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Continuous random variables

There are many other ways in which the Gaussian distribution may be used. We now illustrate
some of the uses in more complicated examples.

Example 6.9. Sawmill A produces boards whose lengths are Gaussian distributed
with mean 209.4 cm and standard deviation 5.0 cm. A board is accepted if it is
longer than 200 cm but is rejected otherwise. Show that 3% of boards are rejected.
Sawmill B produces boards of the same standard deviation but of mean length 210.1
cm. Find the proportion of boards rejected if they are drawn at random from the
outputs of A and B in the ratio 3 : 1. ,

2
Solution. Let X = length of boards from A, so that X ∼ N(209.4, 5 ) and
( ) ( )
200 − µ 200 − 209.4
P(X < 200) = Φ = Φ = Φ(−1.88).
σ 5

But, since Φ(−z) = 1 − Φ(z) we have

P(X < 200) = 1 − Φ(1.88) = 1 − 0.9699 = 0.0301,

i.e. 3.0% of boards are rejected.
2
Now let Y = length of boards from B, so that Y ∼ N(210.1, 5 ) and
( )
200 − 210.1
P(Y < 200) = Φ = Φ(−2.02) = 1 − Φ(2.02) = 1 − 0.9783 = 0.0217.
5

Therefore, when taken alone, only 2.2% of boards from B are rejected. If, however, boards
are drawn at random from A and B in the ratio 3 : 1 then the proportion rejected is
1
(3 × 0.030 + 1 × 0.022) = 0.028 = 2.8%.
4

We may sometimes work backwards to derive the mean and standard deviation of a population
that is known to be Gaussian distributed.

Example 6.10. The time taken for a computer ’packet’ to travel from Odessa
(Ukraine) to Odessa (USA) is Gaussian distributed. 6.8% of the packets take over
200 ms to make the journey, and 3.0% take under 140 ms. Find the mean and standard
deviation of the distribution. ,

2
Solution. Let X = journey time in ms; we are told that X ∼ N(µ, σ ) where µ and σ are
unknown. Since 6.8% of journey times are longer than 200 ms,
( )
200 − µ
P(X > 200) = 1 − Φ = 0.068,
σ

from which we find
( )
200 − µ
Φ = 1 − 0.068 = 0.932.
σ
We have therefore
200 − µ
= 1.49. (6.26)
σ
Also, 3.0% of journey times are under 140 ms, so

( )
140 − µ
P(X < 140) = Φ = 0.030.
σ

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