Lecture Notes 6

Continuous Probability Distribution
(Lecture Six)

Dr. Lari Arjomand

Introduction:

In lecture number four we said that a continuous random variable is a variable which can take on any value over a given interval. Continuous variables are measured, not counted. Items such as height, weight and time are continuous and can take on fractional values. For example, a basketball player may be 6.8432 feet tall.
There are many continuous probability distributions, such as, uniform distribution, normal distribution, the t distribution, the chi-square distribution, exponential distribution, and F distribution. In this lecture note, we will concentrate on the uniform distribution, and normal distribution.

Uniform (or Rectangular) Distribution:

Among the continuous probability distribution, the uniform distribution is the simplest one of all. The following figure shows an example of a uniform distribution. In a uniform distribution, the area under the curve is equal to the product of the length and the height of the rectangle and equals to one.

Figure 1

where: a=lower limit of the range or interval, and b=upper limit of the range or interval.

Note that in the above graph, since area of the rectangle = (length)(height) =1, and since length = (b - a), thus we can write: (b - a)(height) = 1 or height = f(X) = 1/(b - a).
The following equations are used to find the mean and standard deviation of a uniform distribution:

Example:

There are many cases in which we may be able to apply the uniform distribution. As an example, suppose that the research department of a steel factory believes that one of the company's rolling machines is producing sheets of steel of different thickness. The thickness is a uniform random variable with values between 150 and 200 millimeters. Any sheets less than 160 millimeters thick must be scrapped because they are unacceptable to the buyers. We want to calculate the mean and the standard deviation of the X (the tickness of the sheet produced by this machine), and the fraction of steel sheet produced by this machine that have to be scrapped. The following figure displays the uniform distribution for this example.

Figure 2

Note that for continuous distribution, probability is calculated by finding the area under the function over a specific interval. In other words, for continuous distributions, there is no probability at any one point. The probability of X>= b or of X<= a is zero because there is no area above b or below a, and area between a and b is equal to one, see figure 1.

The probability of the variables falling between any two points, such as c and d in figure 2, are calculated as follows:
P (c <= x <="d)" c)/(b a))=?
In this example c=a=150, d=160, and b=200, therefore:
Mean = (a + b)/2 = (150 + 200)/2 = 175 millimeters, standard deviation is the square root of 208.3, which is equal to 14.43 millimeters, and P(c <= x <="d)" 150)/(200 150)="1/5" thus, of all the sheets made by this machine, 20% of the production must be scrapped.)=....

Normal Distribution or Normal Curve:

Normal distribution is probably one of the most important and widely used continuous distribution. It is known as a normal random variable, and its probability distribution is called a normal distribution. The following are the characteristics of the normal distribution:

Characteristics of the Normal Distribution:

1. It is bell shaped and is symmetrical about its mean.
2. It is asymptotic to the axis, i.e., it extends indefinitely in either direction from the mean.
3. It is a continuous distribution.
4. It is a family of curves, i.e., every unique pair of mean and standard deviation defines a different normal distribution. Thus, the normal distribution is completely described by two parameters: mean and standard deviation. See the following figure.
5. Total area under the curve sums to 1, i.e., the area of the distribution on each side of the mean is 0.5.
6. It is unimodal, i.e., values mound up only in the center of the curve.
7. The probability that a random variable will have a value between any two points is equal to the area under the curve between those points.

Figure 3

Note that the integral calculus is used to find the area under the normal distribution curve. However, this can be avoided by transforming all normal distribution to fit the standard normal distribution. This conversion is done by rescalling the normal distribution axis from its true units (time, weight, dollars, and...) to a standard measure called Z score or Z value. A Z score is the number of standard deviations that a value, X, is away from the mean. If the value of X is greater than the mean, the Z score is positive; if the value of X is less than the mean, the Z score is negative. The Z score or equation is as follows:

Z = (X - Mean) /Standard deviation

A standard Z table can be used to find probabilities for any normal curve problem that has been converted to Z scores. For the table, refer to the text. The Z distribution is a normal distribution with a mean of 0 and a standard deviation of 1.
The following steps are helpfull when working with the normal curve problems:
1. Graph the normal distribution, and shade the area related to the probability you want to find.
2. Convert the boundaries of the shaded area from X values to the standard normal random variable Z values using the Z formula above.
3. Use the standard Z table to find the probabilities or the areas related to the Z values in step 2.

Example One:

Graduate Management Aptitude Test (GMAT) scores are widely used by graduate schools of business as an entrance requirement. Suppose that in one particular year, the mean score for the GMAT was 476, with a standard deviation of 107. Assuming that the GMAT scores are normally distributed, answer the following questions:

Question 1. What is the probability that a randomly selected score from this GMAT falls between 476 and 650? <= x <="650)" the following figure shows a graphic representation of this problem.

Figure 4

Applying the Z equation, we get: Z = (650 - 476)/107 = 1.62. The Z value of 1.62 indicates that the GMAT score of 650 is 1.62 standard deviation above the mean. The standard normal table gives the probability of value falling between 650 and the mean. The whole number and tenths place portion of the Z score appear in the first column of the table. Across the top of the table are the values of the hundredths place portion of the Z score. Thus the answer is that 0.4474 or 44.74% of the scores on the GMAT fall between a score of 650 and 476.

