Calculus Applied to Probability and Statistics
by
Stefan Waner and Steven R. Costenoble

This Section: 2. Probability Density Functions: Uniform, Exponential, Normal, and Beta

1. Continuous Random Variables and Histograms

Section 2 Exercises

3. Mean, Median, Variance and Standard Deviation

Calculus and Probability Main Page

"Real World" Page

We have seen that a histogram is a convenient way to picture the probability distribution associated with a continuous random variable X and that if we use subdivisions of 1 unit, the probability P(c X d) is given by the area under the histogram between X = c and X = d. But we have also seen that it is difficult to calculate probabilities for ranges of X that are not a whole number of subdivisions. The following example--based on an example in the previous section--introduces the solution to this problem.

A survey finds the following probability distribution for the age of a rented car.

Age (Years)	0-1	1-2	2-3	3-4	4-5	5-6	6-7
Probability	0.20	0.28	0.20	0.15	0.10	0.05	0.02

The histogram of this distribution is shown on the left of the figure below, and it suggests a curve something like the one given on the right.

This curve is the graph of some function f, which we call a probability density function. We take the domain of f to be [0,+), since this is the possible range of values X can take (in principle). In general, a probability distribution function will have some (possibly unbounded) interval as its domain.

Suppose now that as in the previous section, we wanted to calculate the probability that a rented car is between 0 and 4 years old. Referring to the table,

P(0 X 4) = 0.20 + 0.28 + 0.20 + 0.15 = 0.83.

Referring to the following figure, notice that we can obtain the same result by adding the areas of the corresponding bars, since each bar has a width of 1 unit.

Ideally, our probability density curve should have the property that the area under it for 0 X 4 is the same, that is,
P(0 X 4) = _{0 ⁴}f(x) dx = 0.83.

(This area is shown in the figure as well.)

Now what happens if we want to find P(2 X 3.5)? In the previous section we estimated this by taking half of the rectangle between 3 and 4 (see the next figure).

Instead, we could use the definite integral

P(2 X 3.5) = _2^3.5f(x) dx.
.

Before We Go On ...

Although we haven't given you a formula for f(x), we would like f(x) to behave as described above. Here is something else we would like: Since a car has probability 1 of having an age between 0 and +, we want

P(0 X < +) =_0⁺f(x) dx = 1.

This example motivates the following.

Probability Density Function A probability density function (or probability distribution function) is a function f defined on an interval (a, b) and having the following properties. (a) f(x) 0 for every x (b) abf(x) dx = 1 We allow a, b, or both to be infinite, as in the above example. This would make the integral in (b) an improper one.

Probability Associated with a Continuous Random Variable A continuous random variable X is specified by a probability density function f. The probability P(c X d) is specified by P(c X d) = cdf(x) dx.

In particular,

P(X = c) = P(c X c) =_c^cf(x) dx = 0,

showing once again that there is a zero probability that X will assume any specified value.

For what constant k is f(x) = k e^x a probability density function on [0,1]?

Solution

We need to choose a k that makes requirements (a) and (b) of the definition true. Since e ^x > 0 for all x, all we need for (a) is to make sure that we choose k 0. For (b), we calculate

_0¹ ke ^x dx
= [ke ^x]¹₀
= k(1 1/e) = k(e1)/e..

Since this must equal 1, we get

k(e1)/e = 1,
so
k = e/(e1) 1.582.

Therefore, the function

f(x) = [ e/(e1) ]e ^x

is a probability density function on [0,1].

Uniform Density Function

A uniform density function f is a density function that is constant, making it the simplest kind of density function. Since we require f(x) = k for some constant k, requirement (b) in the definition of a probability density function tells us that

1 = _a^bf(x) dx = _a^bk dx = k(ba).

Thus we must have

k = 1/(ba).

In other words, a uniform density function must have the following form.

Uniform Density Function The uniform density function on the interval [a, b] is the constant function defined by f(x) = 1/(ba). Its graph is a horizontal line:

Note

If f(x) is a uniform density function as above, then f(x) is independent of x, and so the probability P(c X d) depends only on the width dc. In fact,

P(c X d) = (dc)(ba)

for a c d b (why?)

Suppose that you spin the dial shown in the figure so that it comes to rest at a random position. Model this with a suitable distribution, and use it to find the probability that the dial will land somewhere between 5û and 300û.



Solution

We take X to be the angle at which the pointer comes to rest, so we use the interval [0, 360]. Since all angles are equally likely, the probability density function should not depend on x and therefore should be constant. That is, we take f to be uniform.

f(x) = 1/(ba)
= 1/(3600)= 1/360.

Thus,

P(5 X 300) = _5³⁰⁰ 1/360 dx
= (300 5)/360 = 295/360 0.8194.

