Question

STATISTICS AND PROBABILITYThe average number of family income is $55,000 with a standard deviation of $15,000. If the poverty level is $20,000, what percentage of the population lives in poverty? A new tax law is expected to benefit middle income families that have an income between $50,000 to $60,000. What percentage of the population will benefit from the law?

Answer 1

To answer this question we will compute the z-score.

Recall that the z-score can be computed using the following formula:

`\begin{gathered} z=(x-\mu)/(\sigma), \\ \text{where }\mu\text{ is the mean, }\sigma\text{ is the standard deviation, and x is the observed value.} \end{gathered}`

Substituting μ=55000, σ=15000, and x=20000 we get:

`z=(20000-55000)/(15000)=-(35000)/(15000)=-2.3\bar{3}.`

Now, the percentage of the population that lives in poverty is:

`P(z<-2.3\bar{3})=0.0098153\cdot100percent=0.98153\text{percent}.`

Now, we know that the new tax law is expected to benefit middle-income families that have an income between $50,000 to $60,000, using the z-score formula we get:

`\begin{gathered} z=(50000-55000)/(15000)=-(5000)/(15000)=-0.3\bar{3}, \\ z=(60000-55000)/(15000)=(5000)/(15000)=0.3\bar{3}\text{.} \end{gathered}`

Then, the percentage of the population that will be benefited from the law is:

[tex]P(-0.3\bar{3}

Answer:

The percentage of the population that lives in poverty is 0.98153%.

The percentage of the population that will be benefited from the law is 26.112%.