To calculate the z-score for the 10th patient (with a wait time of 36 minutes) in the sample of offices without a wait-tracking system, we use the formula:

$z = \frac{x - \mu}{\sigma}$

where:

• $x$ = 36 (the data value for which we are calculating the z-score),
• $\mu$ = sample mean,
• $\sigma$ = sample standard deviation.

Step 1: Calculate the sample mean $\mu$

The given data set for offices without a wait-tracking system is:

$20, 56, 12, 35, 26, 48, 16, 10, 27, 36$

The mean is calculated as:

$\mu = \frac{20 + 56 + 12 + 35 + 26 + 48 + 16 + 10 + 27 + 36}{10}$

$\mu = \frac{286}{10} = 28.60$

Step 2: Calculate the sample standard deviation $\sigma$

The formula for sample standard deviation is:

$\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n-1}}$

First, calculate the squared differences from the mean:

$(20 - 28.60)^2 = (-8.60)^2 = 73.96$ $(56 - 28.60)^2 = (27.40)^2 = 750.76$ $(12 - 28.60)^2 = (-16.60)^2 = 275.56$ $(35 - 28.60)^2 = (6.40)^2 = 40.96$ $(26 - 28.60)^2 = (-2.60)^2 = 6.76$ $(48 - 28.60)^2 = (19.40)^2 = 376.36$ $(16 - 28.60)^2 = (-12.60)^2 = 158.76$ $(10 - 28.60)^2 = (-18.60)^2 = 345.96$ $(27 - 28.60)^2 = (-1.60)^2 = 2.56$ $(36 - 28.60)^2 = (7.40)^2 = 54.76$

Summing these squared differences:

$73.96 + 750.76 + 275.56 + 40.96 + 6.76 + 376.36 + 158.76 + 345.96 + 2.56 + 54.76 = 2086.40$

Dividing by $n - 1 = 10 - 1 = 9$:

$\sigma = \sqrt{\frac{2086.40}{9}} = \sqrt{231.82} \approx 15.23$

Step 3: Calculate the z-score

$z = \frac{36 - 28.60}{15.23}$

$z = \frac{7.40}{15.23} \approx 0.49$

Final Answer:

$\text{z-score} \approx \mathbf{0.49}$

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Do you want a more detailed breakdown or explanation? Here are some related questions:

1. How is a z-score interpreted in statistics?
2. What does a higher z-score indicate in this context?
3. How can we determine if the wait-tracking system significantly reduces waiting times?
4. How do we compare two different groups statistically (e.g., t-test)?
5. What are other measures of dispersion besides standard deviation?

Tip: A z-score of approximately 0.49 means the wait time of 36 minutes is 0.49 standard deviations above the mean, which is relatively close to the average.