the mean household income for a certain state is $71,000 with a standard deviation of $10,000 assuming a normal distribution about 95% of all households have an income over what amount?

About 95% of households have an income over $54,550

Step-by-step explanation:

95% is a probability of around 0.95

So we start by calculating the z-score with a probability of 0.95

We can use the standard normal distribution table for this

the z-score is -1.645

Mathematically;

z-score = (x-mean)/SD

-1.645 = (x-71,000)/10,000

10,000 * -1.645 = x-71,000

-16450 = x - 71000

x = 71,000 - 16450

x = 54,550

the answer is -3.5 + (-6.2)