STA301 Highlighted Handouts 185-231 by VUBuddy

This probability is P(2) + P(3) + P(4) + P(5) = .132 +.309 + .360 + .168 = .969. Therefore, 96.9% of the probability distribution lies within 2 standard deviations of the mean. This percentage is CONSISTENT with both the Chebychev’s rule and the Empirical Rule.

d) Fewer than two successful ventures out of five implies that x = 0 or x = 1. Since both these values of x lie outside the interval µ + 2σ, we know from the Empirical Rule that such a result is unlikely (with approximate probability of only .05).The exact probability, P(x < 1), is P(0) + P(1) = .002 + .029 = .031. Consequently, in a single experiment where we invest in five business ventures, we would not expect to observe fewer than two successful ones. The key question: What is the significance of the Chebychev’s Inequality and the Empirical Rule? The answer to this question is that both these rules assist us in having a certain IDEA regarding amount of data lying between the mean minus a certain number of standard deviations and mean plus that same number of standard deviations. Given any data-set, the moment we compute the mean and standard deviation, we HAVE an idea regarding the two points (i.e. mean minus two standard deviations, and mean plus two standard deviations) between which the BULK of our data lies. If our data-set is hump-shaped, we obtain this idea through the Empirical Rule, and if we don’t have any reason to believe that our data-set is hump-shaped, then we obtain this idea through the Chebychev’s Rule We now begin the discussion of CONTINUOUS RANDOM VARIABLES – quantities that are measurable. As stated in the very first lecture, continuous variables result from measurement, and can therefore take any value within a certain range. For example, the height of a normal Pakistani adult male may take any value between 5 feet 4 inches and 6 feet. The temperature at a place, the amount of rainfall, time to failure for an electronic system, etc. are all examples of continuous random variable. Formally speaking, a continuous random variable can be defined as follows:

A random variable X is defined to be continuous if it can assume every possible value in an interval [a, b], a < b, where a and b may be –∞ and +∞ respectively. The function f(x) is called the probability density function, abbreviated to p.d.f., or simply density function of the random variable X. A continuous probability distribution looks something like this:

f(x) · X · 0.1 · 0.2 · 0.3 · 0.4 · p(x) · x

µ µ + 2σ

(5.54)

µ + 2σ

(1.46)

A p.d.f. has the following properties:

iii) The probability that X takes on a value in the interval [c, d], c < d is given by:

P(c < x < d)

which is the area under the curve y = f(x) between X = c and X = d, as shown in the following figure:

The TOTAL area under the curve is 1. In other words: f(x) a non-negative function, The integration takes place over all possible values of the random variable X between the specified limits, and The probabilities are given by appropriate areas under the curve. Since

It should therefore be noted that the probability of a continuous random variable X taking any particular value k is always zero. That is why probability for a continuous random variable is measurable only over a given interval. Further, since for a continuous random variable X, P(X = x) = 0 for every x, the following four probabilities are regarded as the same: P(c < X < d), P(c < X < d), P(c < X < d) and P(c < X d).

They may be different for a discrete random variable. The values (expressed as intervals) of a continuous random variable and their associated probabilities can be expressed by means of a formula. We now discuss the distribution function of a continuous random variable.

A random variable X may also be defined as continuous if its distribution function F(x) is continuous and is differentiable everywhere except at isolated points in the given range. In contrast with the graph of the distribution function of a discrete variable, the graph of F(x) in the case of a continuous variable has no jumps or steps but is a continuous function for all x-values, as shown in the following figure:

c · d · f(x) · P(c < x < d) · i) · f(x) > 0, for all x · ii) · = 1 · dx · x · f

d

c

dx · x · f · ,0 · dx · x · f · k · X · P

k

k Since F(x) is a non-decreasing function of x, we have i) f(x) > 0,

ii) ( )

dx · x · f · x · F

x for all x.

The relationship between f(x) and F(x) is as follows: f(x) is obtained by finding the derivative of F(x), i.e.

EXAMPLE

a) Find the value of k so that the function f(x) defined as follows, may be a density function f(x) = kx, 0 < x < 2 = 0, elsewhere

b) Compute P(X = 1).

c) Compute P(X > 1).

d) Compute the distribution function F(x).

e)

SOLUTION

a) The function f(x) will be a density function, if i) f(x) > 0 for every x, and ii)

The first condition is satisfied when k > 0. The second condition will be satisfied, if

We had

X · F(x) · F(a) · F(b) · x · f · dx · x · F · d · X · P · X

dx x f

,1 · dx · x · f · i.e. if · dx · x · f · dx · x · f · dx · x · f · i.e. if · dx · dx · kx · dx · i.e. if · k · x · k · This gives k = 1/2