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 acl reconstruction surgery at high-volume hospitals (with more than 300 such surgeries per year).

(a) What is the probability that your acl surgery at a high-volume hospital requires a time more than two standard deviations above the mean?
(b) What is the probability that your acl surgery at a high-volume 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?

Answer 1

To find the probability that the ACL surgery at a high-volume hospital requires a time more than two standard deviations above the mean, calculate the z-score and use a standard normal distribution table or calculator. The probability is approximately 0.0228. To find the probability that the ACL surgery at a high-volume hospital is completed in less than 100 minutes, calculate the z-score and use a standard normal distribution table or calculator. The probability is approximately 0.019. The probability of a completed ACL surgery at a high-volume hospital taking 153.13 minutes or less is equal to 95%.

Step-by-step explanation:

To solve this problem, we need to use the properties of the normal distribution. First, let's calculate the z-score for two standard deviations above the mean:

z = (x - μ) / σ

where x is the observation, μ is the mean, and σ is the standard deviation. In this case, x = μ + 2σ = 129 + 2(14) = 157 minutes:

z = (157 - 129) / 14 = 2

Next, we use a standard normal distribution table or calculator to find the probability associated with a z-score of 2. The probability of a value being greater than two standard deviations above the mean is approximately 0.0228.

For part (b), we need to calculate the z-score for a time of 100 minutes:

z = (100 - 129) / 14 = -2.07

The probability of a value being less than 100 minutes can be found using the standard normal distribution table or calculator. The probability is approximately 0.019.

For part (c), we can use the z-score formula again to find the value of x that corresponds to a probability of 95%, which is located in the right tail of the distribution:

z = (x - 129) / 14

Using a standard normal distribution table or calculator, we find that the z-score for a probability of 95% is approximately 1.645. Substituting this value back into the equation:

1.645 = (x - 129) / 14

x - 129 = 1.645 * 14

x = 129 + 1.645 * 14

x ≈ 153.13

Therefore, the probability of a completed ACL surgery at a high-volume hospital taking 153.13 minutes or less is equal to 95%.