Your local grocery store sells 5 lb. bags of potatoes. However, the 5 lb. bags d
ID: 3046346 • Letter: Y
Question
Your local grocery store sells 5 lb. bags of potatoes. However, the 5 lb. bags don’t weigh exactly 5 lbs. If we let Xi be the weight of a randomly selected 5 lb bag of potatoes, historical data indicates that Xi~N(5.32, 0.13). The local warehouse store sells 10 lb bags of potatoes, which also do not weigh exactly 10 lbs. If Y is the weight of a randomly selected 10 lb. bag, historical data indicates that Y~N(10.23, 0.17). We randomly select two 5 lb bags of potatoes and one 10 lb. bag of potatoes.
a) Define T = Xi + Xi. What is the distribution of T? (1 point)
b) Define W = T – Y. What is the distribution of W? (1 point)
c) What is the probability that the sum of the weights of the two 5 lb. bags of potatoes exceeds the weight of one 10 lb. bag of potatoes? (2 points)
Explanation / Answer
Xi : Weight of 5 lb bag of potatoes
Xi ~ N ( 5.32 , 0.13)
E(Xi) = mean = 5.32
Var(Xi) = 0.13
Yi : Weight of 5 lb bag of potatoes
Yi ~ N( 10.23,0.17)
E(Yi) =mean = 10.23
Var(Yi) =0.17
a) T = Xi +Xi = 2Xi
E(T) = E(2*Xi) =2 * E(Xi) = 2* 5.32 = 10.64
Var(T) =Var(2*Xi) =4 * Var(T) = 4*0.13 =0.52
The distribution of T is
T ~ N( 10.64, 0.52)
b) W = T - Y
E(W) = mean = E(T-Y) = E(T) - E(Y) = 10.64 - 10.23 = 0.39
Var(W) = Var(T-Y) = Var(T) + Var(Y) ( Since T and Y are independent)
Var(W) = 0.52 + 0.17 = 0.69
The distribution of W is
W ~ N( 0.39 , 0.69)
c) P( Sum of the weights of the two 5 lb bags of potatoes is greater than weights of one 10 lb bags of potatoes)
= P( 2X > Y) = P(T>Y) = P(T-Y>0) = P(W > 0)
P(W>0) = P( (W-mean(W))/S.D(W) > (0-0.39)/0.69)
= P( Z > -0.5652) where Z = ( W - mean(W))/ S.D(W) ~ N(0,1)
From normal probability table
P(Z > -0.5662) = 0.714
P( Sum of the weights of the two 5 lb bags of potatoes is greater than weights of one 10 lb bags of potatoes) = 0.714
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