A new sample of 225 employed adults is chosen. Find the probability that less than 7.1% of the individuals in this sample hold multiple jobs. Round the answer to at least four decimal places. The probability that less than 7.1% of the individuals in this sample hold multiple jobs is

The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

Step-by-step explanation:

Let X = number of individuals in the United States who held multiple jobs.

The probability that an individual holds multiple jobs is, p = 0.13.

The sample of employed individuals selected is of size, n = 225.

An individual holding multiple jobs is independent of the others.

The random variable X follows a Binomial distribution with parameters n = 225 and p = 0.13.

But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion p.

Conditions of Normal approximation to Binomial are:

Check the conditions as follows:

The distribution of the proportion of individuals who hold multiple jobs is,

Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:

*Use a z-table.

Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

it would be 6xy and -16xy.

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3