bearing in mind that parallel lines have the same exact slope, then any parallel line to the one above will have the same slope as that one.
Consider the linear system of equations. y = –x + 9 y = 0.5x – 6 If the solution is (a,–1), what is a? a =
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Answer:
a) The expected value is 2.680642
b) The minimun number of newspapers the manager should order is 6.
Step-by-step explanation:
a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.
X is a random variable of Poisson distribution with mean , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values
- PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
- PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)
we obtain:
PY(0) = ε^(-3) = 0.04978..
PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936
PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404
PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404
PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768
E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) = 0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642
The store is <em>expected</em> to sell 2.680642 newspapers
b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05
we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231
FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262
FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608
FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489
So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.
I hope this helped you!