Find the following z values for the standard normal variable Z. I got all the answers but B and I am pretty sure I'm right and the answer in the back of the book is wrong.Can you verify?
a. P(Z ≤ z) = 0.1010 =-1.28
b. P(z ≤ Z ≤ 0) = 0.1456 =-1.06 It says I am wrong though
c. P(Z > z ) = 0.8186 =-.91
d. P(0.29 ≤ Z ≤ z) = 0.3513 =1.81
On
Are you saying that P(-1.06 < Z < 0) = .1456? Can't be right. Suggest you draw a sketch. The answer to P(-1.06 < Z < 0) is about .36. Under a normal curve, about 68% of the area is between -1 and +1; 34% on each side of 0. That leaves about 16% for each of the tails (one to the left of -1, the other to the right of +1).
Where you write $\Pr(Z\le z) = 0.1010 = -1.28$, that is certainly not correct since $0.1010$ is not equal to $-1.28$. I presume you meant that if $\Pr(Z\le z)=0.1010$ then $z=-1.28$, which is correct up to rounding error. The same thing applies to your other examples. Don't do it that way. Math professors look at assignments where things like that are written and they think are, to put it politely, deficient in understanding.
I'm getting $$\Pr(Z\le -1.055493)=0.1456$$ which is numerically close to what you have but it implies that $$\Pr(-1.055493\le Z\le 0) = 0.5-0.1456 = 0.0544.$$ Similar considerations affect your other answers.