15. The number of hot dogs sold at a football stand has the following probabilit

Question

15. The number of hot dogs sold at a football stand has the following probability distribution: r Probability 800 0.05 900 0.10 1000 0.25 1100 0.35 200 0.10 1300 0.10 1400 0.05 The hot dog vendor pays 30 cents for each hot dog and sells it for 45 cents. Thus for every hot dog sold he makes a profit of 15 cents, and for every hot dog unsold he loses 30 cents. What are the expected value and variance of his profit if the number of hot dogs he orders is: (a) 1100; (b) 1200; and (c) 1300? If he wants to maximize his expected profits, how many hot dogs should he order?

Explanation / Answer

a)

expected value =E(X)=0.05*(800*15-(1100-800)*30)+0.1*(900*15-(1100-900)*30)+0.25*(1000*15-(1100-1000)*30)+0.6*(1100*15)=13800 cents

E(X2 )==0.05*(800*15-(1100-800)*30)2+0.1*(900*15-(1100-900)*30)2+0.25*(1000*15-(1100-1000)*30)2+0.6*(1100*15)2 =205425000

Var(X) =E(X2 )-(E(X))2 =14985000 cents2

b)

expected value =E(X)=0.05*(800*15-(1200-800)*30)+0.1*(900*15-(1200-900)*30)+0.25*(1000*15-(1200-1000)*30)+0.35*(1100*15-(1200-1100)*30)+0.25*(1200*15)=11925 cents

E(X2 )=0.05*(800*15-(1200-800)*30)2+0.1*(900*15-(1200-900)*30)2+0.25*(1000*15-(1200-1000)*30)2+0.35*(1100*15-(1200-1100)*30)2+0.25*(1200*15)2=167062500

Var(X) =E(X2 )-(E(X))2 =24856875 cents2

c)

expected value =E(X)=0.05*(800*15-(1300-800)*30)+0.1*(900*15-(1300-900)*30)+0.25*(1000*15-(1300-1000)*30)+0.35*(1100*15-(1300-1100)*30)+0.10*(1200*15-(1300-1200)*30)+0.15*(1300*15)=9600 cents

E(X2 )=0.05*(800*15-(1300-800)*30)2+0.1*(900*15-(1300-900)*30)2+0.25*(1000*15-(1300-1000)*30)2+0.35*(1100*15-(1300-1100)*30)2+0.10*(1200*15-(1300-1200)*30)2+0.15*(1300*15)2=127800000

Var(X) =E(X2 )-(E(X))2 =35640000

from above we can see that making 1100 hot dog maximizes the profit

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