# Solved: Beer bottles are filled so that they contain an average of 330 ml of beer in each bottle. Suppose tha

#### ByDr. Raju Chaudhari

Oct 12, 2020

Beer bottles are filled so that they contain an average of 330 ml of beer in each bottle. Suppose that the amount of beer in a bottle is normally distributed with a standard deviation of 4 ml. Use Table 1.

a. What is the probability that a randomly selected bottle will have less than 325 ml of beer? (Round your intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.)
b. What is the probability that a randomly selected 6-pack of beer will have a mean amount less than 325 ml? (Round your intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.)
c. What is the probability that a randomly selected 12-pack of beer will have a mean amount less than 325 ml? (Round your intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 5 decimal places.)

#### Solution

Let $X$ denote the amount of beer in a bottle. Given that $X\sim N(330, 4^2)$.

That is $\mu = 330$ and $\sigma = 4$.

a. The probability that a randomly selected bottle will have less than 325 ml of beer is

\begin{aligned} P(X< 325) &=P\bigg(\frac{X-\mu}{\sigma} < \frac{325-330}{4}\bigg)\\ &=P\bigg(Z< -1.25\bigg)\\ &= P(Z< -1.25)\\ &=0.1056 \end{aligned}

b. A sample size is $n = 6$.

Then $\overline{X}\sim N(\mu, \sigma^2/n)$. That is $\overline{X}\sim N(330, 1.633^2)$

The probability that a randomly selected 6-pack of beer will have a mean amount less than 325 ml is

\begin{aligned} P(\overline{X}< 325) &=P\bigg(\frac{\overline{X}-\mu}{\sigma/\sqrt{n}} < \frac{325-330}{4/\sqrt{6}}\bigg)\\ &=P\bigg(Z< -3.06\bigg)\\ &= P(Z< -3.06)\\ &=0.0011 \end{aligned}

c. A sample size is $n = 12$.

Then $\overline{X}\sim N(\mu, \sigma^2/n)$.

That is $\overline{X}\sim N(330, 1.1547^2)$

The probability that a randomly selected 12-pack of beer will have a mean amount less than 325 ml is

\begin{aligned} P(\overline{X}< 325) &=P\bigg(\frac{\overline{X}-\mu}{\sigma/\sqrt{n}} < \frac{325-330}{4/\sqrt{12}}\bigg)\\ &=P\bigg(Z< -4.33\bigg)\\ &= P(Z< -4.33)\\ &=0 \end{aligned}