SOLUTION: Really Big Medical Center (RBMC) is testing patients who have spent time in grassy areas this summer and fall for Lyme disease. Suppose that the probability of someone with exposur

Question 1200618: Really Big Medical Center (RBMC) is testing patients who have spent time in grassy areas this summer and fall for Lyme disease. Suppose that the probability of someone with exposure to tall grass having Lyme is .3 (or 30 percent). In a sample of 15 patients, what is the standard deviation of the number of patients with Lyme disease? Round your answer to 1 decimal place.

Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!
**1. Identify the Distribution**
* This scenario follows a binomial distribution:
* There are a fixed number of trials (n = 15 patients)
* Each trial has two possible outcomes (having Lyme disease or not)
* The probability of success (having Lyme disease) is constant (p = 0.3)
* The trials are independent of each other.
**2. Calculate the Standard Deviation**
* The standard deviation (σ) of a binomial distribution is given by:
σ = √(n * p * (1 - p))
where:
* n = number of trials (15 patients)
* p = probability of success (0.3)
* (1 - p) = probability of failure (0.7)
* Calculate:
σ = √(15 * 0.3 * 0.7)
σ = √(3.15)
σ ≈ 1.8
**Therefore, the standard deviation of the number of patients with Lyme disease in the sample is approximately 1.8.**

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