Question 647522
A)
i)
By use of TI or table: invNorm(.1003,0,1) For P({{z<x}}}) = .1003   x = -1.28

So, our z-score is -1.28. 

ii)Z-score = (x-mean)/SD  

-1.28 = (x-50)/5

-6.40 = x -50

44.50 = x

B) Use of TI:

P({{{57.5<x<65}}}) = normalcdf(57.5,65,50,5) = .0654 

You cannot use a TI though, so convert 57.5 and 65 into z-scores Z1,Z2. Then with Z1 and Z2 take P({{{z<Z2}}}) - P({{{z<Z1}}}) by use of table.

Let's do that:

Z1 = (57.5-50)/5 = 7.5/5 = 1.5
Z2 = (65-50)/5 = 15/5 = 3

P({{{1.5<z<3}}}) = P({{{z<3}}}) - P({{{z<1.5}}}) = .9986 - .9332 = .0654

Or 6.54%

C) Z-score = (35-50)/5 = -15/5 = -3. 

P({{{z<-3}}}) = 1-P({{{z<3}}}) = 1- .9986 = .0014 

.14%

D)

34% of the data leaves the left tail to be (100-34)/2 = 33%  and equally the right tail to be 33%

So to find the right most value, we need to find P({{{z<x}}})=.67.

By looking at at a table, you'll find that the z-score that achieves this is .44. 

Next we need to find the left endpoint. P({{{z<x}}}) = .33. Similarly, by looking at a table we find that the z-score that achieves this is -.44

Convert z-scores to raw scores:

.44 = (x-50)/5

2.2 = x-50

x = 52.2

-.44 = (x-50)/5 

-2.2 = x-50

x= 48.8 

The one below the mean is 48.8.