Problem 73 - Applied Statistics and Probability for Engineers

Question

An article in Knee Surgery Sports Traumatol Arthrosc ["Effect of Provider Volume on Resource Utilization for Surgical Procedures" (2005, Vol. 13, pp. 273-279)] showed a mean time of 129 minutes and a standard deviation of 14 minutes for anterior cruciate ligament (ACL) reconstruction surgery at highvolume hospitals (with more than 300 such surgeries per year). (a) What is the probability that your ACL surgery at a highvolume hospital requires a time more than two standard deviations above the mean? (b) What is the probability that your ACL surgery at a highvolume hospital is completed in less than 100 minutes? (c) The probability of a completed ACL surgery at a high-volume hospital is equal to \(95 \%\) at what time? (d) If your surgery requires 199 minutes, what do you conclude about the volume of such surgeries at your hospital? Explain.

Step-by-Step Solution

Answer

(a) P > 157 min ≈ 0.0228; (b) P < 100 min ≈ 0.0192; (c) 95% completed by ≈152 min; (d) 199 min is highly abnormal if high-volume.

Step 1: Understanding the Normal Distribution

In this exercise, we are given a mean time of 129 minutes and a standard deviation of 14 minutes. We will assume the surgery times are normally distributed. This allows us to use the properties of the normal distribution to determine probabilities and percentiles for various surgery times.

Step 2: Calculating for Part (a)

To find the probability that surgery takes more than two standard deviations above the mean, we first calculate this value: \[\text{Time} = 129 + 2 \times 14 = 129 + 28 = 157 \, \text{minutes}\]Next, we find the probability that a surgery takes more than 157 minutes. Using the standard normal distribution, we calculate the z-score:\[z = \frac{157 - 129}{14} = 2\]The probability of a z-score being greater than 2 in a standard normal distribution is approximately 0.0228. Thus, the probability of surgery time exceeding 157 minutes is 0.0228.

Step 3: Calculating for Part (b)

For a surgery completed in less than 100 minutes, we calculate the z-score:\[z = \frac{100 - 129}{14} = \frac{-29}{14} \approx -2.071\]Using the standard normal distribution table, the probability of a z-score less than -2.071 is approximately 0.0192. Therefore, the probability of a surgery taking less than 100 minutes is 0.0192.

Step 4: Solving Part (c) for 95% Completion

To find the time by which 95% of surgeries are completed, we identify the z-score corresponding to the 95th percentile, which is approximately 1.645. We use the formula:\[\text{Time} = 129 + 1.645 \times 14 \]Calculating this, we get:\[\text{Time} = 129 + 23.03 \approx 152.03 \, \text{minutes}\]Thus, 95% of surgeries are completed by approximately 152 minutes.

Step 5: Analyzing Part (d) for 199 Minutes

A surgery time of 199 minutes corresponds to a z-score calculated as follows:\[z = \frac{199 - 129}{14} = 5\]This z-score of 5 is extremely high, lying far in the tail of the normal distribution. It is highly unlikely (almost unprecedented) for such a time to occur in high-volume hospitals based on the provided data. This suggests that the hospital may not be a high-volume center, or there might have been unusual circumstances affecting the surgery duration.

Key Concepts

Probability Calculation

Probability calculation in the context of a normal distribution involves determining the likelihood of a certain range of values occurring.
In scenarios like surgery times, where we assume a normal distribution, calculating probabilities helps us understand the likelihood of surgeries taking specific amounts of time.
For example:

• When calculating the probability of surgery times more than two standard deviations above the mean, we first establish the specific time (mean + 2 standard deviations) and then find the corresponding probability using a z-score, which represents the number of standard deviations away from the mean.
• Similarly, for probabilities of surgeries under a specific time, we find the z-score for that time and check its likelihood using the standard normal distribution.

Determining these probabilities often requires looking up values from a standard normal distribution table or using statistical software.

Standard Deviation

Standard deviation is a measure that indicates the typical distance between each data point and the mean in a dataset.
It reflects the spread or dispersion of the data points. In a normal distribution:

• A smaller standard deviation means data points are close to the mean, indicating low variability.
• A larger standard deviation means data points are spread out over a wider range, showing high variability.

In our exercise, the standard deviation is 14 minutes, which helps us determine how surgery times vary around the average of 129 minutes.
It's key to calculating probabilities associated with different surgery times as it standardizes the deviation of any particular value (like 100 minutes or 157 minutes) from the mean.

Z-score

A z-score is a statistical measure that describes a value's position relative to the mean of a group of values, expressed in terms of standard deviations.
It's a crucial concept for analyzing normal distributions because it helps standardize different data points on the same scale.
In relation to surgery time:

• We calculate the z-score by subtracting the mean from the value and dividing the result by the standard deviation. Thus, for a surgery time of 157 minutes, the z-score is \( rac{157 - 129}{14} = 2\).
• Z-scores help us quickly determine how unusual or common a particular data point is within a normal distribution, facilitating the probability calculation.

A higher absolute z-score indicates a lower probability of occurrence within the given distribution.

Percentiles

Percentiles are values below which a given percentage of data falls, providing a way to understand the distribution of values relative to each other.
They are extremely helpful in contexts like setting thresholds or performance metrics.
In the example of ACL surgery:

• The 95th percentile is the value below which 95% of surgery times fall. Calculating this involves finding a specific time that corresponds to a z-score of 1.645, resulting in a time where 95% of surgeries are completed by.
• Percentiles offer a more intuitive way for interpreting data compared to probabilities, as they directly relate outcomes to entire distributions.

Analyzing percentiles is particularly useful in quality control and benchmarking fields.