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Perhaps the simplest of all the continuos probability distributions is the uniform

distribution. The frequency function has a rectangular shape, as shown in Figure

6.2, Note that the possible values of x consist of all points on the real line between

point c and point d. The height f(x) is constant in that interval and equals 1/(d-c).

Therefore, the total area under f(x) is given by

Total area of rectangle=(base0(height)= (d-c) (l/d-c)=1

f(x)

1

d - c

C d

The uniform probability distribution provides a model for continuous random

variables that are evenly distributed over a certain interval. That is, a uniform

random variable is one that is just likely to assume a value in one interval as it is to

assume a value in any other interval of equal size. There is no clustering of valuesaround any value; instead, there is an even spread over the entire region of

possible values.

The uniform distribution is sometimes referred to as the randomness

distribution, since one way of generating a uniform randomly selected on the

horizontal axis between the points c and d. If we were to repeat this experiment

infinitely often, we would create a uniform probability distribution like that shown in

Figure 6.2. The random selection of points on a line can also be used to generate

random numbers such as those in Table I of Appendix B. Recall that random

numbers are selected in such a way that every digit has an equal probability of

selection. Therefore, random numbers are realizations of a uniform variable.(Random numbers were used to draw random samples in Section 4.6)

The formulas for the uniform probability distribution and its mean and

standard deviation are shown in the box on page 262.

PROBABILITY DISTRIBUTION, MEAN AND STANDARD DEVIATION OF A UNIFORM

RANDOM VARIABLE x

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F(x)= 1 (c x d)

d c

= c + d = d c

2 12

Suppose the interval a < x < b lies within the domain of x; i.e., it falls within

the larger interval c < x < d. Then the probability that x assumes a value within the

interval a < x < b is the area of the rectangle over the interval---namely,

b a

d c

Suppose the research department of a steel manufacturer believes that one of the

companys rolling machines is producing sheets of steel of varying thickness. The

thickness is a uniform random variable with values between 150 and 200millimeters. Any sheet less than 160 millimeters thick must be scrapped because

they are unacceptable to buyers.

a. Calculate the mean and standard deviation of x, the thickness of the sheets

produced by this machine. Then graph the probability distribution and show

the mean on the horizontal axis. Also show 1 and 2 standard deviation

intervals around the mean.

b. Calculate the fraction of steel sheets produced by this machine that have to

be scrapped.

a. To calculate the mean and standard deviation for x, we substitute 150 and

200 millimeters for c and d. respectively, in the formulas. Thus,

= c + d = 150 + 200 = 174 millimeters

2 2

and

= d c = 200 150 = 50 = 14.43

12 12 3.464

The uniform probability distribution is

f(x) = 1 = 1 = 1

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d c 200 150 50

THE UNIFORM DISTRIBUTION

The graph of this function is shown in Figure 6.3. The mean and the 1 and 2

standard deviation intervals around the mean are shown on the horizontal axis.

f(x)

1

50

0 150 160 170 180 190 200

- 2 - + +

2

b. To find the fraction of steel sheets produced by the machine that have to

be scrapped, we must find the probability that x, the thickness, is less

than 160 millimeters. As indicated in Figure 6.4, we need to calculate thearea under the frequency function f(x) between the points x = 150 and x

= 160. This is the area rectangle w/ base 160 150 = 10 and height 1/50.

The fraction that has to be scrapped is then

P(x < 160) = (Base)(Height) = (10) (1/50) = 1/5

That is, 20% of all the sheets made by this machine must be scrapped.

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f(x)

1

50

0 150 160 170 180 190 200