The second option is your answer 2x2+8x-24
Answer:
the dependant variable is the time
Hi there what you need is lagrange multipliers for constrained minimisation. It works like this,
V(X)=α2σ2X¯1+β2\sigma2X¯2
Now we want to minimise this subject to α+β=1 or α−β−1=0.
We proceed by writing a function of alpha and beta (the paramters you want to change to minimse the variance of X, but we also introduce another parameter that multiplies the sum to zero constraint. Thus we want to minimise
f(α,β,λ)=α2σ2X¯1+β2σ2X¯2+λ(\alpha−β−1).
We partially differentiate this function w.r.t each parameter and set each partial derivative equal to zero. This gives;
∂f∂α=2ασ2X¯1+λ=0
∂f∂β=2βσ2X¯2+λ=0
∂f∂λ=α+β−1=0
Setting the first two partial derivatives equal we get
α=βσ2X¯2σ2X¯1
Substituting 1−α into this expression for beta and re-arranging for alpha gives the result for alpha. Repeating the same steps but isolating beta gives the beta result.
Lagrange multipliers and constrained minimisation crop up often in stats problems. I hope this helps!And gosh that was a lot to type!xd
Answer:
Step-by-step explanation:
Given that 10000 times a fair coin is flipped. Since fair coin prob for head of tail =0.5
If tail comes it will be taken as 1. Hence for the sum to be less than 5000, no of heads <5000
Since trials are large we approximate to normal with mean = np = 5000 and variance = np(1-p) = 2500
Thus no of heads x is normal with (5000, 50)
Since this prob is less than 0.5 only minimum amount to be bet i.e. 1 dollar
If sum of coins <5100, then Z =
P(Z<2) = 0.9500
This time we can bet maximum of 100 dollars as probability is very high nearer to 1.