A scientist claims that 4%4% of viruses are airborne.If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 849849 viruses would be greater than 3%3%? Round your answer to four decimal places.

To solve this problem, we need to use the concept of normal distribution and the Central Limit Theorem. Here are the steps:

1. First, we need to find the mean and standard deviation of the distribution. The mean (μ) is the expected proportion of airborne viruses, which is 4% or 0.04. The standard deviation (σ) can be calculated using the formula for the standard deviation of a proportion, which is sqrt[p(1-p)/n], where p is the proportion (0.04) and n is the sample size (849).

2. Plugging the values into the formula, we get σ = sqrt[0.04(1-0.04)/849] = 0.0068.

3. Next, we need to find the z-score for 3% or 0.03. The z-score is calculated as (X - μ) / σ, where X is the value we are interested in. So, the z-score is (0.03 - 0.04) / 0.0068 = -1.47.

4. The z-score tells us how many standard deviations our value is from the mean. A negative z-score indicates that our value is below the mean.

5. To find the probability that the proportion of airborne viruses is greater than 3%, we need to find the area to the right of our z-score in the standard normal distribution. But standard normal distribution tables typically give the area to the left. So, we find the area to the left and subtract it from 1.

6. Looking up -1.47 in the z-table, we find that the area to the left is 0.0708.

7. Subtracting this from 1 gives us 1 - 0.0708 = 0.9292.

So, if the scientist's claim is accurate, the probability that the proportion of airborne viruses in a sample of 849 viruses would be greater than 3% is approximately 0.9292, or 92.92%.