Respuesta :
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:
[tex]np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10[/tex]
The distribution of the proportion of individuals who hold multiple jobs is,
[tex]p\sim N(p, \frac{p(1-p)}{n})[/tex]
Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:
[tex]P(p<0.071)=P(\frac{p-\mu}{\sigma}<\frac{0.071-0.13}{\sqrt{\frac{0.13(1-0.13)}{225}}})\\=P(Z<-2.63)\\=1-P(Z<2.63)\\=1-0.9957\\=0.0043[/tex]
*Use a z-table.
Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.
