A bottler of soft drinks packages cans in six-packs. Suppose that the fill per c

Question

A bottler of soft drinks packages cans in six-packs. Suppose that the fill per can has an approximate normal distribution with a mean of 12 fluid ounces and a standard deviation of 0.2 fluid ounces. a. What is the distribution of the total fill for a case of 24 cans? b. What is the probability that the fill for a case is less than 286 fluid ounces? c. If a six-pack of soda can be considered a random sample of size n=6 from the population, what is the probability that the average fill per can for a six-pack of soda is less than 11.8 fluid ounces?

Explanation / Answer

a) Given Mean = 12, SD = 0.2 and also Can follows N(Mean=12, SD = 0.2)

This is the sum of 24 normal random variables

i.e. X follows N(24*12, sqrt(24) *0.2) = N(288, 0.98)

b) The distribution of the total fill for a case of 24 cans is

P(Case < 286) = P(Z < (286-288) / 0.98) = P(Z < -2.0408) = 0.02063 = 2.063%

c)

Given n= 6

Sample mean (average) Xbar ~ N(mean =12, SD=0.2/sqrt(n))

i.e. Xbar ~ N(mean =12, SD=0.2/sqrt(6))

i.e. Xbar ~ N(mean =12, SD=0.08165)

P(the average fill per can for a si-pack of soda is less than 11.8 fluid ounces) is P( Xbar < 11.8)

= P(Z < (11.8 - 12) / 0.08165)
= P(Z < -2.4494)
P(Z < -2.45) = 0.00715

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