To find the z-scores for the given wait times in both cases, we follow these steps:

The z-score formula is:

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

Where:

• $x$ is the value for which we are calculating the z-score (the observed wait time),
• $\mu$ is the mean of the data set,
• $\sigma$ is the standard deviation of the data set.

(a) For offices without a wait-tracking system:

We are given the wait time for the 10th patient: 32 minutes.

Step 1: Find the mean ($\mu$) of the wait times without a wait-tracking system.

The wait times are: $22, 65, 10, 34, 23, 47, 15, 12, 25, 32$

To calculate the mean: $\mu = \frac{22 + 65 + 10 + 34 + 23 + 47 + 15 + 12 + 25 + 32}{10}$

Step 2: Calculate the standard deviation ($\sigma$).

The formula for the standard deviation is: $\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n}}$ Where:

• $x_i$ are the individual wait times,
• $n$ is the number of observations.

Step 3: Use the z-score formula for $x = 32$.

(b) For offices with a wait-tracking system:

We are given the wait time for the 6th patient: 32 minutes.

Step 1: Find the mean ($\mu$) of the wait times with a wait-tracking system.

The wait times are: $12, 8, 31, 16, 10, 32, 9, 17, 9, 6$

Step 2: Calculate the standard deviation ($\sigma$).

Step 3: Use the z-score formula for $x = 32$.

Let me calculate both z-scores.Here are the results:

(a) For offices without a wait-tracking system:

• Mean ($\mu$): 28.5 minutes
• Standard deviation ($\sigma$): 16.99 minutes
• Z-score for the 10th patient (wait time = 32 minutes): $z = 0.21$

(b) For offices with a wait-tracking system:

• Mean ($\mu$): 15.0 minutes
• Standard deviation ($\sigma$): 9.35 minutes
• Z-score for the 6th patient (wait time = 32 minutes): $z = 1.82$

Would you like more details or have any other questions?

Here are 5 related questions:

1. What does a z-score tell us in terms of data distribution?
2. How can we interpret the z-score of 1.82 in the context of this problem?
3. How would adding more data points impact the mean and standard deviation?
4. What is the relationship between the mean and the median in a normally distributed dataset?
5. How can z-scores help identify outliers in a dataset?

Tip: A z-score closer to 0 indicates that the data point is near the mean, while higher absolute values indicate the point is further away from the mean.