1.1 A bottling company uses a filling machine to fill plastic bottles with a popular cola. The bottles are supposed to contain 300 ml. In fact, the contents vary according to a normal distribution with mean of 298 ml and standard deviation of 3 ml. 1.1.1. What is the probability that an individual bottle contains more than 295 ml? [5] 1.1.2 What is the probability that the mean contents of the bottles in a six-pack is between 295 ml and 299 ml? [6]

To solve these problems, we need to understand the concepts of normal distribution and how to apply them to find probabilities. Let's break down each question.

### 1.1.1. Probability that an individual bottle contains more than 295 ml

Given:
- Mean (\(\mu\)) = 298 ml
- Standard deviation (\(\sigma\)) = 3 ml
- We need to find the probability that the bottle contains more than 295 ml.

The probability can be found using the standard normal distribution (Z-distribution). We first convert the problem into a Z-score.

The Z-score formula is:
\[
Z = \frac{X - \mu}{\sigma}
\]
Where:
- \(X\) is the value we are interested in (295 ml)
- \(\mu\) is the mean (298 ml)
- \(\sigma\) is the standard deviation (3 ml)

Substituting the values:
\[
Z = \frac{295 - 298}{3} = \frac{-3}{3} = -1
\]

Now, we need to find the probability that \(Z > -1\).

Using Z-tables or a normal distribution calculator:
\[
P(Z > -1) = 1 - P(Z \leq -1) = 1 - 0.1587 = 0.8413
\]

So, the probability that an individual bottle contains more than 295 ml is **0.8413**.

### 1.1.2. Probability that the mean contents of a six-pack is between 295 ml and 299 ml

Given:
- Mean (\(\mu\)) = 298 ml
- Standard deviation (\(\sigma\)) = 3 ml
- Sample size (\(n\)) = 6

The standard deviation of the sample mean (\(\sigma_{\text{mean}}\)) is given by:
\[
\sigma_{\text{mean}} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{6}} \approx 1.2247 \text{ ml}
\]

Now, we want to find the probability that the sample mean is between 295 ml and 299 ml.

First, we calculate the Z-scores for 295 ml and 299 ml.

For 295 ml:
\[
Z_{295} = \frac{295 - 298}{1.2247} \approx \frac{-3}{1.2247} \approx -2.45
\]

For 299 ml:
\[
Z_{299} = \frac{299 - 298}{1.2247} \approx \frac{1}{1.2247} \approx 0.82
\]

Now, find the probabilities corresponding to these Z-scores:

\[
P(Z_{295}) \approx P(Z \leq -2.45) = 0.0071
\]
\[
P(Z_{299}) \approx P(Z \leq 0.82) = 0.7939
\]

The probability that the mean content is between 295 ml and 299 ml is:
\[
P(295 < \text{Mean} < 299) = P(Z \leq 0.82) - P(Z \leq -2.45) = 0.7939 - 0.0071 = 0.7868
\]

So, the probability that the mean contents of a six-pack is between 295 ml and 299 ml is **0.7868**.

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Do you want details on these steps or have any questions?

**5 Relative Questions:**
1. What is the Z-score and how is it used in probability calculations?
2. How does sample size affect the standard deviation of the mean?
3. How can the normal distribution be applied to quality control in manufacturing?
4. What would happen to the probabilities if the standard deviation increased?
5. How would the probability change if the mean content was exactly 300 ml?

**Tip:** Understanding how to use Z-scores is crucial for solving problems related to normal distribution, especially when determining probabilities for specific ranges or values.

Explore probability calculations using normal distribution for a bottling company's filling machine problem. Learn how to find probabilities for individual bottle contents and mean contents of a six-pack based on given mean and standard deviation. Suitable for university-level statistics and probability students.

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