8.1 question 11

\(n=364\), \(Y_{365}= 100\), \(\sigma = \sqrt{n/4}=\sqrt{n}/2\)

a >= 100

  • .5
my_std_dev <- sqrt(364)/2
pnorm(100-100,mean=0, sd=my_std_dev)
## [1] 0.5

b >= 110

  • .147
pnorm(110-100,mean=0,lower.tail= FALSE, sd=my_std_dev)
## [1] 0.1472537

c >= 120

  • .018
pnorm(120-100,mean=0,lower.tail= FALSE, sd=my_std_dev)
## [1] 0.01801584

Question 2A Solve for the mean and variance of the binomial distribution using MGF

  • link where I found solution

  • from wiki page the MGF for binomial is \[ (1-p+pe^{t})^{n} \]

  • the mean is equivalent to the expected value of the MGF(x’) evaluated at t=0

\[ M_x'(t)=n[1-p+p^{et}]^{n-1}(pe^t) \]

  • Evaluated at 0 \[M_x'(t)= n[1-p+p]*p(1)= n*p\]

  • To find the vairance
    • M_x’‘(0)- M_x’(0)^2

\[M_x''(t)= n[1-p+pe^t]^{n-1}*(pe^t)+(pe^t)(n)(n-1)[1-p+pe^t]^{n-2}(pe^t) \]

  • evaluated at 0 \[M_x''(t)= n[1]^{n-1}p+np(n-1)(1)^{n-2}p(1)\]

\[M_x''(t)= np+n^2p^2-np^2 - (np)^2\] \[np-np^2\]

\[q=(1-p).......Variance=np(1-p)=npq\]

2b Solve for the mean and variance of the Exponential distribution using MGF

link where I found solution

  • from wiki MGF = \(\frac{\lambda}{\lambda-t}\)

mx’(t)= \(\frac{\lambda}{(\lambda-t)^2}\)

  • evaluated at 0

\(mean=\frac{1}{\lambda}\)

  • now solve for vairance at t=0 at 2nd derivative of MGF \[mx''(t)= -2\lambda(\lambda-t)^{-3}\] \[mx''(t)= \frac{2\lambda}{(\lambda-t)^3}\] \[Mx''x(0)= \frac{2}{\lambda^2}\]
  • Variance= \(M_x''(0)- M_x'(0)^2\) \[ \frac{2}{\lambda^2}- (\frac{1}{\lambda})^2 \] \[Variance= \frac{1}{\lambda^2}\]