A manufacturer knows that their items have a normally distributed length, with a mean of 9.8 inches, and standard deviation of 0.8 inches. If one item is chosen at random, what is the probability that it is less than 11.9 inches long?

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### Understanding Normal Distribution in Manufacturing

When assessing the quality and dimensions of manufactured items, it is important to understand the concept of normal distribution. We know that a manufacturer has determined that the lengths of their items are normally distributed. This means that the lengths follow a bell-shaped curve when plotted, with most items clustering around a central mean value and progressively fewer items occurring as the length deviates from this mean.

Here’s an important scenario:

#### Scenario
A manufacturer knows that their items have a normally distributed length, with a **mean** of 9.8 inches, and a **standard deviation** of 0.8 inches.

**Question:**
If one item is chosen at random, what is the probability that it is less than 11.9 inches long?

#### Concept Explanation
To determine this probability, we can utilize the properties of the normal distribution. The **mean (µ)** of the distribution is 9.8 inches, and the **standard deviation (σ)** is 0.8 inches. The probability question requires us to find the likelihood that a randomly chosen item will have a length less than 11.9 inches.

#### Steps to Solution
1. **Convert the value to a Z-score**: The Z-score represents the number of standard deviations a particular value is from the mean. The formula for the Z-score is:
   \[
   Z = \frac{(X - \mu)}{\sigma}
   \]
   Where:
   - \(X\) is the value (11.9 inches)
   - \(\mu\) is the mean (9.8 inches)
   - \(\sigma\) is the standard deviation (0.8 inches)

2. **Calculate the Z-score for 11.9 inches**:
   \[
   Z = \frac{(11.9 - 9.8)}{0.8} = \frac{2.1}{0.8} = 2.625
   \]

3. **Use Z-tables to find the probability**: Once we have the Z-score, we use standard normal distribution tables (Z-tables) or a calculator to find the probability that Z is less than 2.625.

4. **Interpret the result**: The Z-score of 2.625 corresponds to a cumulative probability of approximately 0.9956 (or 99.56%).

Therefore, the probability that a randomly chosen item is less than
Transcribed Image Text:### Understanding Normal Distribution in Manufacturing When assessing the quality and dimensions of manufactured items, it is important to understand the concept of normal distribution. We know that a manufacturer has determined that the lengths of their items are normally distributed. This means that the lengths follow a bell-shaped curve when plotted, with most items clustering around a central mean value and progressively fewer items occurring as the length deviates from this mean. Here’s an important scenario: #### Scenario A manufacturer knows that their items have a normally distributed length, with a **mean** of 9.8 inches, and a **standard deviation** of 0.8 inches. **Question:** If one item is chosen at random, what is the probability that it is less than 11.9 inches long? #### Concept Explanation To determine this probability, we can utilize the properties of the normal distribution. The **mean (µ)** of the distribution is 9.8 inches, and the **standard deviation (σ)** is 0.8 inches. The probability question requires us to find the likelihood that a randomly chosen item will have a length less than 11.9 inches. #### Steps to Solution 1. **Convert the value to a Z-score**: The Z-score represents the number of standard deviations a particular value is from the mean. The formula for the Z-score is: \[ Z = \frac{(X - \mu)}{\sigma} \] Where: - \(X\) is the value (11.9 inches) - \(\mu\) is the mean (9.8 inches) - \(\sigma\) is the standard deviation (0.8 inches) 2. **Calculate the Z-score for 11.9 inches**: \[ Z = \frac{(11.9 - 9.8)}{0.8} = \frac{2.1}{0.8} = 2.625 \] 3. **Use Z-tables to find the probability**: Once we have the Z-score, we use standard normal distribution tables (Z-tables) or a calculator to find the probability that Z is less than 2.625. 4. **Interpret the result**: The Z-score of 2.625 corresponds to a cumulative probability of approximately 0.9956 (or 99.56%). Therefore, the probability that a randomly chosen item is less than
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