At a grocery store check-out, define X k as the time it takes for the k th custo

Question

At a grocery store check-out, define Xk as the time it takes for the kth customer to check out. Suppose the mean check-out time E(Xk) is five minutes and that Var(Xk) = 4 min2 . Assuming that each customer starts to check out as soon as his predecessor has finished checking out, use the normal approximation to calculate the probability that 100 customers were served by this store check-out in less than nine hours. (Assume X1, X2, ..., X100 are identically distributed, so that the Central Limit Theorem is relevant.)

Explanation / Answer

for mean time for 100 customers =100*5 =500 minutes

and std deviation =(4*100)1/2 =20

there fore probability that 100 customers were served by this store check-out in less than nine hours or 540 minutes =P(X<540)=P(Z<(540-500)/20)=P(Z<2)= 0.97725

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