1. ## Last Probability Question!

Sorry for all the questions,
I've been having trouble with these homework questions all week.

Heres the last question:

μ: $60k σ:$5k

Assume that for the standard normal population the probability to not exceed 1.960 is 0.975

A.) Find the value V such that 97.5% of households have the income above (greater than V)

B.) Find the value U such that 2.5% of households have the income below (less than) U

C.) What is the proportion of households with the annual income between U and V

I have tried this a couple times, but everytime I check the answer it comes out wrong

Any help would be greatly appreciated
Thanks

2. Originally Posted by JohhnySD
Sorry for all the questions,
I've been having trouble with these homework questions all week.

Heres the last question:

μ: $60k σ:$5k

Assume that for the standard normal population the probability to not exceed 1.960 is 0.975

A.) Find the value V such that 97.5% of households have the income above (greater than V)

B.) Find the value U such that 2.5% of households have the income below (less than) U

C.) What is the proportion of households with the annual income between U and V

I have tried this a couple times, but everytime I check the answer it comes out wrong

Any help would be greatly appreciated
Thanks
A.) Pr(X > v) = 0.975 => Pr(X < v) = 0.025 => Pr(X > -v) = 0.025 => Pr(X < -v) = 0.975.

But Pr(Z < 1.960) = 0.975 (given) and $Z = \frac{X -\mu}{\sigma} \Rightarrow 1.960 = \frac{-v - 60}{5} \Rightarrow v = ....$

You can do the others. Given the definitions of u and v in A.) and B.) I would have thought the the answer to C.) is obvious.