A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed into a can is known to have a normal distribution with a mean of 5 milliliters (ml) and a standard deviation of 4 ml. Answer the following questions based on this information. What proportion of the paint cans contain between 4.80 ml and 5.40 ml of the dye?

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## Distribution of Dye in Paint Production

### Problem Statement
A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed into a can is known to have a normal distribution with a mean of 5 milliliters (ml) and a standard deviation of 4 ml.

**Question**: What proportion of the paint cans contain between 4.80 ml and 5.40 ml of the dye?

### Normal Curve Areas Table
The table provided shows the area under a standard normal distribution curve to the left of a given z-score. This area represents the cumulative probability for that z-score.

**Diagram**: 
- A bell curve is depicted with the mean situated at \( z = 0 \).
- The x-axis represents the z-scores.

**Table Interpretation**:
- The rows represent the z-value's integer and first decimal place.
- The columns (.00 to .09) represent the second decimal place of the z-score.
- The table values below indicate the proportion of the distribution less than the given z-score.

### Source
Data is sourced from a statistical table by A. Hald, *Statistical Tables and Formulas* (1952).

### Multiple Choice Options
- 0.4758
- 0.5328
- 0.1915
- 0.3413
- 0.1498

Use the normal distribution table to find the required cumulative probabilities to determine the correct answer.

### Explanation of Steps
1. Calculate the z-scores for 4.80 ml and 5.40 ml.
2. Use the table to find cumulative probabilities for these z-scores.
3. Subtract to find the proportion of cans within the specified range.

This interactive exercise will enhance understanding of normal distribution and cumulative probabilities in practical applications like paint production.
Transcribed Image Text:## Distribution of Dye in Paint Production ### Problem Statement A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed into a can is known to have a normal distribution with a mean of 5 milliliters (ml) and a standard deviation of 4 ml. **Question**: What proportion of the paint cans contain between 4.80 ml and 5.40 ml of the dye? ### Normal Curve Areas Table The table provided shows the area under a standard normal distribution curve to the left of a given z-score. This area represents the cumulative probability for that z-score. **Diagram**: - A bell curve is depicted with the mean situated at \( z = 0 \). - The x-axis represents the z-scores. **Table Interpretation**: - The rows represent the z-value's integer and first decimal place. - The columns (.00 to .09) represent the second decimal place of the z-score. - The table values below indicate the proportion of the distribution less than the given z-score. ### Source Data is sourced from a statistical table by A. Hald, *Statistical Tables and Formulas* (1952). ### Multiple Choice Options - 0.4758 - 0.5328 - 0.1915 - 0.3413 - 0.1498 Use the normal distribution table to find the required cumulative probabilities to determine the correct answer. ### Explanation of Steps 1. Calculate the z-scores for 4.80 ml and 5.40 ml. 2. Use the table to find cumulative probabilities for these z-scores. 3. Subtract to find the proportion of cans within the specified range. This interactive exercise will enhance understanding of normal distribution and cumulative probabilities in practical applications like paint production.
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