Question 2. What is the probability of receiving a score greater than 750 on a GMAT test that has a mean of 476 and a standard deviation of 107? i.e., P(X >= 750) = ?. This problem is asking for determining the area of the upper tail of the distribution. The Z score is: Z = ( 750 - 476)/107 = 2.56. From the table, the probability for this Z score is 0.4948. This is the probability of a GMAT with a score between 476 and 750. The rule is that when we want to find the probability in either tail, we must substract the table value from 0.50. Thus, the answer to this problem is: 0.5 - 0.4948 = 0.0052 or 0.52%. Note that P(X >= 750) is the same as P(X >750), because, in continuous distribution, the area under an exact number such as X=750 is zero. The following figure shows a graphic representation of this problem.

Figure 5

Question 3. What is the probability of receiving a score of 540 or less on a GMAT test that has a mean of 476 and a standard deviation of 107? i.e., P(X <= 540)="?." we are asked to determine the area under the curve for all values less than or equal to 540. the z score is: z="(540" 476)/107="0.6." from the table, the probability for this z score is 0.2257 which is the probability of getting a score between the mean (476) and 540. the rule is that when we want to find the probability between two values of x on either side of the mean, we just add the two areas together. Thus, the answer to this problem is: 0.5 + 0.2257 = 0.73 or 73%. The following figure shows a graphic representation of this problem.

Figure 6

Question 4. What is the probability of receiving a score between 440 and 330 on a GMAT test that has a mean of 476 and a standard deviation of 107? i.e., P(330

Figure 7

In this problem, the two values fall on the same side of the mean. The Z scores are: Z1 = (330 - 476)/107 = -1.36, and Z2 = (440 - 476)/107 = -0.34. The probability associated with Z = -1.36 is 0.4131, and the probability associated with Z = -0.34 is 0.1331. The rule is that when we want to find the probability between two values of X on one side of the mean, we just subtract the smaller area from the larger area to get the probability between the two values. Thus, the answer to this problem is: 0.4131 - 0.1331 = 0.28 or 28%.

Example Two:

Suppose that a tire factory wants to set a mileage guarantee on its new model called LA 50 tire. Life tests indicated that the mean mileage is 47,900, and standard deviation of the normally distributed distribution of mileage is 2,050 miles. The factory wants to set the guaranteed mileage so that no more than 5% of the tires will have to be replaced. What guaranteed mileage should the factory announce? i.e., P(X <= ?)="5%.<br"> In this problem, the mean and standard deviation are given, but X and Z are unknown. The problem is to solve for an X value that has 5% or 0.05 of the X values less than that value. If 0.05 of the values are less than X, then 0.45 lie between X and the mean (0.5 - 0.05), see the following graph.

Figure 8

Refer to the standard normal distribution table and search the body of the table for 0.45. Since the exact number is not found in the table, search for the closest number to 0.45. There are two values equidistant from 0.45-- 0.4505 and 0.4495. Move to the left from these values, and read the Z scores in the margin, which are: 1.65 and 1.64. Take the average of these two Z scores, i.e., (1.65 + 1.64)/2 = 1.645. Plug this number and the values of the mean and the standard deviation into the Z equation, you get:
Z =(X - mean)/standard deviation or -1.645 =(X - 47,900)/2,050 = 44,528 miles.
Thus, the factory should set the guaranteed mileage at 44,528 miles if the objective is not to replace more than 5% of the tires.

The Normal Approximation to the Binomial Distribution:

In lecture note number 5 we talked about the binomial probability distribution, which is a discrete distribution. You remember that we said as sample sizes get larger, binomial distribution approach the normal distribution in shape regardless of the value of p (probability of success). For large sample values, the binomial distribution is cumbersome to analyze without a computer. Fortunately, the normal distribution is a good approximation for binomial distribution problems for large values of n. The commonly accepted guidelines for using the normal approximation to the binomial probability distribution is when (n x p) and [n(1 - p)] are both greater than 5.

Example:

Suppose that the management of a restaurant claimed that 70% of their customers returned for another meal. In a week in which 80 new (first-time) customers dined at the restaurant, what is the probability that 60 or more of the customers will return for another meal?, ie., P(X >= 60) =?.

The solution to this problemcan can be illustrated as follows:
First, the two guidelines that (n x p) and [n(1 - p)] should be greater than 5 are satisfied: (n x p) = (80 x 0.70) = 56 > 5, and [n(1 - p)] = 80(1 - 0.70) = 24 > 5. Second, we need to find the mean and the standard deviation of the binomial distribution. The mean is equal to (n x p) = (80 x 0.70) = 56 and standard deviation is square root of [(n x p)(1 - p)], i.e., square root of 16.8, which is equal to 4.0988. Using the Z equation we get, Z = (X - mean)/standard deviation = (59.5 - 56)/4.0988 = 0.85. From the table, the probability for this Z score is 0.3023 which is the probability between the mean (56) and 60. We must substract this table value 0.3023 from 0.5 in order to get the answer, i.e., P(X >= 60) = 0.5 -0.3023 = 0.1977. Therefore, the probability is 19.77% that 60 or more of the 80 first-time customers will return to the restaurant for another meal. See the following graph.

Figure 9

Correction Factor:

The value 0.5 is added or subtracted, depending on the problem, to the value of X when a binomial probability distribution is being approximated by a normal distribution. This correction ensures that most of the binomial problem's information is correctly transferred to the normal curve analysis. This correction is called the correction for continuity. The decision as to how to correct for continuity depends on the equality sign and the direction of the desired outcomes of the binomial distribution. The following table shows some rules of thumb that can help in the application of the correction for continuity, see the above example.

Value Being Determined..............................Correction
X >................................................+0.50
X > =..............................................-0.50
X <.................................................-0.50
X <=............................................+0.50
<= X <="...................................-0.50" & +0.50
X =.............................................-0.50 & +0.50

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