Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PROBABILITY

SawmillAproduces boards whose lengths are Gaussian distributed with mean209.4 cm

and standard deviation5.0 cm. A board is accepted if it is longer than200 cmbut is

rejected otherwise. Show that3%of boards are rejected.

SawmillBproduces boards of the same standard deviation but of mean length210.1 cm.

Find the proportion of boards rejected if they are drawn at random from the outputs ofA

andBin the ratio3:1.

LetX= length of boards fromA,sothatX∼N(209. 4 ,(5.0)^2 )and

Pr(X<200) = Φ

(

200 −μ

σ

)

=Φ

(

200 − 209. 4

5. 0

)

=Φ(− 1 .88).

But, since Φ(−z)=1−Φ(z) we have, using table 30.3,

Pr(X<200) = 1−Φ(1.88) = 1− 0 .9699 = 0. 0301 ,

i.e. 3.0% of boards are rejected.
Now letY= length of boards fromB,sothatY∼N(210. 1 ,(5.0)^2 )and

Pr(Y<200) = Φ

(

200 − 210. 1

5. 0

)

=Φ(− 2 .02)

=1−Φ(2.02)

=1− 0 .9783 = 0. 0217.

Therefore, when taken alone, only 2.2% of boards fromBare rejected. If, however, boards
are drawn at random fromAandBin the ratio 3 : 1 then the proportion rejected is

1

4 (3×^0 .030 + 1×^0 .022) = 0.028 = 2.8%.

We may sometimes work backwards to derive the mean and standard deviation

of a population that is known to be Gaussian distributed.

The time taken for a computer ‘packet’ to travel from Cambridge UK to Cambridge MA

is Gaussian distributed.6.8%of the packets take over200 msto make the journey, and

3.0%take under140 ms. Find the mean and standard deviation of the distribution.

LetX= journey time in ms; we are told thatX∼N(μ, σ^2 )whereμandσare unknown.
Since 6.8% of journey times are longer than 200 ms,

Pr(X>200) = 1−Φ

(

200 −μ

σ

)

=0. 068 ,

from which we find

Φ

(

200 −μ

σ

)

=1− 0 .068 = 0. 932.

Using table 30.3, we have therefore

200 −μ

σ

=1. 49. (30.112)

Also, 3.0% of journey times are under 140 ms, so

Pr(X<140) = Φ

(

140 −μ

σ

)

=0. 030.