An article considered the use of a uniform distribution with

An article considered the use of a uniform distribution with A = 0.20 and B = 4.25 for the diameter X of a certain type of weld (mm) (a) Determine the pdf of X. (Round your answers to three decimal places.) 0.20

Solution

a)

f(x) = 1/(b-a) = 1/(4.25-0.2) = 0.24691358

Thus,

f(x) = 0.24691358, 0.20<x<4.25
0, otherwise [ANSWER]

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a)

The correct graph is the UPPER RIGHT GRAPH.

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b)

Here,

P = width of interval/(b-a)

Thus, from 3 to 4.25,

P(x>3) = (4.25-3)/(4.25-0.2) = 0.308641975 [ANSWER]

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c)

Within 2 mm means the width of that interval is 2 mm *2= 4 mm.

Thus,

P(within 2mm) = 4/(4.25-0.2) = 0.987654321 [ANSWER]
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d)

The width of this interval is 3. Thus,

P(a<x<a+3) = 3/(4.25-0.2) = 0.740740741 [ANSWER]