91影视

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

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.

$$\begin{gathered} P\left(x猢�12.19\right)=1-P\left(x<12.19\right) \\ =1-P\left(\frac{x-渭}{蟽}<\frac{12.19-渭}{蟽}\right) \\ =1-P\left(z<\frac{12.19-12}{0.11}\right) \\ =1-P\left(z<1.73\right) \end{gathered}$$

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

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

b.

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

The sample mean weight follows a normal distribution with a mean equal to ${渭}_{\stackrel{炉}{x}}=渭$ and a standard deviation equal to ${蟽}_{\stackrel{炉}{x}}=\frac{蟽}{\sqrt{n}}$.

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.

$$\begin{gathered} P\left(\stackrel{炉}{x}猢�12.19\right)=1-P\left(\stackrel{炉}{x}<12.19\right) \\ =1-P\left(\frac{\stackrel{炉}{x}-渭}{\frac{蟽}{\sqrt{n}}}<\frac{12.19-渭}{\frac{蟽}{\sqrt{n}}}\right) \\ =1-P\left(z<\frac{12.19-12.00}{\frac{0.11}{\sqrt{36}}}\right) \\ =1-P\left(z<10.36\right) \end{gathered}$$

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.

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.