To determine the percentage of data that lies between 5.3 cm and 5.7 cm in a normally distributed sample, we can use the properties of the normal distribution. Specifically, we can standardize the values into z-scores and refer to the standard normal distribution table (or use a calculator).

Step 1: Convert the values to z-scores

The formula to convert a value $x$ to a z-score is: $z = \frac{x - \mu}{\sigma}$ Where:

• $\mu$ is the mean (5.5 cm),
• $\sigma$ is the standard deviation (0.1 cm),
• $x$ is the value of interest.

For $x = 5.3$ cm: $z_{5.3} = \frac{5.3 - 5.5}{0.1} = \frac{-0.2}{0.1} = -2$

For $x = 5.7$ cm: $z_{5.7} = \frac{5.7 - 5.5}{0.1} = \frac{0.2}{0.1} = 2$

Step 2: Look up the z-scores in the standard normal distribution table

From the standard normal distribution table:

• The area to the left of $z = 2$ is approximately 0.9772 (97.72%).
• The area to the left of $z = -2$ is approximately 0.0228 (2.28%).

Step 3: Find the percentage of data between 5.3 cm and 5.7 cm

The percentage of data between the two values is the difference between these areas: $\text{Percentage} = 0.9772 - 0.0228 = 0.9544$ Thus, about 95.44% of the data lies between 5.3 cm and 5.7 cm.

Conclusion:

Approximately 95.44% of the screws will have a length between 5.3 cm and 5.7 cm.

Would you like more details or have any questions about this?

Here are 5 related questions for deeper understanding:

1. What is the probability of finding a screw shorter than 5.3 cm?
2. How would you calculate the percentage of screws longer than 5.7 cm?
3. What if the standard deviation changed to 0.05 cm—how would that affect the percentage?
4. Can you explain how the z-score relates to the normal distribution?
5. How would the result change if the data was not normally distributed?

Tip: Z-scores help standardize different data points, making it easier to compare data from different distributions.