1. approximations of distributions

In a multiple choice test have 200 questions, each with four possible answers, of which only one is correct. What is the probability that a student hit between 25 and 30 of 80 questions among 200 of which he knows nothing?

2. You've identified what you need to do in the title (approximations of distributions). Where are you stuck?

3. Originally Posted by Apprentice123
In a multiple choice test have 200 questions, each with four possible answers, of which only one is correct. What is the probability that a student hit between 25 and 30 of 80 questions among 200 of which he knows nothing?
I'm confused by the wording. Is the number 200 irrelevant, and we're just considering the 80 questions for which the student guesses? Also, is "between 25 and 30" inclusive, as in {25,26,27,28,29,30} or exclusive as in {26,27,28,29}? I would assume inclusive, but it's nice if it's specified.

4. My attempt

p = 25% = 0,25
P(25 < X < 30)
P(X>25) = P(X > 25,5) by approximations of distributions
Z1 = 1,42 -> 0,922196
P(X<30) = P(X < 29,5) by approximations of distributions
Z2 = 2,45 -> 0,992857

Z2 - Z1 = 0,070661

5. You have tried to do: P(25 < X < 30) = P(X<30) - P(X>25)

it should be
$P(25 < X < 30) = P(X < 30) - P(X \leq 25)$

I assume the question actually wants $P(25 \leq X \leq 30)$ which can be worked out as follows

$X \sim Bi(80,0.25)$

We will say this is approximated by
$Y \sim N(20,15)$

$P(X \leq 30) \approx P(Y < 30.5) = \Phi(\frac{10.5}{\sqrt{15}})$ = 0.9967
$P(X \leq 24) \approx P(Y < 24.5) = \Phi(\frac{4.5}{\sqrt{15}})$ = 0.877361

.9967-.877361 = .119286

I assume the difference between me and the textbook is due to rounding.

6. You not used approximations of distributions ?

7. indeed i did...

Originally Posted by Me
We will say this is approximated by
...

I have made minor changes to the original post in case it wasn't clear.

8. Ohh yes. Thank you