Question

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

**Answer:**

Given,

Mean = 12.1

Standard deviation = 0.15

a)

P(X < 12) = P((x-u)/s < (12 - 12.1)/0.15)

= P(z < -0.67)

= 0.2514289 [since from z table]

= 0.2514

b)

P(X < 12) = P((x-u)/(s/sqrt(n)) < (12 - 12.1)/(0.15/sqrt(6)))

= P(z < -1.63)

= 0.0515507 [since from z table]

= 0.0516

c)

P(X < 12) = P((x-u)/(s/sqrt(n)) < (12 - 12.1)/(0.15/sqrt(40)))

= P(z < -4.22)

= 0 [since from z table]

= 0

d)

Here the probability decreases as there is increase in sample size due to that standard error decrease with increase in sample size.

e)

Here we can say , No due to that sample size is < 30.

f)

Here for low 2.5% , z = -1.96

consider,

z = (x-u)/s

substitute values

- 1.96 = (12 - u)/0.15

u = 12.294

g)

Here for low 9% , z = -1.34

consider,

z = (x-u)/s

substitute values

- 1.34 = (12 - 12.1)/s

s = 0.075

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