Question

1. Approximately 8.26% of patients in the Seasonal Effect data set contracted an SSI. If this happens completely randomly, calculate the probability that three randomly selected patients all contract an SSI in each of the following two ways:
   1. Using counting rules and probability trees (10%)
2. 2. Using the binomial distribution (hint: n=3, p=0.0826) (10%)

2. Approximately 8.26% of patients in the Seasonal Effect data set contracted an SSI. If this happens completely randomly, in a random selection of 10, calculate the following probabilities, from a binomial distribution with parameters n=10 and p=0.0826. Show all work.
   1. Identify the complement of {X≥1} and use the rule of complements to calculate the probability that at least 1 patient contracts an SSI, P(X≥1) (10%)
   2. probability that more than 1 but less than 5 patients contract an SSI, P(1<X<5) (10%) (Hint: use the cumulative probability function in excel).

3. In the Seasonal Effect data set, an average of 20 patients develop an SSI each month. For a randomly selected month in the year, calculate the following probabilities using the Poisson distribution. Show all work.
   1. Exactly 20 patients develop an SSI in the month, P(X=20) (10%)
   2. Use the cumulative distribution to calculate the probability that less than 10 patients develop an SSI in the month, P(X≤10) (10%)

4. The average duration of surgery (in hours) in all patients in the Seasonal Effect data set is approximately 3.581 with a standard deviation of approximately 1.946. The duration of surgery values seems to follow a normal distribution. Estimate the percentage of surgeries that took longer than 6 hours using the normal probability distribution, P(Duration>6). Show all work. (20%) (Hint: do not try to calculate the probability using the probability density function by hand. Use google sheets or Microsoft excel). (Note: the actual percentage of surgeries that took longer than 6 hours is 10.24%.)

Answer 1

Solution:-

4) The percentage of surgeries that took longer than 6 hours using the normal probability distribution is 10.69%.

Mean = 3.581, S.D = 1.946

x = 6

By applying normal distribution:-

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

z = 1.2431

P(z > 1.2431) = 0.1069

P(z > 1.2431) = 10.69%