Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.

Mean = $\frac{3}{2}$, Variance = $\frac{3}{4}$, Standard Deviation = $\frac{\sqrt{3}}{2}$

The correct answer is Option (3) → Mean = $\frac{3}{2}$, Variance = $\frac{3}{4}$, Standard Deviation = $\frac{\sqrt{3}}{2}$

When a coin is tossed three times, then the sample space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.

Let random variable X denote the number of tails, so X can take values 0, 1, 2, 3.

$\text{P(X = 0) = P(no tails) = P(HHH)}=\frac{1}{8}$

$\text{P(X = 1) = P(one tail) = P(HHT, HTH, THH)}=\frac{3}{8}$

$\text{P(X = 2) = P(two tails) = P(HTT, THT, TTH)}=\frac{3}{8}$

$\text{P(X = 3) = P(three tails) = P(TTT)}=\frac{1}{8}$

We construct the following table:

∴ Mean = $μ=Σp_ix_i=\frac{3}{2}$

Variance = $σ^2 = Σp_i{x_i}^2 - μ^2 = 3 - (\frac{3}{2})^2 =3-\frac{9}{4}=\frac{3}{4}$

Standard deviation = $σ =\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$