Question

Of the 120 families, approximately___pay more than $5710 annually for day car per child.

Answer 1

1) Considering a Normal Distribution, then we can write out the following:

$P(X>5710)=P(X-\mu>5710-6000)=P(\frac{X-\mu}{\sigma}>\frac{5710-6000}{1000})$

Note that we're dealing with probabilities.

2) Let's find out the Z-score resorting to a table, we get:

$Z=\frac{x-\mu}{\sigma}=\frac{5710-6000}{1000}=-0.29$

2.2) So we can infer from 1 and 2:

$P(X>5710)=P(Z>-0.29)=0.6141$

Notice that this distribution refers to 120 families