Question

Coke CansAssume that cans of Coke are filled so that the actual amounts have a mean of 12.00 oz and a standard deviation of 0.11 oz.

a.Find the probability that a single can of Coke has at least 12.19 oz.

b.The 36 cans of Coke in Data Set 26 “Cola Weights and Volumes”.inAppendix B have amean of 12.19 oz. Find the probability that 36random cans of Coke have a mean of at least12.19 oz.

c. Given the result from part (b), is it reasonable to believe that the cans are actually filled with a mean equal to 12.00 oz? If the mean is not equal to 12.00 oz, are consumers being cheated?

Short answer

a.The probability that a single can has at least 12.19 oz of coke is 0.0418.

b.The probability that the sample mean amount present in a sample of 36 cans is at least 12.19 oz is 0.000.

c. It is reasonable to believe that the cans are filled with a mean amount of 12.00 oz coke.

The consumers are not being cheated because there are cases when the mean amount in the cans is even greater than the mean of 12.00 oz.

Step-by-step solution

Given information

The population of the amounts of coke that can be filled in coke cans is normally distributed with a mean $\left(\mu \right)$ equal to 12.00 oz and a standard deviation $\left(\sigma \right)$ equal to 0.11 oz.

Required probabilities

a.

Let X denote the amount of coke filled in cans.

The probability that a single can has at least 12.19 oz of coke is computed using the standard normal table, as shown below.

$$\begin{gathered} P\left(x⩾12.19\right)=1-P\left(x<12.19\right) \\ =1-P\left(\frac{x-\mu }{\sigma }<\frac{12.19-\mu }{\sigma }\right) \\ =1-P\left(z<\frac{12.19-12}{0.11}\right) \\ =1-P\left(z<1.73\right) \end{gathered}$$

$$\begin{gathered} =1-0.9582 \\ =0.0418 \end{gathered}$$

Therefore, the probability that a single can has at least 12.19 oz of coke is 0.0418.

b.

Let $\overline{x}$denote the sample mean amount of coke filled in cans.

The sample mean weight follows a normal distribution with a mean equal to ${\mu }_{\overline{x}}=\mu$ and a standard deviation equal to ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

The sample size is equal to n=36.

The probability that the sample mean amount present in a sample of 36 cans is at least 12.19 ozis computed using the standard normal table, as shown below.

$$\begin{gathered} P\left(\overline{x}⩾12.19\right)=1-P\left(\overline{x}<12.19\right) \\ =1-P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{12.19-\mu }{\frac{\sigma }{\sqrt{n}}}\right) \\ =1-P\left(z<\frac{12.19-12.00}{\frac{0.11}{\sqrt{36}}}\right) \\ =1-P\left(z<10.36\right) \end{gathered}$$

Therefore, the probability that the sample mean amount present in a sample of 36 cans is at least 12.19 ozis equal to 0.000.

Analysis of the correct capacity of the cans

c.

Here, the probability that the cans have a mean amount greater than 2.19 oz exists.

Thus, it is reasonable to believe that the cans are filled with a mean amount of 12.00 oz coke.

Also, asthere are cases when the mean amount in the cans is even greater than the mean of 12.00 oz, the consumers are not being cheated.