Question

Soda six-packs Most soda cans list the volume of soda as 12 fluid ounces. As with all process, some variation occurs when filling soda cans. Suppose that a company knows this and tries to over-fill cans a bit, so that the actual volume of soda in a can follows a normal distribution with mean 12.1 fluid ounces and standard deviation .15 fluid ounces. a) What proportion of soda cans filled by this process will contain less than 12 fluid ounces? (In other words, what is the probability that a randomly selected soda can weighs less than 12 fluid ounces?) b) Determine the probability that a six-pack of soda cans would contain a sample mean volume of less than 12 fluid ounces. (Assume that a six-pack consists of six randomly selected cans.) c) Determine the probability that a random sample of 40 cans would contain a sample mean volume of less than 12 fluid ounces. d) Comment on how the probabilities change from a) to b) to c). Explain why this makes intuitive sense. e) Would the calculations in b) and/or c) be valid even if the distribution of can volumes were skewed rather than normal? Explain. f) Suppose that the company wants to change the mean volume so that only 2.5% of cans contain less than 12 fluid ounces, while leaving the standard deviation at .15 fluid ounces. What mean volume should they use? g) Now suppose that the company wants to change the standard deviation of the volumes so that only 9% of cans contain less than 12 fluid ounces, while leaving the mean at 12.1 fluid ounces. What standard deviation should they use?

Answer 1

bit, so Ciuen, The actual volune of Cans a the actual volume of Soda in a a nommal die taibution can that with follows fluid ounces 12.1 mean stan dard deviation ai5 tuid and Ounces x NUE12.1 , 2- C0.15)) and 2varian = Mean= 12.1 (0.15)2 and S=stand ard deuiation a Here we want to find the plx<12) = ? P(x<12)=P1. P X- 12-12.1 O.15 PC212-12.1 O.15 = P(2<-0.6667) (Faom standard nosmal paobability table) = O-2525 Now, ne know that i xNc2) then X N(u,) whese, Sample mean. =Population mean Variance =standard deuiation n b Here, n=6, X N(12-1, 0.15 we wantto find and Heat PCX 12)=? P( 12) P く12-12.1 015A16 - P(z-16323g) a-15NG = -O512

c Hert n=40 X N XN(12.1,(o.152) Here wle want to tind the Same porobabilty as opart (b) only ditfesence is gien distaibulion 1s bit change toom distaibuton in past (b) P (Y 12) P(2 12-12.1 = 9l2-42164) OO000124 As we not ice that fhe distaibution is dihferert in all in Pant a) x N2.105) in part b) 3 parts N2-1,6) in past N(12.1,15 Just see the differenee in 40 Variance and vanonce her decacases standard deuictron also decreases f Poobabilities also decoreases -for part a) ast b past c) 12

> # part a) > pnorm(12,12.1,0.15) [1] 0.2524925 > # obtain the same proportion as above by using standard normal dostribution > z1 <- (12-12.1)/0.15 > pnorm(z1,0,1) [1] 0.2524925 > # part b) > pnorm(12,12.1,0.15/sqrt(6)) [1] 0.05123522 # obtain the same proportion as above by using standard normal dostribution > z2 (12-12.1)/(0.15/sqrt(6)) pnorm(z2,0,1) [1] 0.05123522 # part c) pnorm(12,12.1,0.15/sqrt (40)) [1] 1.24133e-05 > # obtain the same proportion as above by us ing standard normal dostribution > z3 <- (12-12.1)/(0.15/sqrt(40)) > pnorm(z3,0,1) [1] 1.24133e-05