Answer in Statistics and Probability for Kyei William Frimpong #119572

Let “X=”  the number of days of consecutive days of snow beginning on April 1: “Xsim Po(lambda)”

“P(X=x)={e^{-lambda}cdot lambda^xover x!}”

Given “lambda=0.6”

Let “Y=” the amount that the insurance company must pay. The possible values for “Y” would be:

“GHS 0_X=0”

“GHS 1,000:X=1”

“GHS 2,000:Xu22652”

Then

“P(Y=0)=P(X=0)={e^{-0.6}cdot (0.6)^0over 0!} n=e^{-0.6}”

“P(Y=1000)=P(X=1)={e^{-0.6}cdot (0.6)^1over 1!} n=0.6e^{-0.6}”

“P(Y=2000)=P(Xu22652)=1u2212P(X=0)u2212P(X=1)=”

“=1-e^{-0.6}-0.6e^{-0.6}=1-1.6e^{-0.6}”

“E(Y)=0u22c5e^{-0.6}+1000u22c50.6e^{-0.6}+2000u22c5(1u22121.6e^{-0.6})=”

“=2000-2600e^{-0.6}approx573.089746”

“E(Y^2 )=(0)^2u22c5e^{-0.6}+(1000)^2u22c50.6e^{-0.6}+”

“+(2000)^2u22c5(1u22121.6e^{-0.6})=”

“=4000000u22125800000e^{-0.6} nn u2248816892.510655”

“Var(Y)=u03c3^ n2n =E(Y^ n2n )u2212(E(Y))^ n2n u2248”

“u2248816892.510655u2212(573.089746)^ 2u2248488460.653685”

“u03c3=sqrt{sigma^2} u2248 488460.653685u2248699”

The standard deviation of the amount that the insurance company will have to pay is “GHS 699.”