Question

The Yankee Stadium food menu includes dirty fries and loaded fries. Let D be the weight, in ounces, of a randomly selected order of dirty fries. Let L be the weight, in ounces, of a randomly selected order loaded fries. Suppose D is normally distributed with a mean weight of 5.2 ounces and a standard deviation of 0.4 ounces. Suppose L is normally distributed with a mean weight of 5.4 ounces and a standar deviation of 0.5 ounces. L and D are independent. a) Calculate the expected value of L D. (in ounces) 10.6 b) Calculate the standard deviation of L D. (in ounces) 0.640 c) What is the probability that an order of loaded fries weighs less than 5 ounces? 0.212 d) What is the probability that an order of loaded fries weighs between 5 and 6 ounces? 0.673 e) What is the 95th percentile of D? f) What is the probability that LD is greater than 11 ounces? g) Suppose we purchase an order of loaded fries and an order of dirty fries. What is the probability that each of the orders weighs more than 5 ounces? (L>5 and D 5) h)h. What is the probability that D is greater than L?

Answer 1

page 1. Let D be weight in ounces and Dis normally distributed with mean 5.2 ounces and standard deviation of 0.4 ounces. also, L is normally distributed with mean weight 5.4 ounces and standard deviation of 0.5 ounces 1. DUN(5.2 (0.4) ²). LON(5.4, (0.5)) I and I are independent. @ we have to calculate E(L+D) : .. we know that, suppose , XUNC , 02) and YON Calls 902) By additive property Xtr NN(41+2, 01702) Using this property, we can write, L+D N N((5.2 +5.4),((0.4)+(-))) LTD NN (1006, 0.61) E(L+D) = 10.6 6 standard deviation of LHD. o = var(1+0) - 0.41 standard deviation = Joul = 0.640.3 © we have to calculate P(<5) LUN(5.4, (0.512) standardized the variate Lusing zu L- u

Ipage 2 P(125) =P L -1 5 -5.4 0.5 = P(Z -0.8) = P(Z 70.8) By using standard normal probability table P(27 0.8) = 0.2119. 1P (245) = 0.2119 Now.computing P(5<L<6) = ? P(5 (226) = 1/5-504 , L-d (6-5.41 O 0.5 = P(-0.8C241.2) 55.4 =P(2(1.2)-P (Z (-0.8) 31-P(Z >1.2) - P(Z 70-8) By using standard normal probo (table - 1-0.1151-0.2119 = 0.673 P (54L(6) = 0.673) @ we have to find 95th percentile of D e P(D<d) = 0.95 DON(5.2.(0.6)) PL 44.) -0.95 -P[2 .2] -0.95

Ipage 3 P[22 ) d-5. 21 0.4 1-0.95 0.9 (zonco.n) (ZNN (0,12 P(27 d - 5.2 = 0.05 0.6 By using inverse probability table of normal distsibution d - 5.2 = 1064 d= 0.656+ 5.4 Id=6.056 95th percentile of Dis 6.056 Now, finding P(l+D >11). we know that L+D ON (10.6,0.6) P(l+D > 11) - P Pilto)-10.6, 11-10.6 L 10.41 Tool - = P(2 70. 62) By using standard normal probability table, = 0.2676 (PILID > 11) = 0.2676 I we have to find (135 and D >5) P(175 and D >5)=P(435) .P (D75). since land D are independent, first we find P(L> 5)=P(L-4 y a o 0.5

Ipages = P(27-0.8) = P( Zco.8). - I-P(Z > 0.8) = 1-0.2119 P(LS) = 0.7881] similarly for P(D25) -PL D-Y 0.4 = P(27-0.5) - 1-P (2) 0.5) -1- 0.3085 P (D> 55 0.6915 P(L>5 and D75) - P(L75)xP. (D >5) 30.78 81 x 0.6915 P (L75 and D5 -0.5449 6 P(071) = ? DON (5.2, (0.6) 2) & LINN (5.4, (0.582) D-L ON (5.2-5. 4, (0.6) ²+ (0.5)) D-LON6o. 2,0.41) ..PDXL) - P (D-L> 0) =PICO-L -(-0. 2 0 (-0.2) I 10.4T Joul I - P(Z > 0.03) (By using standard nooral P(DYL=0.4880 Aprobo table