Question

The number of emails that I get in one hour has such density function:
x 2 1 4 5 3 6 and more 0
f(x) 0.17 0.23 0.08 0.04 0.14 0.02 ?
a) find the probability, that there will be no letter for the next hour;
b) find cumulative distributional function.

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c:c} x & 0 & 1 & 2 & 3 & 4 & 5 & 6\text{ and more} \\ \hline f(x) & a & 0.23 & 0.17& 0.14 & 0.08 & 0.04 & 0.02 \\ \end{array}$

a) $\sum \limits _{x} P(X=x)=1 \ \ \Rightarrow \ \ a+0.23+0.17+0.14+0.08+0.04+0.02=1 \ \ \Rightarrow \ \ a=0.32$

Answer: the probability, that there will be no letter for the next hour, is equal to 0.32

b)

The cumulative distribution function is given by $F(x)=P(X\leq x)$

$P(X\leq x)=\sum \limits _{y\leq x}P(X=y)= \sum \limits _{y\leq x-1}P(X=y)+P(X=x)=P(X\leq x-1)+P(X=x)$

$F(x)=F(x-1)+P(X=x) \ \ (\text{for }x>0)$

Now we can calculate $F(x):$

$F(0)=0.32$

$F(1)=0.32+0.23=0.55$

$F(2)=0.55+0.17=0.72$

$F(3)=0.72+0.14=0.86$

$F(4)=0.86+0.08=0.94$

$F(5)=0.94+0.04=0.98$

$F(6 \text{ and more}) =0.98+0.02=1$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c:c} x & 0 & 1 & 2 & 3 & 4 & 5 & 6\text{ and more} \\ \hline F(x) & 0.32 & 0.55 & 0.72& 0.86 & 0.94 & 0.98 & 1 \\ \end{array}$