Before We Go On ...

Check the following probabilities. Why are these the answers you expect?

P(0 X 90) = 1/4
P(90 X 180) = 1/4
P(0 X 180) = 1/2
P(0 X 270) = 3/4
P(0 X 120) = 1/3

Exponential Density Function

You are an investment analyst, and recent surveys show that troubled saving and loan (S&L) institutions are failing continuously at 5% per year. What is the probability that a troubled S&L will fail sometime within the next x years?

To answer the question, suppose that you started with 100 troubled S&Ls. Since they are failing continuously at a rate of 5% per year, the number left after x years is given by the decay equation

number left = 100e^0.05x,

number that fail = total number number left
= 100 100e^0.05x
= 100(1 e^0.05x).

Thus, the percentage that will have failed by that time--and hence the probability that we are asking for--is given by

P = 100(1e ^0.05x)/100 = 1 e ^0.05x.

Now let X be the number of years a randomly chosen troubled S&L will take to fail. We have just calculated the probability that X is between 0 and x. In other words,

P(0 X x) = 1 e ^0.05x.

But we also know that

P(0 X x) = _0^x f(t) dt

for a suitable probability density function. Thus,

_0^x f(t) dt = 1 e ^0.05x.

The Fundamental Theorem of Calculus tells us that the derivative of the left side is f(x). Thus,

f(x) = _0^x f(t) dt = (1 e ^0.05x) = 0.05 e ^0.05x,

which is the probability density function we were seeking.

Question

Does this function satisfy the mathematical conditions necessary for it to be a probability density function?

Answer

First, the domain of f is [0, +), since x refers to the number of years from now. Checking requirements (a) and (b) for a probability density function,

(a) 0.05e^0.05x ³ 0,

(b) _0⁺0.05e ^0.05x dx = _{M+ o^M}0.05e ^0.05x dx

= _M+ [ 0.05e ^0.05x]₀^M
= _M+ (e⁰ 0.05e ^0.05M)
= 10 = 1.

There is nothing special about the number 0.05. Any function of the form

f(x) = ae ^ax

with a a positive constant is a probability density function. A density function of this form is referred to as an exponential density function.

Exponential Density Function An exponential density function is a function of the form f(x) = aeax (a a positive constant) with domain [0 +). Its graph is shown in the figure.

Continuing the example in the text, what is the probability that a given troubled S&L will fail between 2 and 4 years from now? What is the probability that it will last 5 or more years?

Solution

These probabilities are given by integrals.

P(2 X 4) = _2⁴0.05e ^0.05xdx
[0.05e ^0.05x]₂⁴
= e ^0.2 + e ^0.1
0.086
P(X 5) = _5⁺ 0.05e ^0.05x dx
= _{M+ 5^M} 0.05e ^0.05x dx
= _M+ [0.05e ^0.05x]₅^M
= _M+ (e ^0.25 e ^0.05M)
= e ^0.25 0.779

So there is an 8.6% chance that a given S&L will fail between 2 and 4 years from now, and a 77.9% chance that it will last 5 or more years.

Before We Go On ...

We could also calculate P(X 5) = 1 P(0 X 5) and not have to calculate an improper integral. Try this for practice.

Plutonium 239 decays continuously at a rate of 0.00284% per year. If X is the time a randomly chosen plutonium atom will decay, write down the associated probability density function, and use it to compute the probability that a plutonium atom will decay between 100 and 500 years from now.

Solution

Using the discussion on failing S&Ls as our guide, we see that a = 0.0000284, so that the probability density function is

f(x) = 0.0000284e ^0.0000284x.
For the second part of the question,
P(100 X 500) = _100⁵⁰⁰(0.0000284e ^0.0000284x) dx
0.011.

Thus, there is a 1.1% chance that a plutonium atom will decay sometime during the given 400 year period.

Normal Density Function

Perhaps the most interesting class of probability density functions are the normal density functions, defined as follows.

Normal Density Function A normal density function is a function of the form f(x) = , with domain (, +). The quantity µ is called the mean and can be any real number, while is called the standard deviation and can be any positive real number. The graph of a normal density function is shown in the following figure.

You can check the following properties using calculus and a little algebra.

Properties of a Normal Density Curve (1) It is "bell-shaped" with the peak occurring at x = µ. (2) It is symmetric about the vertical line x = µ. (3) It is concave down in the range µ x µ + . (4) It is concave up outside that range, with inflection points at x = µ and x = µ+.

The normal density function applies in many situations that involve measurement and testing. For instance, repeated imprecise measurements of the length of a single object, a measurement made on many items from an assembly line, and collections of SAT scores tend to be distributed normally. It is for this reason that the normal density curve is so important in quality control and in assessing the results of standardized tests.

