Assume that the family income in a certain town is distributed normally with mean $60000 and a standard deviation of $10000. If the poverty level is $30000, what percentage of the family live above the poverty line?

Recall that F(z)=(1/(2pi)^1/2)integral(e^(-x^2/2))dt from z to -infinity

Question

Assume that the family income in a certain town is distributed normally with mean $60000 and a standard deviation of $10000. If the poverty level is $30000, what percentage of the family live above the poverty line?

Recall that F(z)=(1/(2pi)^1/2)integral(e^(-x^2/2))dt from z to -infinity

anonymous · Answer

Let $X$ be the variable to represent the family income. You want to find $P(X>30,000)$.

First transform to a standard normal distribution:
$$P(X>30,000)=P\left(\frac{X-60,000}{10,000}>\frac{30,000-60,000}{10,000}\right)=P(Z>-3)$$

anonymous · Answer

the answer choices are
a) F(5/3)
b)F(-3)
c)1-F(3)
d)1-F(-5/3)
e)F(3)

anonymous · Answer

Then
$$\large \begin{align*}P(Z>-3)&=1-P(Z<-3)\&=1-F(-3)\
&=1-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-3}\exp\left(-\dfrac{t^2}{2}\right)~dt
\end{align*}$$
If I remember correctly, $P(Z>-3)$ is very close to 1 as a probability...

anonymous · Answer

Ah, they want you to use the symmetry of the standard normal distribution here. $F(3)=1-F(-3)$.

anonymous · Answer

Oh my god thank you so much!!!

anonymous · Answer

yw