Applied Statistics and Probability for Engineers

4-2 PROBABILITY DISTRIBUTIONS AND PROBABILITY DENSITY FUNCTIONS 101

As another example,

EXAMPLE 4-2 Let the continuous random variable Xdenote the diameter of a hole drilled in a sheet metal

component. The target diameter is 12.5 millimeters. Most random disturbances to the process

result in larger diameters. Historical data show that the distribution of Xcan be modeled by a

probability density function

If a part with a diameter larger than 12.60 millimeters is scrapped, what proportion of

parts is scrapped? The density function and the requested probability are shown in Fig. 4-5. A

part is scrapped if Now,

What proportion of parts is between 12.5 and 12.6 millimeters? Now,

Because the total area under f(x) equals 1, we can also calculate

EXERCISES FOR SECTION 4-2

1 P 1 X12.6 2  1 0.1350.865.

P 1 12.5X12.6 2 

P 1 12.5X12.6 2  

12.6

12.5

f 1 x 2 dxe^201 x12.5^2 `

12.6

12.5

0.865

P 1 X12.60 2  



12.6

f 1 x 2 dx 



12.6

20 e^201 x12.5^2 dxe^201 x12.5^2 `



12.6

0.135

X12.60.

f 1 x 2  20 e^201 x12.5^2 , x12.5.

P 15 X 202 

20

5

f 1 x 2 dx0.75

Figure 4-5 Probability density function for

Example 4-2.

12.5

f(x)

12.6 x

4-1. Suppose that for Determine the fol-

lowing probabilities:

(a) (b)

(c) (d)

(e)

4-2. Suppose that for

(a) Determine xsuch that

(b) Determine xsuch that P 1 Xx 2 0.10.

P 1 xX 2 0.10.

f 1 x 2 ex 0 x.

P 13 X 2

P 1 X 32 P 1 X 42

P 11 X 2 P 11 X2.5 2

f 1 x 2 ex 0 x. 4-3. Suppose that for Determine the

following probabilities:

(a) (b)

(c) (d)

(e)

4-4. Suppose that Determine the

following probabilities:

(a)P 11 X 2 (b)P 12 X 52

f 1 x 2 e^1 x^42 for 4x.

P 1 X3.5 or X4.5 2

P 14 X 52 P 1 X4.5 2

P 1 X 42 P 1 X3.5 2

f 1 x 2 x 8 3 x5.

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