TAOCP 1.2.10 Exercise 5
Figure 11 corresponds to the coin-tossing experiment of Eq.
Section 1.2.10: Analysis of an Algorithm
Exercise 5. [M13] What are the mean and standard deviation of the distribution in Fig. 11?
Verified: yes
Solve time: 35s
Solution
Figure 11 corresponds to the coin-tossing experiment of Eq. (18) with $n=12$ and $p=\frac35$. Hence $q=1-p=\frac25$.
By Eq. (19), the mean number of heads is
$$ pn=\frac35\cdot 12=\frac{36}{5}=7.2. $$
The variance is
$$ pqn=\frac35\cdot\frac25\cdot 12 =\frac{72}{25}. $$
Therefore the standard deviation is
$$ \sqrt{\frac{72}{25}} =\frac{\sqrt{72}}{5} =\frac{6\sqrt2}{5} \approx 1.697056. $$
Thus the distribution in Fig. 11 has mean
$$ \boxed{7.2} $$
and standard deviation
$$ \boxed{\frac{6\sqrt2}{5}\approx 1.697056}. $$