A beverage company advertises that their cans of a certain soft drink contain 12

Question

A beverage company advertises that their cans of a certain soft drink contain 12 ounces of soda. A study by the company estimates that the filling machine has a standard deviation of 0.39 ounces. The company is being accused of underfilling its cars of soft drink. For the following questions, support that the company is telling the truth. (a) Based on the given information, can you calculate the probability that a single randornly selected can contains less than 11.88 ounces? If so, do it. If not, explain why you cannot. No because they didn't tell us it was (b) A quality control inspector takes in SRS of 40 cans of the company's soda, measures the contents, and calculates the sample mean. What are the mean and the standard deviation of the sampling distribution for the samples of size n = 42? mux = 12 sigma x = sigma/ n = 39/ 40 = 0617 (c) The inspector in part (b) obtains a sample mean of 11.88 ounces. Calculate the probability that a random sample of 40 cans produces a sample mean fill of 11.88 ounces or less. (d) Based on your answer to part (c). what would you conclude about whether the company is under filling its cans of soda? Justify your answer.

Explanation / Answer

let X be the random variable denoting the contain of soda in a soft drink can

the assumption is that X follows a normal distribution.

by question mean of X is E[X]=12 ounces and standard deviation of X is=sigma=0.39 ounces

hence X~N(12,0.392)

a) probability that a randomly selected can contains soda less than 11.88 ounces is

P[X<11.88]=P[(X-12)/0.39<(11.88-12)/0.39]=P[Z<-0.30769]   where Z~N(0,1)

                =0.379159 [answer] [using MINITAB]

b) let the sample mean of n=40 cans be Xbar

so mean of Xbar is E[Xbar]=E[X1+X2+..+X40]/40=40*12/40=12 [answer]

and standard deviation of Xbar is=sigma/sqrt(n)=0.39/sqrt(40)=0.0617 [answer]

hence Xbar~N(12,0.6172)

c) the probability that the sample mean is 11.88 ounces or less is

P[Xbar<11.88]=P(Xbar-12)/0.617<(11.88-12)/0.617]=P[Z<-0.19449]=0.422896   [answer]

d) hence from part c) we have that a can is underfilled in 42.2896% of times

so the can is not underfilled in (100-42.2896)%=57.7104% of times

hence the can is not underfilled has more probability than the can is underfilled

hence the conlusion is that the statement: "the company is underfilling its cans of soda" is not true [answer]

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