Answer to Question #108228 in Statistics and Probability for mazen

Question

A coin is tossed 3 times. Let Z is a number of tails. Find:
a) a probability density function f(z);
b) a cumulative distributional function F(z);
c) a mean E(Z) and a variance D(Z).

Answer 1

(a)

We can consider all outcomes for this experiment: $\{ttt, thh, hth, hht , tth, tht, htt, hhh\}$

For $Z=0 : \ \{hhh\}$

$Z=1:\ \{thh, hth, hht\}$

$Z=2: \ \{ tth, tht,htt \}$

$Z=3: \ \{hhh\}$

Therefore, we have the probability distribution of $Z$ :

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} z & 0 & 1 & 2 & 3 \\ \hline P(z) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \end{array}$

Probability density function:

(b)

$F(z)=P(Z\leq z)$

$F(0)=P(Z\leq 0)=P(0)=\frac{1}{8}, \ F(1)=P(Z\leq1)=P(0)+P(1)=\frac{1}{2}$

$F(2)=P(Z\leq 2)=P(0)+P(1)+P(2)=\frac{7}{8}, \ F(3)=P(Z\leq 3)=1$

$F(z)=\begin{cases} \frac{1}{8}, \ z=0 \\ \frac{1}{2}, \ z=1 \\ \frac{7}{8}, \ z=2 \\ 1 , \ z=3 \end{cases}$

(c)

Mean value is $E(Z)=\sum zP(z)=0\times P(0)+1\times P(1)+2\times P(2)+3\times P(3)=0+\frac{3}{8}+\frac{6}{8} +\frac{3}8=\frac{12}{8}=1.5$

$D^2(Z)=[\sum z^2 P(z)]-E^2(Z)=[0^2\times P(0)+1^2\times P(1)+2^2\times P(2)+3^2\times P(3)-1.5^2=0+\frac{3}{8}+\frac{12}{8}+\frac{9}{8}-2.25=\frac{24}{8}-2.25=0.75$

Variance is $D(z)=\sqrt{0.75}\approx 0.866$