Final answer:
The probability that the proportion of airborne viruses in the sample would be greater than 6% is approximately 0.5596.
Step-by-step explanation:
To find the probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6%, we can use the normal distribution. First, let's find the mean and standard deviation of the proportion of airborne viruses. The mean is 7% (0.07) and the standard deviation can be calculated using the formula: sqrt((p(1-p))/n), where p is the proportion and n is the sample size.
Using the given information, the standard deviation is sqrt((0.07(1-0.07))/418) ≈ 0.0064.
Next, we convert the 6% threshold to a z-score using the formula (x - mean) / standard deviation. So, the z-score is (0.06 - 0.07) / 0.0064 ≈ -0.156.
To find the probability that the proportion is greater than 6%, we look up the z-score (-0.156) in the standard normal distribution table and subtract the corresponding probability from 1. The probability is approximately 0.5596.