’s IQ falls between 90 and 110?
𝑃(90 ≤ 𝑥 ≤ 110) = 𝑃 (90−10015 ≤ 𝑧 ≤110−10015 ) = 𝑃(−0.67 ≤ 𝑧 ≤ 0.67) = 0.4792
For example, suppose that we are interested in finding the cutoff for the top 1% of IQ’s. In other words, we want to find the IQ such that there is only probability 0.01 that an individual’s IQ will be higher than this value.
We need to search through the z-table for the value of 𝑧0 such that 𝑃(𝑧 ≥ 𝑧0) = 0.01. We can use the z-table to see that 𝑃(𝑧 ≥ 2.33) = 0.01. Thus, we simply use the z-statistic conversion to find the value of IQ score 𝑥 that is equivalent to a z-score of 2.33.
𝑃(𝑧 ≥ 2.33) = 0.01 ⇒ 𝑃 (𝑥−𝜇𝜎 ≥ 2.33) = 0.01 ⇒ 𝑃 (𝑥−10015 ≥ 2.33) = 0.01
⇒ 𝑃(𝑥 ≥ 2.33(15) + 100) = 0.01 ⇒ 𝑃(𝑥 ≥ 134.95) = 0.01
In essence, we knew using the z-table that 𝑃(𝑧 ≥ 2.33) = 0.01, so we just needed to find the value of 𝑥 to match this z-score. Doing so gives that 𝑃(𝑥 ≥ 134.95) = 0.01, so a cutoff of 𝑥 =
EXERCISES
1. A company manufactures apple juice uses a machine that is set to fill bottles with 16 ounces of apple juice. However, there is some variation in the amount of juice that is actually dispensed. Specifically, the amount dispensed is normally distributed with a mean of 16 ounce and a standard deviation of 0.5 ounces. What proportion of bottles will have more than 17 ounces dispensed in them?
2. The weekly amount spent on vehicle maintenance by a company is normally distributed with a mean of $400 and a standard deviation of $20. Suppose that the company budgets $450 per week for maintenance.
a. What is the probability that the maintenance expenses will be within the budget? b. How much should be budgeted so that the budget is satisfied 90% of the time?
3. A machine operation produces bearings with diameters that are normally distributed with mean 3.0005 inches and standard deviation 0.0010 inches. Specifications require that any bearings whose diameters are not in the interval [2.998, 3.002] be scrapped. What fraction of bearings will be scrapped?
4. Scores on an exam are normally distributed with a mean of 78 and a variance of 36.
a. What is the probability that a student taking the exam scores higher than 72? b. Students scoring in the top 25% must receive a grade of A. What is the cutoff
score to earn an A?
c. Anyone with a grade in the bottom 10% of the class must receive a grade of F. What is the cutoff score to earn an F?
5. The time needed to take an exam in a course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. If the class has 90 students, approximately how many of them will require between 60 and 75 minutes to finish?
6. The SAT and ACT college entrance exams are taken by thousands of students each year. SAT mathematics scores are normally distributed with mean 480 and standard deviation 100. ACT math scores are normally distributed with mean 18 and standard deviation 6.
a. An engineering school sets a minimum math SAT of 550. What fraction of students will satisfy this requirement?
