Probabilities for Normal Random Variables in General

CONTINUOUS RANDOM VARIABLES The Binomial random variable is called a discrete random variable because it takes on discrete values 0, 1, 2, 3, ..,n. Consider the probability histogram of X, Binomial n =3, p =.7. The areas under the histogram are related to the probability distribution.

P(X=2) =

P(X 2) =

The total area under this probability histogram is p(0) + p(1) + p(2) + p(3) = 1 A continuous random variable X is one which represents measurements that (theoretically) can be made to any degree of accuracy. For example suppose X = the weight (in kg) of a randomly chosen newborn baby. Depending on the accuracy of our scale the weight X of a randomly selected baby could be recorded either as 3 or 3.3 or 3.26 or 3.258 etc. The probability histogram for such a variable has to be formed in a very different way than for a discrete random variable. For example, in the case of “X=the birth weight of a newborn”, we could take a very large sample from the population of newborns, measure the sample birth weights very accurately ( with many decimal of accuracy) and form a histogram of the birth weights using classes of small width. If the vertical scale is adjusted so the total area under the histogram is one then area under the histogram can be used to calculate (approximately) the probabilities. The larger the sample and the more accurate our measurements, the more accurate will be these probabilities. In the illustration below a srs of 812 babies was used to form a probability histogram. (70)

Birth Weight (kg) of 812 Newborns

Frequency

30

20

10

0 1

2

3

4

5

WT(kg)

Let X be the birth weight of a randomly chosen newborn. P(X3) = Shaded Area The larger the sample, the smaller we can make these rectangles, and the “smoother” will be the resulting histogram. Thus a “model” of the distribution could be obtained by fitting a curve to such a histogram and using the area under the curve to calculate the probabilities ( this curve is called a Probability Density Function).

Thus P(X3) is the area under the curve to the left of 3. Thus total area under the curve must be 1. Note: As we know there are many different shapes among various populations (e.g. left skewed, right skewed, symmetric etc.). In the class of bell-shaped curves there is a specific one which is called normal curve ( there is a mathematical formula which defined it exactly). If a population can be modeled by this certain bell –shaped curve the population is said to have a Normal ( or Gaussian) Distribution. (71)

The Normal Distribution A Normal ( or Gaussian) Population is one that can be modeled by a certain bellshaped curve called the normal curve. The population of weights of newborn babies described above is an example of such a population. In describing such a population, we need to know two quantities,  and  .  stands for the population mean and  stands for the population standard deviation. In our example  is the mean (average) weight of all newborn babies in the population and  measures the spread of the population values about the mean  .

A Normal (or Gaussian) random variable X represents a randomly chosen measurement sampled from this population. Probabilities about X are found by finding the appropriate areas under the curve i.e. P(Xx) = the area under the normal curve to the left of x. Note: (a) The total area under the curve is 1. (b) For a normal random variable P(X=x) is always 0. Practically speaking this means that if we can measure observations very accurately, the chances of finding a newborn weighing exactly 3.0000000000kg, say, is very small. Thus for all practical purposes (X=3) =0. One consequence of this is that P(Xx) = P(X1.5) =

(b) (i) P(Z  -1.5) = (ii) P(Z-1.5) =

(c) P(-1.5  Z  1.5) =

(d) P(-1.5 < Z < 2.21) =

(73)

Probabilities for Normal Random Variables in General Example: The heart rate of patients suffering from heart disease is normally distributed with a mean of 97 beats per minute and a standard deviation of 18 beats per minute. For a randomly chosen patient, find the probability the heart rate is (a) below 80 (b) more than 140, (c) between 55 and 90. Let X = the heart rate of a randomly selected patient;  =97,  = 18.

(a) P(X< 80) =

(b) P(X>140) =

(c) P(55