Can anyone help?make sure you simplify
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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 <em>X</em> = number of individuals in the United States who held multiple jobs.
The probability that an individual holds multiple jobs is, <em>p</em> = 0.13.
The sample of employed individuals selected is of size, <em>n</em> = 225.
An individual holding multiple jobs is independent of the others.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 225 and <em>p</em> = 0.13.
But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion <em>p</em>.
Conditions of Normal approximation to Binomial are:
- np ≥ 10
- n (1 - p) ≥ 10
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 <em>z</em>-table.
Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.