Question

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

Answer 1

Answer:

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:

• np ≥ 10
• n (1 - p) ≥ 10

Check the conditions as follows:

$np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10$

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

$p\sim N(p, \frac{p(1-p)}{n})$

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

$P(p$

*Use a z-table.

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