Question

1. A box of donuts at a bakery usually contains 12 donuts, but sometimes contains 11 or 13. Suppose that there is a 15% chance that a box contains 11 donuts, a 60% chance that a box contains 12 donuts, and a 25% chance that a box contains 13 donuts. Approximate the probability that you will get at least 365 donuts if you buy 30 boxes.

Answer 1

Let X; - Then + X30 Hexe 30 boxes are independently bought- number of donuts in the ith box bought i= 1,2,---, 30. total number of denuts in 30 boxes is sa x,+x2t we are to find P(57,365) rid x where that X, Xagoragazo given can be defined as, no no of donuts in a hox bought / picked pandomly. It is given that X=51, with probability NOW it is 11 12 131 11 The pmf of x is nal / 100 for (n) X 12 བ 2.5 x=13 otherwise 2 13 E(X) =u I XER P[Xan 2 x x Plxan 2011 15 100 25 + 12* + 13x 12.1 E (x2 E x² P[xan nal 11 х 15 100 2 12 t 2 13 X 25 + 160 196.8 2. var(x) = 62 € (x2) - E(X) 0:39

ind central limit theorem (CLT): If x,,x2,..., tu x where E(X) al and vas (X) = 62 exist finitely, then In-u ANCOI) where xn as no - Exi ial Thas for large an, the distribution of Intu 6/5 can be well approximated by a NC0, 1) distribution- Here we =30, is quite lange. so the required probability be well approximated as (57365) = 8 [ Tn3 365 I have an = 15 =P[ Xn-u can and n=30. 365 -M n shan ap [NCO > 365/viru /n where El-( 30.5 slun 2.1-2 ( 34 12.!) P(x) =P[NCO, 1) < u y as n=30, M = 12.1 8220:39 [0.39 -1- (0.584) 0.721 17 0.279 This an approximation of the required probability is m.279.
The detailed solution is given in the pictures above.

Please go through them carefully specially the notations.

I have used central limit theorem here. You may consult any standard book or internet for that.

Hope the solution helps. Thank you.