Let z ∼ norm(0, 1). use the normal table and also r's "pnorm" function to find (a) p r(z ≤ 1.45) (b) p r(z > −1.28) (c) p r(−0.674 ≤ z < 1.036) (d) p r(z > 0.836)

If z~N(0,1), and using R, 
(a) P(z<=1.45)=pnorm(1.45)=0.9264707
(b) P(z>-1.28)=pnorm(-1.28,lower.tail=FALSE)=0.8997274
(c) P(-0.674<=z<1.036)=pnorm(1.036)-pnorm(-0.674)=0.5997433
(d) P(z>0.836)=pnorm(0.836,lower.tail=FALSE)=0.2015775

Using normal probability tables (find one on the Internet if you don't have a book) use the normal table of left probabilities, i.e. the shaded part is on the left.
(a) Go to line 1.4, and column 0.5 to find the value 0.9265
(b) Go to line -1.2 and column 0.8 to find 0.1003, subtract from 1 to get the right tail (i.e. z>-1.28), or 1-0.1003=0.8997
(c) The probability between two values of z is found be the difference of the left tails
Here P(-0.674)=0.2502 (interpolate between P(-0.67) and P(-0.68).
P(1.036)=0.8499 (interpolate between P(1.03)=0.8485 and P(1.04)=0.8499
So P(-0.674<=z<1.036)=0.8499-0.2502=0.5997
(d) P(z<0.836)=interpolate between P(z<0.83)=0.7967 and P(z<0.84) =0.7995 to get 0.7894
P(z>0.836)=1-P(z<0.836)=1-0.7984=0.2016

What class is this? i have no idea and i'm not going to give you the wrong answer. do you have a graph that goes with all the questions