Variable 2: Desempeño Laboral

Help learners recall that random variables map outcomes of independent random events into numbers, and that there are two general types: discrete (which takes on a specific and countable number of possible values) and continuous (which takes on possible values in a continuum). An example of continuous random variable is the weight of a randomly-chosen rock from a pile of rocks since the exact weight of the rock (to the smallest of measures) is impossible to pinpoint as weight can be

subjected to varying levels of accuracy, and we are only limited by the precision of our instruments. We feel that, in principle, weight could be any numerical value. For example, the weight of this randomly-chosen rock could be 245.4 grams, but it could also be 245.38 grams, or 245.382 grams, and so on.

Ask learners to remember the heights data they provided in Lesson 1 of Chapter 1, or the measured height they obtained in Lesson 3 of Chapter 1 (for calculating the Body Mass Index). Tell them to assume that H is the height of a student chosen at random in the class. Is H a continuous random variable or is it discrete? What is it about the experiment that makes H continuous or discrete? What could they do to change it? (Tell them that, in theory, H falls on a continuum of values, but if the data could be “discretized” to be a countable number of data, i.e., if the heights were to be put up to one decimal place in centimeters, then the possible values would be countable. Another way is to define H = 1 if the height is less than 167 cm, H=2 if between 167 and 180 cm, and H=3 if 180 cm or more. Then H would be discrete with possible values of 1, 2, and 3.).

Define Y as the top bicycling speed of a randomly-chosen learner in class. Ask learners if Y is a continuous random variable. (Speed is a common continuous variable, and the value is chosen by a random process so Y is a continuous random variable since there is always another possible value between any two speed values.

It would not be possible to count all possible speeds that Y could be.)

Define Z to be the number of people in the city/municipality/town who will vote in the next presidential election. Ask learners if Y is a continuous random variable.

Their answer should be no since even though there may be a huge number of voters in the city/municipality/town, the number of voters is countable (and finite).

Let T be the total time spent on Facebook by a randomly-chosen learner from the time he/she registered on FB. Then T is a continuous random variable.

B. Motivation: Probability Density Function as Analog of Probability Mass Function

In the previous lesson, you showed learners that for discrete random variable X, that takes on a finite or countably infinite number of possible values, there is a

probability distribution (or probability mass function) P(X = x) for all of the possible values of X

Now, inform learners that for continuous random variables, the probability that X takes on any particular value x is zero (0). That is, finding P(X = x) for a continuous random variable X is always known to be zero (0).

Instead, we will need to find the probability that X falls in some interval (a, b), that is, we'll need to find P(a < X < b). We can use a curve called the probability density function f(x), for this purpose.

Even though a fast food chain would claim that a hamburger weighs 100 grams, a randomly- selected hamburger might weigh 98 grams while another might weigh 103 grams. What is the probability that a randomly-selected hamburger weighs between 95 and 105 grams? That is, if we let X denote the weight (in grams) of a randomly selected hamburger, what is P(95 < X < 105)?

Tell learners to assume that we selected 100 hamburgers and created a histogram of the resulting weights. Perhaps, the histogram might look something like this:

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weight

If we decreased the length of the class intervals on the histogram, then, the histogram would look something like this:

and if we pushed this further and decreased the intervals even more (and selected even more hamburgers, say 1000 or 10,000), the intervals would eventually get so small that we could represent the probability distribution of X, not as a histogram, but as a curve (by connecting the "dots" at the tops of the tiny rectangles) that, in this case, might look like this:

Tell learners that such a curve is denoted as f(x) and is called a (continuous) probability density function.

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weight

Help learners recall that when a histogram is drawn, the area of each rectangle equals the relative frequency (or percentage) of the corresponding class, and the area of the entire histogram equals one (1).

