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Let's break down and solve the two parts of this problem:

### (a) Probability that the ketchup amount is between 1 and 1.2 ounces

We are given:
- The amount of ketchup dispensed follows a normal distribution with mean \(\mu = 1.05\) fluid ounces and standard deviation \(\sigma = 0.08\) fluid ounces.
- The goal is to calculate the probability that the amount of ketchup dispensed is between 1 and 1.2 fluid ounces.

We can standardize this range using the **z-score formula**:
\[
z = \frac{x - \mu}{\sigma}
\]
where \(x\) is the value we are standardizing.

For \(x = 1.0\) (lower bound):
\[
z_1 = \frac{1.0 - 1.05}{0.08} = \frac{-0.05}{0.08} = -0.625
\]

For \(x = 1.2\) (upper bound):
\[
z_2 = \frac{1.2 - 1.05}{0.08} = \frac{0.15}{0.08} = 1.875
\]

Now, we use the standard normal distribution table (or a calculator) to find the probabilities corresponding to these z-scores:
- For \(z_1 = -0.625\), the cumulative probability is approximately 0.2660.
- For \(z_2 = 1.875\), the cumulative probability is approximately 0.9696.

Thus, the probability that the amount of ketchup is between 1.0 and 1.2 ounces is:
\[
P(1.0 < X < 1.2) = P(z_2) - P(z_1) = 0.9696 - 0.2660 = 0.7036
\]

Converting this to a percentage:
\[
P = 0.7036 \times 100 = 70.36\%
\]

### (b) Adjusting the standard deviation to ensure 99% of ketchup is between 1 and 1.2 ounces

In this case, we are given:
- The new mean is \(\mu = 1.1\) ounces.
- We need to adjust the standard deviation so that 99% of the ketchup dispensed is between 1 and 1.2 ounces.

To ensure 99% of the ketchup is within this range, the middle 99% of the normal distribution needs to fall between 1 and 1.2 ounces, which leaves 0.5% in each tail.

From the z-table, the z-scores that correspond to the middle 99% (i.e., leaving 0.5% in each tail) are approximately \(\pm 2.576\).

Now, we calculate the required standard deviation using the z-score formula, solving for \(\sigma\):
\[
z = \frac{x - \mu}{\sigma}
\]
For the lower bound \(x = 1.0\) and \(z = -2.576\):
\[
-2.576 = \frac{1.0 - 1.1}{\sigma} = \frac{-0.1}{\sigma}
\]
Solving for \(\sigma\):
\[
\sigma = \frac{-0.1}{-2.576} = 0.0388
\]

Thus, the standard deviation needs to be reduced to approximately **0.0388 fluid ounces**.

### Final answers:
- (a) 70.36% of the time, the machine will dispense between 1.0 and 1.2 ounces of ketchup.
- (b) The standard deviation needs to be reduced to **0.0388 fluid ounces**.

Do you want more details on the calculations or have any questions?

Here are 5 related questions:
1. How do you interpret z-scores in a normal distribution?
2. What are the common z-scores corresponding to confidence intervals in statistics?
3. How can you find probabilities for a normal distribution without a table?
4. What changes occur to the spread of a normal distribution when the standard deviation decreases?
5. Why might a business want to control standard deviations in automated processes?

**Tip**: In many applications, reducing the standard deviation leads to more precise outcomes, but it may require adjustments to machinery or processes.

A fast-food restaurant has just installed a new automatic ketchup dispenser for use in preparing its burgers. The amount of ketchup dispensed by the machine follows an approximately normal distribution with a mean of 1.05 fluid ounces and a standard deviation of 0.08 fluid ounces. (a) If the restaurant's goal is to put between 1 and 1.2 ounces of ketchup on each burger, about what percent of the time will this happen? (b) Suppose that the manager adjusts the machine's settings so that the mean amount of ketchup dispensed is 1.1 ounces. How much does the machine’s standard deviation have to be reduced to ensure that at least 99% of the restaurant’s burgers have between 1 and 1.2 ounces of ketchup on them?

Explore the probability of dispensing between 1 and 1.2 ounces of ketchup from an automatic machine. Learn about adjusting standard deviation to achieve 99% precision with real-world examples.

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