Factories located in Colorado and Missouri work around the clock producing apparatuses. Sergio is comparing monthly production of apparatuses at the two factories. Round answers to 2 decimal places.

Colorado: n = 20 months, mean = 
524
524 apparatuses/month, StDev = 
15.5
15.5 apparatuses/month

Missouri: n = 25 months, mean = 
592.5
592.5 apparatuses/month, StDev = 
17
17 apparatuses/month



a) What is the standardized z-score for the 552 apparatuses production output for the factory in Colorado?  
 



b) What is the standardized z-score for the 552 apparatuses production output for the factory in Missouri?

To calculate the standardized \( z \)-score, the formula is:

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

Where:
- \( X \) is the value we are standardizing (552 in both cases),
- \( \mu \) is the mean of the data,
- \( \sigma \) is the standard deviation of the data.

### Part a) Colorado Factory

Given:
- \( X = 552 \) (the output),
- \( \mu = 524 \) (mean for Colorado),
- \( \sigma = 15.5 \) (standard deviation for Colorado).

Plug the values into the formula:

\[
z = \frac{552 - 524}{15.5} = \frac{28}{15.5}
\]

Let's compute that:

\[
z \approx 1.81
\]

So, the \( z \)-score for 552 apparatuses in Colorado is approximately **1.81**.

### Part b) Missouri Factory

Given:
- \( X = 552 \) (the output),
- \( \mu = 592.5 \) (mean for Missouri),
- \( \sigma = 17 \) (standard deviation for Missouri).

Plug the values into the formula:

\[
z = \frac{552 - 592.5}{17} = \frac{-40.5}{17}
\]

Let's compute that:

\[
z \approx -2.38
\]

So, the \( z \)-score for 552 apparatuses in Missouri is approximately **-2.38**.

### Summary:
- The \( z \)-score for Colorado is **1.81**.
- The \( z \)-score for Missouri is **-2.38**.

Would you like more details on z-scores or have any other questions?

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**Related Questions:**
1. What does a positive z-score indicate about the Colorado factory’s production?
2. What does a negative z-score indicate about the Missouri factory’s production?
3. How does the variance in production between the two factories affect the z-scores?
4. If the output in Missouri were 610 apparatuses, what would the z-score be?
5. What is the probability of getting a z-score greater than 1.81 in a normal distribution?

**Tip:** A z-score shows how many standard deviations a value is from the mean.

Factories located in Colorado and Missouri work around the clock producing apparatuses. Sergio is comparing monthly production of apparatuses at the two factories. Round answers to 2 decimal places.

Colorado: n = 20 months, mean = 524 apparatuses/month, StDev = 15.5 apparatuses/month

Missouri: n = 25 months, mean = 592.5 apparatuses/month, StDev = 17 apparatuses/month

(a) What is the standardized z-score for the 552 apparatuses production output for the factory in Colorado?
(b) What is the standardized z-score for the 552 apparatuses production output for the factory in Missouri?

Learn how to calculate the standardized z-scores for factory production in Colorado and Missouri. With means of 524 and 592.5 apparatuses/month, and standard deviations of 15.5 and 17, respectively, this problem guides you through the process of standardizing production outputs. Suitable for Grades 10-12.

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