For A) We know it should be 50,50 since E[x_1] is 100. So P=.5
For B) Here is an estimate using chevyshev’s inequality
\(P(Y_{365}>110)=P(Y_{365}>100+10) \le Var(Y_{365}) /(Var(Y_{365}) + 10^2) = 91.25/(91.25+100)\) I think I did it right; feel free to correct me.
Anyway; lets do this in R.
o=sqrt(365*.25)
A = 1-pnorm(0,mean=0,sd=o,lower.tail=TRUE)
B = 1-pnorm(10,mean=0,sd=o,lower.tail=TRUE)
C = 1-pnorm(20,mean=0,sd=o,lower.tail=TRUE)
answers = c(A,B,C)
questions = c('A','B','C')
rbind(questions,answers)## [,1] [,2] [,3]
## questions "A" "B" "C"
## answers "0.5" "0.147584879761555" "0.0181435485458444"
To make an MGF of a specific distribution; plug it in like so. (fx(x) being the distribution)
The MGF for the binomial distribution is given as..
To find mean; we take the first derivative of M(t) with respect to t and let t = 0.
\(\mu = M'(o) = n(1-p+pe^0)^{n-1}(pe^0)\) ==>
\(n(1-p+p)^{n-1}p(1)\) ==>
\(\mu=np\)
To find variance; we take the 2nd derivative of M(t) - the 1st derivative with respect to t and let t = 0.
\(M''(0) = n(1-p+pe^0)^{n-1}(pe^0)+(pe^0)n(n-1)(1-p+pe^0)^{n-2}pe^0\)
\(n(1)^{n-1}p+pn(n-1)(1)^{n-2}p(1)\)
\(np+n^2p^2-np^2\) Now that we have the 2nd derivative, deduct the 1st derivitive\(^2\) to find \(\sigma^2\)
\(\sigma^2 = np+n^2p^2-np^2-(np)^2\) ==>
\(np-np^2\) ==>
\(\sigma^2 = np(1-p)\)
Same process as finding the mean and variance of the binomial distribution. First we take the MGF of an exponential distribution; then we derive 1st and 2nd derivatives, 1st derivative = mean : 2nd-1st derivative\(^2\) = variance.
Exponential mgf : \(\lambda \over \lambda-t\) (for \(t<\lambda\))
\(M'(0) =\) \(\lambda \over (\lambda-t)^2 =\) \(1\over \lambda\) So this one is our mean \(\mu =\) \(1\over \lambda\)
\(M''(0) =\) \(2\lambda \over (\lambda-t)^3 =\) \(2\over \lambda^2\) And we use both of these to find variance
\(\sigma^2 =\)\(2\over\lambda^2 -\)\(1\over\lambda^2\) = \(1\over\lambda^2\)
Citations : http://www.maths.qmul.ac.uk/~bb/MS_Lectures_5and6.pdf
https://onlinecourses.science.psu.edu/stat414/node/73