Let, Y be the number of meat pies a food stand sells at a local footy match. Fro
ID: 3225410 • Letter: L
Question
Let, Y be the number of meat pies a food stand sells at a local footy match. From previous matches, it is known that E [K] = 90 and Var (Y) = 16. It takes 2 hours to prepare the ingredients for all the meat pics and 6 minutes to assemble each individual pic. Let T be the total time in hours spent preparing all the meat pics sold at the match. (a) Write an expression for T in terms of Y. (b) Find a lower bound for the probability that the total time to prepare all the meat pies lies between 9.8 and 12.2 hours.Explanation / Answer
E(Y) = 90 and Var (Y) = 16
(a) T = f(Y)
T = 2 + 0.1 Y
(b) E [T] = E [ 2 + 0.1 Y ] = 2 + 0.1 E(Y) = 2 + 0.1 * 90 = 11 hours
Var (T) = Var ( 2 + 0.1 Y) = b2 Var (Y) = 0.12 * 16 = 0.16 Hrs2
Std. Dev (T) = 0.4 Hrs2
Pr ( 9.8 <= T <= 12.2) = ?
Z = ( 12.2 - 11)/ 0.4 = 3
Z = ( 9.8 - 11)/ 0.4 = -3
Chebychev's Inequality
P [ - k < < + k ] >= 1 - 1/k2 >= 1 - 1/32 = 8/9 = 0.889
so lower bound for probability = 0.889
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