Normal Approximation to the Binomial Distribution Help Needed

#1
SOLVED - Normal Approximation to the Binomial Distribution Help Needed

A multiple choice test consists of a series of questions, each with four possible choices.
a) If there are 60 questions, estimate the probability that a student guessing blindly on each question will get at least 30 right.
b) How many questions are needed in order to be 99% confident that a student who guesses blindly will score no more than 35%?

For a, I took this to ask P(X>=30;n=60,p=.25)
Which would translate to X~N(15,11.25)
So, P(X_N>=29.5)
which is a z-score of 1.29,
So 1-Phi(1.29)=1-.9015=.0985
But this seems too high to me. I think that it should be a very small probability, not nearly .1.

And for b, I am completely lost. I tried P(x/n<=.35;n,p=.25)=.99 so that X~N(.25n,.1875n), but that gave me n=1.48, which makes literally no sense.
What am I doing wrong? I've never quite understood this.
 
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Dason

Ambassador to the humans
#2
Remember when you compute a z-score you want to divide by the standard deviation - not the variance.
 
#3
Oh, right!
So with that z=4.32, so 1-Phi(4.32) is about 8*10^-6, according to my calculator, which seems much more right to me. Thanks!
When I try b with this, I get something like 2.33=(.1n+.5)/Sqrt(.1875n), which tells me that n=.2731, which manages to make less sense than the previous answer.
Is there something wrong with the setup of this?
EDIT:
So I looked at b again, and realized that the equation has two solutions, and the other, n=92, makes a lot more sense. Thanks!
 
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