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 553 viruses would be greater than 5%? Round your answer to four decimal places.

The probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

The standard deviation of this sampling distribution of sample proportion is:

The sample size is, n = 553 > 30. Thus, the Central limit theorem is applicable in this case.

Compute the mean and standard deviation as follows:

Compute the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% as follows:

Thus, the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.

hi there! my name is zalgo and i am here to you out today. the answer is pretty simple (in a way). this equation equals to 25.2!

i hope this answer !

"stay and stay proud! " - zalgo

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all