The darkness of the print is measured quantitatively using an index. If the index is greater than or equal to 2.0 then the darkness is acceptable. Anything less than 2.0 means the print is too light and not acceptable. Assume that the machines print at an average darkness of 2.2 with a standard deviation of 0.20.
(a)
State four properties of the Normal Distribution Function. (4)
(b)
What percentage of printing jobs will be acceptable? (4)
(c)
If the mean cannot be adjusted, but the standard deviation can, what must be the new standard deviation such that a minimum of 95% of jobs will be acceptable? (7)

Let's break down the problem step by step.

### (a) **State four properties of the Normal Distribution Function**.

1. **Symmetry**: The normal distribution is symmetric about its mean, meaning the left side is a mirror image of the right side.
2. **Mean, Median, and Mode are Equal**: In a normal distribution, the mean, median, and mode all lie at the center of the distribution.
3. **Bell-shaped Curve**: The graph of a normal distribution is bell-shaped, with most data points clustered around the mean, and fewer data points further from the mean.
4. **Empirical Rule (68-95-99.7 Rule)**: Approximately 68% of the data falls within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations.

---

### (b) **What percentage of printing jobs will be acceptable?**

To find the percentage of acceptable printing jobs, we need to calculate the proportion of jobs where the darkness index is greater than or equal to 2.0. The machine prints at an average darkness of 2.2 with a standard deviation of 0.20. We can use the **Z-score** formula to standardize the 2.0 darkness level:

\[
Z = \frac{X - \mu}{\sigma}
\]

Where:
- \(X\) is the value of interest (2.0)
- \(\mu\) is the mean (2.2)
- \(\sigma\) is the standard deviation (0.20)

Substituting the values:

\[
Z = \frac{2.0 - 2.2}{0.20} = \frac{-0.2}{0.20} = -1
\]

Now, using the Z-score table (or normal distribution function), the area to the left of \(Z = -1\) is approximately 0.1587, meaning that 15.87% of the jobs are below a darkness index of 2.0.

Thus, the percentage of acceptable jobs (those with darkness \( \geq 2.0 \)) is:

\[
100\% - 15.87\% = 84.13\%
\]

So, **84.13% of the printing jobs will be acceptable**.

---

### (c) **If the mean cannot be adjusted, but the standard deviation can, what must be the new standard deviation such that a minimum of 95% of jobs will be acceptable?**

We need to find the standard deviation such that 95% of the jobs have a darkness index of 2.0 or more. This means that the cutoff point for the lower 5% will correspond to a Z-score of -1.645 (from the Z-table for 5%).

Using the Z-score formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

We need \( Z = -1.645 \), \( X = 2.0 \), and \( \mu = 2.2 \). Substituting these into the formula:

\[
-1.645 = \frac{2.0 - 2.2}{\sigma}
\]

This simplifies to:

\[
-1.645 = \frac{-0.2}{\sigma}
\]

Solving for \( \sigma \):

\[
\sigma = \frac{-0.2}{-1.645} \approx 0.1216
\]

Therefore, the new standard deviation must be approximately **0.1216** to ensure that 95% of the jobs are acceptable.

---

### Summary of Results:
- (a) Properties of the Normal Distribution Function:
  1. Symmetry
  2. Mean = Median = Mode
  3. Bell-shaped curve
  4. Empirical Rule (68-95-99.7 Rule)
- (b) Percentage of acceptable printing jobs: **84.13%**
- (c) The new standard deviation must be approximately **0.1216**.

---

Would you like further clarification or additional details on any part of the solution? Here are 5 related questions:

1. How does the Empirical Rule apply to different standard deviations in a normal distribution?
2. What are Z-scores and how are they used in normal distribution problems?
3. Can you explain why we used a Z-score of -1.645 in part (c)?
4. How would the answer change if the mean could also be adjusted?
5. What other real-world processes might benefit from this kind of normal distribution analysis?

**Tip**: To quickly find the percentage above or below a certain value in a normal distribution, always standardize the value using the Z-score formula and use Z-tables or a calculator.

The darkness of the print is measured quantitatively using an index. If the index is greater than or equal to 2.0 then the darkness is acceptable. Anything less than 2.0 means the print is too light and not acceptable. Assume that the machines print at an average darkness of 2.2 with a standard deviation of 0.20.
(a) State four properties of the Normal Distribution Function. (4)
(b) What percentage of printing jobs will be acceptable? (4)
(c) If the mean cannot be adjusted, but the standard deviation can, what must be the new standard deviation such that a minimum of 95% of jobs will be acceptable? (7)

Discover how to calculate acceptable printing jobs using normal distribution with a mean of 2.2 and standard deviation of 0.20. Learn how to adjust standard deviation to meet a 95% acceptance rate and explore key properties of the Normal Distribution Function. Suitable for Grades 10-12.

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