An Orthopedic Surgeon Sees About 60 New Patients Per Month

An Orthopedic Surgeon Sees About60 New Patients Per Month The Rest Of

An orthopedic surgeon sees about 60 new patients per month. He knows that, on average, about 30% of these patients will require surgical intervention. The questions relate to the probabilities of different numbers of patients requiring surgery out of the 60 new patients seen monthly.

Specifically, the tasks are:

  1. Calculate the probability that 50 or more patients out of 60 will require surgical intervention.
  2. Calculate the probability that 15 or fewer patients will require surgical intervention.
  3. Calculate the probability that between 40 and 50 patients (inclusive) will require surgical intervention.

Paper For Above instruction

The scenario presented involves analyzing probabilities related to the number of patients requiring surgical intervention among a fixed number of new patients seen by an orthopedic surgeon each month. Given the data, the problem is well-approximated by a binomial distribution due to the fixed number of trials, binary outcomes (requiring surgery or not), constant probability, and independence of each patient's outcome. In this context, the total number of patients requiring surgery, denoted by the random variable X, follows a binomial distribution with parameters n = 60 (number of trials) and p = 0.3 (probability of success, i.e., requiring surgery for each patient).

The binomial probability mass function (PMF) is given by:

P(X = k) = C(n, k) p^k (1 - p)^{n - k}

where C(n, k) is the binomial coefficient, representing the number of ways to choose k successes from n trials.

To compute the probabilities of the specified ranges, the cumulative distribution function (CDF) of the binomial distribution is employed, either directly via statistical software, calculators, or approximation methods such as the normal distribution approximation for large n.

Part A: Probability that 50 or more patients require surgery

The probability that at least 50 out of 60 patients will require surgery is:

P(X ≥ 50) = 1 - P(X ≤ 49)

Using the binomial distribution, calculating P(X ≤ 49) exactly may be done via statistical software or binomial tables. Alternatively, for large n, a normal approximation can be used.

Approximate the binomial distribution with a normal distribution with mean μ = np = 60 0.3 = 18 and standard deviation σ = sqrt(np(1-p)) = sqrt(60 0.3 * 0.7) ≈ 3.86.

Applying the continuity correction for P(X ≥ 50):

P(X ≥ 50) ≈ P(Y ≥ 49.5)

where Y ~ N(μ = 18, σ ≈ 3.86). Calculating the z-score:

Z = (49.5 - 18) / 3.86 ≈ 8.56

Consulting standard normal distribution tables, Z ≈ 8.56 yields a probability effectively zero (

Part B: Probability that 15 or fewer patients require surgery

Similarly, to find P(X ≤ 15), the normal approximation with continuity correction is employed:

P(X ≤ 15) ≈ P(Y ≤ 15.5)

Calculating z-score:

Z = (15.5 - 18) / 3.86 ≈ -0.66

From standard normal tables, P(Z ≤ -0.66) ≈ 0.2546.

Therefore, the probability that 15 or fewer patients will require surgical intervention is approximately 25.46%.

Part C: Probability that between 40 and 50 patients (inclusive) require surgery

This probability is:

P(40 ≤ X ≤ 50) = P(X ≤ 50) - P(X ≤ 39)

Using the normal approximation with continuity correction:

  • For P(X ≤ 50):

P(X ≤ 50) ≈ P(Y ≤ 50.5), Z = (50.5 - 18)/3.86 ≈ 8.83

This probability is essentially 1, given the high z-score.

  • For P(X ≤ 39):

P(X ≤ 39) ≈ P(Y ≤ 39.5), Z = (39.5 - 18)/3.86 ≈ 6.21

Again, this Z-score corresponds to a probability nearly 1.

Thus, the difference between these probabilities, or the probability that between 40 and 50 patients (inclusive) will require surgery, is significantly high, effectively close to 1. In practical terms, given the mean and standard deviation, this range captures nearly all of the probability mass.

Conclusion

The analysis shows that the likelihood of a very high number of patients requiring surgery (50 or more out of 60) is essentially zero, reflecting the low probability per patient. Conversely, the probability of 15 or fewer patients requiring surgery is about 25%. The chance that between 40 and 50 patients need surgery is extremely high, almost guaranteed, given the parameters of the distribution.

These calculations aid in resource planning and expectation management for orthopedic surgical services. Understanding these probabilities can improve scheduling, staffing, and inventory preparation for surgical procedures.

References

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