Answer:
1. The distribution of X, [tex]P(X=x)= 0.95^x*0.05; x = 0, 1, 2, 3, 4, 5, ...[/tex],
Mean of the distribution, [tex]M_x(t)= \frac{0.05}{(1-0.95e^t)}[/tex],
Variance of X = [tex]\frac{Q}{P^2}[/tex]
2. Probability that there would be an injury in the next 5 working days = 0.2263
3. Probability of winning = 0.0407
Step-by-step explanation:
1. Let us denote the number of days until there's a work place injury by [tex]X[/tex]
then [tex]P =[/tex] Probability of having a work place injury
and [tex]Q =[/tex] Probability of having no work place injury
[tex]X[/tex] derives values from {0, 1, 2, 3, 4, 5,...}
[tex]X = 0[/tex] means that there's a work place injury on the first day
and given that there's a 50% chance of a workplace injury on any given day,
[tex]P(X = 0) = 0.05[/tex]
[tex]X = 1[/tex] means that there's no work place injury on the first day, but on the second day, there is. Hence,
[tex]P(X=1)= 0.95*0.05\\P(X=2)= 0.95^2*0.05\\P(X=x)=0.95^x*0.05;x = 0, 1, 2, 3, 4, 5, ...[/tex]
This is the pmf for geometric random variable
[tex]M_x(t)= E(e^{tx})= \sum^{\infty}_{k=0}e^{tk}P(X = k)=\sum^{\infty}_{k=0}e^{tk}(0.95)^k0.05\\=0.05\sum^{\infty}_{k=0}e^{tk}(0.95)^k=\frac{0.05}{(1-0.95e^t)}[/tex]
[tex]M_X(t) = E(e^{tx}) = \frac{P}{1-Qe^t}[/tex]
[tex]\frac{dMx(t)}{dt}=\frac{d}{dt}[P(1-Qe^t)^{-1}]=P(-1)(1-Qe^t)^{-2}(-Qe^t)\\\frac{d^2Mx(t)}{dt^2}=\frac{d}{dt}[PQe^t(1-Qe^t)^{-2}]\\\\ =PQe^t(1-Qe^t)^{-2}+PQe^t(-2)(1-Qe^t)^{-3}((-Q)e^t)\\[/tex]
The equation becomes,
[tex]\frac{dM_x(t)}{dt} \int\limits_{t=0} = PQ(1-Q)^{-2}=\frac{PQ}{P^2}=\frac{Q}{P} \\ \frac{d^2M_x(t)}{dt^2} \int\limits_{t=0} = PQ*\frac{1}{(1-Q)^2} + 2PQ^{2}*\frac{1}{(1-Q)^3}\\ =\frac{Q}{P} +\frac{2Q^2}{P^2}[/tex]
Let μ[tex]^'_1= \frac{Q}{P}[/tex] and μ[tex]^'_2 = \frac{Q}{P}+\frac{2Q^2}{P^2}[/tex]
[tex]Var(x) =[/tex] μ[tex]^'_2[/tex] - μ[tex]^'_1^2[/tex] =
[tex]\frac{Q}{P}+\frac{2Q^2}{P^2} - (\frac{Q}{P} )^2= \frac{Q}{P} + \frac{Q^2}{P^2}\\ =\frac{QP+Q^2}{P^2} = \frac{Q(P+Q)}{P^2}=\frac{Q}{P^2}[/tex]
since P + Q = 1
2. Probability that there will be an injury within the next 5 working days;
[tex]P(X\leq 5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\=P+QP+Q^2P+Q^3P+Q^4P+Q^5P\\=P(1+Q+Q^2+Q^3+Q^4+Q^5)\\=P(\frac{1-Q^5}{1-Q} )\\=0.05(\frac{1-0.95^5}{1-0.95} )\\=0.05(\frac{1-0.7737}{0.05} )\\=0.2263[/tex]
3. Probability of winning if wager is placed on day 4,
[tex]P(X=4)=(0.95^4)0.05=0.815*0.05=0.0407[/tex]
