A scientist claims that 7%7% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 418418 viruses would be greater than 6%6%? Round your answer to four decimal places.

Probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6% is 0.8051.

Step-by-step explanation:

We are given that a scientist claims that 7% of viruses are airborne.

A sample of 418 viruses is taken.

Let = sample proportion of airborne viruses

The z-score probability distribution for sample proportion is given by;

          Z =  ~ N(0,1)

where, = sample proportion

           p = population proportion of viruses that are airborne = 7%

           n = sample of viruses = 418

Probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6% is given by = P( > 0.06)

   P( > 0.06) = P( > ) = P(Z > -0.86) = P(Z < 0.86)

                                                                                          = 0.8051

Now, in the z table the P(Z x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.86 in the z table which has an area of 0.8051.

Therefore, probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6% is 0.8051.

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: