Distribution

Answer:


Setting:

Let X = error made by the less accurate instrument
Let Y = error made by the more accurate instrument
Let A = average error made by both instrument (X + Y)/2

X ~ N(0, 0.0056h)
Y ~ N(0, 0.0044h)
A ~ N(0, 0.00356h)

E(A)
= E[(X + Y)/2]
= (1/2) * [E(X) + E(Y)]
= (1/2) * (0 + 0)
= 0

V(A)
= V[(X + Y)/2]
= (1/4) * [V(X) + V(Y)]
= (1/4) * [(0.0056h)^2 + (0.0044h)^2]
= 0.00001268 * h^2

SD(A) = sqrt(0.00001268) * h = 0.00356 h


Procedure:

P(their average value is within 0.005h of the height of the tower)
= P(-0.005h ≤ A ≤ 0.005h)
= P{ [-0.005h - E(A)] / SD(A) ≤ Z ≤ [0.005h - E(A)] / SD(A) }
= P{ [-0.005h - 0] / 0.00356h ≤ Z ≤ [0.005h - 0] / 0.00356h }
= P(-1.40 ≤ Z ≤ 1.40)
= P(Z ≤ 1.40) - P(Z ≤ -1.40)
= 0.9192 - 0.0808 (use normal distribution table)
= 0.8384

Conclusion:
The probability that their average value is within 0.005h of the height of the tower is 0.8384.

If
圖片參考：http://upload.wikimedia.org/math/7/0/1/701cfc64a709902a1270c47864119652.png

If
圖片參考：http://upload.wikimedia.org/math/2/e/2/2e26b49dbd44bb0b4650744a65f83097.png
are independent normal random variables, then:

Their sum is normally distributed with
圖片參考：http://upload.wikimedia.org/math/8/1/f/81fcbdfdcfb20f9af35db82829da43b0.png

Their difference is normally distributed with
圖片參考：http://upload.wikimedia.org/math/8/b/2/8b22e1778fd0e4548eda237bad71aa44.png
.
X~(0,0.00003136h^2)
Y~(0,0.00001936h^2)
U=X+Y~(0,0.00005072h^2)
T=1/2V~(0,0.00001268h^2)
the mean of T is 0 and the standard deviation is 0.00356h
the probability that their average value is within 0.005h of the height of the tower
=P(|T|<=0.005h)
=P(-0.005h<=T<=0.005h)
=P(-0.005h/0.00356h<=Z<=0.005h/0.00356h)
=P(-1.404494<=Z<=1.404494)
=F(1.404494)-F(-1.404494)
=0.9192-0.0808
=0.8384