Question

If three coins are tossed, what are the mean, variance, and standard deviation of the number of heads that occur?

Answer 1

We assume that the probabilities of getting heads and tails are the same

$p = q = \frac{1}{2}$

Using the Bernoulli formula, we find the probabilities that there will be 0, 1, 2, 3 and 4 heads, respectively

$P\left( {x = 0} \right) = {q^3} = \frac{1}{{8}}$

$P(x = 1) = C_3^1p{q^2} = \frac{3}{{8}}$

$P(x= 2) = C_3^2{p^2}{q} = \frac{3}{{8}}$

$P\left( {x = 3} \right) = {p^3} = \frac{1}{{8}}$

We have a distribution series

So, the mean is

$M(x) = {\sum x _i}{p_i} = \frac{{0 \cdot 1 + 1 \cdot 3 + 2 \cdot 3 + 3 \cdot 1}}{8} = \frac{{12}}{8} = \frac{3}{2}$

The variance is

$V(x) = M\left( {{x^2}} \right) - {M^2}(x) = \frac{{0 \cdot 1 + 1 \cdot 3 + 4 \cdot 3 + 9 \cdot 1}}{8} - \frac{9}{4} = \frac{3}{4}$

standard deviation is

$\sigma \left( x \right) = \sqrt {V(x)} = \frac{{\sqrt 3 }}{2}$

Answer: $M(x) = \frac{3}{2}$ ; $V(x) = \frac{3}{4}$ ; $\sigma \left( x \right) = \frac{{\sqrt 3 }}{2}$