A scientist claims that 4% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 849 viruses would be greater than 3%? Round your answer to four decimal places.

Step 1 ：First, we need to calculate the standard deviation of the sampling distribution of the sample proportion (σ_p̂) using the formula: \(σ_p̂ = \sqrt{ p(1 - p) / n }\)

Step 2 ：Given values are: \(p = 0.04\), \(n = 849\), \(X = 0.03\)

Step 3 ：Substitute the given values into the formula, we get \(σ_p̂ = \sqrt{ 0.04(1 - 0.04) / 849 } = 0.006725301627347177\)

Step 4 ：Next, we need to calculate the Z-score for the value 0.03 using the formula: \(Z = (X - μ) / σ\)

Step 5 ：Substitute the given values into the formula, we get \(Z = (0.03 - 0.04) / 0.006725301627347177 = -1.4869221566712902\)

Step 6 ：Finally, we need to find the probability that Z is greater than the calculated Z-score using the standard normal distribution. This is equivalent to finding the area to the right of the calculated Z-score under the standard normal curve, which can be found using the survival function (1 - cumulative distribution function) of the standard normal distribution.

Step 7 ：The probability that Z is greater than the calculated Z-score is 0.9315

Step 8 ：Final Answer: The probability that the proportion of airborne viruses in a sample of 849 viruses would be greater than 3% is \(\boxed{0.9315}\)