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Statistical Techniques for Transportation Engineering by Naresh Davergave, Anil Shah, G Rao, Kumar Molugaram

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image

But

|xμ|kf(x)dx=P(|xμ|k)

image

Hence we get

σ2k2(|xμ|k)

image

or

P(|xμ|k)σ2k2 (4.7)

image (4.7)

Also we have

P(|xμ|k)+P(|xμ|k)=1

image

Therefore we get

P(|xμ|k)=1=P(|xμ|k)

From Eq. (4.7), we get

P(|xμ|k)1σ2k2

Hence proved.

Remark

1. P(|xμ|k)1σ

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