In order to use the normal density function to compute probabilities, we need to calculate integrals of the form _a^b f(x) dx. However, the antiderivative of the normal density function cannot be expressed in terms of any commonly used functions. Traditionally, statisticians and others have used tables coupled with transformation techniques to evaluate such integrals. This approach is rapidly becoming obsolete as the technology of hand-held computers and programmable calculators puts the ability to do numerical integration quickly and accurately in everybody's hands (literally). In keeping with this trend, we shall show how to use a graphing calculator to do the necessary calculation in the next example.

Pressure gauges manufactured by Precision Corp. must be checked for accuracy before being placed on the market. To test a pressure gauge, a worker uses it to measure the pressure of a sample of compressed air known to be at a pressure of exactly 50 pounds per square inch. If the gauge reading is off by more than 1% (0.5 pounds), the guage is rejected. Assuming that the reading of a pressure gauge under these circumstances is a normal random variable with mean 50 and standard deviation 0.5, find the percentage of gauges rejected.

Solution

For a gauge to be accepted, its reading X must be 50 to within 1%, in other words, 49.5 X 50.5. Thus, the probability that a gauge will be accepted is P(49.5 X 50.5). X is a normal random variable with µ = 50 and = 0.5. The formula tells us that

P(49.5 X 50.5) = _49.5^50.5 f(x) dx,
where
f(x) = ,

We then calculate the integral numerically. Using a TI-82 or TI-83, for example, you would enter

Y₁=(1/(0.5(2¼)^0.5))e^(-(X50)^2/0.5)

and then enter

fnInt(Y₁,X,49.5,50.5)

Alternatively, you could use the trapezoid rule or Simpson's rule program discussed in the chapter on the integral in Calculus Applied to the Real World.

The numerical calculation yields an answer of approximately 0.6827. In other words, 68.27% of the gauges will be accepted. Thus, the remaining 31.73% of the gauges will be rejected.

Before We Go On ...

In Appendix B of Calculus Applied to the Real World., we have included a graphing calculator program that computes probabilities associated with arbitrary normal distributions, and saves you from the task of having to repeatedly enter different normal density functions.

As we mentioned above, the traditional and still common way of calculating normal probabilities is to use tables. The tables most commonly published are for the standard normal distribution, the one with mean 0 and standard deviation 1. If X is a normal variable with mean µ and standard deviation , the variable Z = (Xµ)/ is a standard normal variable (see the exercises). Thus, to use a table we first write

P(a X b) = P[ (aµ)/( ) Z (bµ)/( ) ]

and then use the table to calculate the latter probability.

The following calculations, true for any normal random variable, are very useful to remember:

P(µ X µ+) 0.6827 (See below.)
P(µ2 X µ+2) 0.9545
P(µ3 X µ+3) 0.9973

Question

Why can we assume that the reading of a pressure gauge is given by a normal distribution? Why is the normal distribution so common in this kind of situation?

Answer

The reason for this is rather deep. There is a theorem in probability theory called the Central Limit Theorem that says that a large class of probability density functions may be approximated by normal density functions. Repeated measurement of the same quantity gives rise to such a function.

Beta Density Function

There are many random variables whose values are percentages or fractions. These variables have density functions defined on [0,1]. A large class of random variables, such as the percentage of new businesses that turn a profit in their first year, the percentage of banks that default in a given year, and the percentage of time a plant's machinery is inactive, can be modeled by a beta density function.

Beta Density Function A beta density function is a function of the form f(x) = (+1)(+2)x(1 x), with domain [0, 1]. The number can be any constant ³ 0. The figure shows the graph of f(x) for several values of .

A utilities industry consultant predicts a cutback in the Canadian utilities industry during 2000-2005 by a percentage specified by a beta distribution with = 0.25. Calculate the probability that Ontario Hydro will downsize by between 10% and 30% during the given five-year period.

Solution

The beta density function with = 0.25 is

f(x) = (+1)(+2)x(1x)
= 2.8125x^0.25(1x)
= 2.8125(x^0.25 x^1.25).

Thus,

P(0.10 X 0.30) = _0.10^0.302.8125(x^0.25 x^1.25)dx

= 2.8125(_0.10^0.30x^0.25 x^1.25)dx

= 2.8125[x^1.25/1.25 x^2.25/2.25] _0.1^0.3
0.2968.

So there is approximately a 30% chance that Ontario Hydro will downsize by between 10% and 30%.

Before We Go On ...

The following figure shows a graphing calculator plot of the density function. Notice that its shape is "in between" those for = 0 and = 0.5 in the preceeding figure.

1. Continuous Random Variables and Histograms

Section 2 Exercises

3. Mean, Median, Variance and Standard Deviation

Calculus and Probability Main Page

"Real World" Page