Finding the probability that a continuous random variable X falls in some interval of values involves finding the area under the curve f(x) sandwiched by endpoints of the interval. In the case of this example, the probability that a randomly-selected

hamburger weighs between 95 grams and 100 grams is then this area:

C. Main Lesson: Probability Density Function

Since you have managed to discuss the idea behind a probability density function for a continuous random variable, you can now formally define it to the class:

Definition: Let X be a random variable, the probability density function f(x) is a function that satisfies the following properties:

(a) the height of the graph, i.e. f(x) must be greater than or equal to zero (0) (i.e nonnegative) for all real numbers x. That is, the graph of the equation must lie on or above the horizontal axis for all possible values of the random variable.

(b) the total area under the curve f(x) is 1

Show to students that the definition of the probability density function of a continuous random variable is analogous to the probability mass function (or

probability distribution) of a discrete random variable, i.e. the first property is about probabilities being nonnegative, and the second property is about probabilities adding up to 1.

Explain to students that we can use the probability density function to find the probability that X lies in some interval from a to b, by considering the area under the graph of the probability density function over the interval a to b. That is:

= the area under the curve f(x) sandwiched by a and b This is also similar to the idea that probability that X lies in some discrete set A, may be obtained by summing the probability for the distinct values in the set A, that is.

We simply changed sums that appeared in the discrete case to “integrals” in the continuous case. You should not assume though that learners know their calculus, but you can tell them this is why they have to “endure” calculus, as it has many uses especially for probability and statistics.

Example 1: Uniform Distribution, also called Rectangular Distribution Suppose that your friend is always late, and that the continuous random variable X represents the time from when you are supposed to meet your friend until he shows up. Suppose that your friend could arrive “a” minute late or up to “b” minutes late with all intervals of equal time between and being equally likely.

For example, if a=10 and b = 60, and your friend is just as likely to be from 10 to 20 minutes late as he is to be 25 to 35 minutes late. The random variable X can be any value in the interval from 10 to 60, that is, because any two intervals of equal length between 10 and 60, inclusive, are equally likely.

The random variable X is said to follow a uniform probability distribution, with probability density function

if a < x < b for some constant c

and suppose f is zero (0) outside the interval a to b.

(i) What is the value of c?

(ii) Suppose that a=0, and b=1, what would be P(0.1<X<0.3} ; P(0.5<X<0.7); and P ( d < X < d +0.2 ) assuming d> a and d +0.2 < b?

Answer: Draw the probability density function

Tell learners that since the total area under the curve (which is the area of the rectangle formed in the graph) should add up to one, then

c (b-a) = 1 Thus, c = 1/(b-a).

If a = 0, b=1, then c=1. In this case, show that the probabilities P(0.1<X<0.3} ; P(0.5<X<0.7); and P ( d < X < d +0.2 ) assuming d> a and d +0.2 < b would all be 0.2 because we would be forming rectangles with a height of 1, and a width of 0.2, whose areas would be 0.2.

Explain to learners that this is why we call this a uniform distribution. Probabilities would be uniformly dependent on the width of the interval.

Example 2: (Triangular Distribution)

Let X be a continuous random variable whose probability density function is if -1 < x < 1

Verify if f is a probability density function, and calculate P(0 < X < ½) Answer:

Draw the probability density function and observe the “triangular distribution.”

Tell learners that since the total area under the curve (which is the sum of two right triangles, each with area of one half—ask learners why) add up to one.

Now to calculate P(0 < X < ½), remind learners about the symmetry at zero (0) (i.e.

that there are two right triangles to the left and right of zero). Tell learners this means that

P(0 < X < 1) = P(-1 < X < 0 ) = ½

Show them that there we can form a smaller right triangle from ½ to 1, whose area is (1/2) x ((1/2) x (1/2) = 1/8 so that we should know that

P (½ < X <1) = 1/8

And thus, we can compute. Remind learners about the symmetry at zero (0) (i.e. that there are two right triangles to the left and right of zero (0)). Tell learners that this means that

P(0 < X < ½) = P(0 < X < 1 P (½ < X <1)) = (1/2) - (1/8) = 3/8

An alternative way to calculate P(0 < X < ½) is to remember the area of a trapezoid, the sum of the bases multiplied to half of the height. A trapezoid can be formed as shown below with bases 1 and ½, and a height of 1/2.

Thus P(0 < X < ½) = ( 1 + (1/2) ) = 3/8^!