I'm try to solve this problem:

A student takes a multiple-choice test with $40$ questions. The probability that the student answers a given question correctly is $0.5$, independent of all other questions.


1) The probability that the student answers more than N questions correctly is greater than $0.10$.

2) The probability that the student answers more than N + 1 questions correctly is less than $0.10$.

Question: Calculate $N$ using a normal approximation with the continuity correction.

I though:

I know that the Binomial $E(x)=40*0.5=20$ also know the variance, should I use $P(X>N)-P(X>N+1)=P(X=N)$? but how about this probability? because the question says >0.1 and <0.1, how to constrain the value of $P(X=N)$,should this be $0.1$? I realy appreciate any help to understand how to lead with the correct answer, beacuse I can use this problem as a model to solve others. Thanks!

Also my paper gave me $5$ options :



c)$32$ I got this, I'm not sure if is ritgh I applied De Movire - Laplace Theorem.




We are given that the number of correctly answered questions $X$ is a binomial random variable with parameters $n = 40$ and $p = 0.5$. We are also given that $$\Pr[X > N] > 0.1, \quad \Pr[X > N+1] < 0.1.$$ Now approximate $X$ as $$Y \sim \operatorname{Normal}(\mu = np = 20, \sigma^2 = np(1-p) = 10),$$ we have $$\Pr[X > N] \approx \Pr[Y > N + 0.5] = \Pr\left[\frac{Y - \mu}{\sigma} > \frac{N + 0.5 - 20}{\sqrt{10}} \right].$$ Similarly, $$\Pr[X > N+1] \approx \Pr\left[\frac{Y - \mu}{\sigma} > \frac{N + 1 + 0.5 - 20}{\sqrt{10}}\right].$$ So we seek $N$ such that the $Z$-score $(19.5-N)/\sqrt{10}$ is at least $0.1$, but the $Z$-score $(18.5-N)/\sqrt{10}$ is less than $0.1$. Since we are given answer choices, it suffices to substitute. Let's start with (C): If $N = 32$, the $Z$-score is $-3.952$, much too small. So we know that this choice of $N$ is too large. Choose (A): If $N = 23$, we get a $Z$-score of $-1.1068$, and $$\Pr[Z \le -1.1068] \approx 0.134191.$$ This looks good! Let's check the other condition: $$\Pr[Z \le -1.42302] \approx 0.0773645.$$ So this works out, and the answer is (A).

To check, we can try (B): The $Z$-scores are $-1.73925$ and $-2.05548$, the first of which, being less than $-1.42302$, will yield a probability less than $0.1$, so we are now certain that (A) is correct.

Now, this makes sense: because $p = 0.5$, the distributions concerned are symmetric about the mean $\mu = np = 20$. And as the standard deviation is $\sigma = \sqrt{10} \approx 3.16228$, we can recall that the $90^{\rm th}$ percentile is about $z_{0.90} \approx 1.282$, so approximately $90\%$ of the probability mass for $X$ will be below $20+(1.282)(3.16228) = 24.05$. This immediately suggests looking at (A) or (B) as the answer choices.

You should memorize the following critical values/percentiles: $$z_{0.90} = 1.282, \quad z_{0.95} = 1.645, \quad z_{0.975} = 1.96,$$ corresponding to two-sided confidence intervals of $80\%$, $90\%$, and $95\%$, or alpha levels of $\alpha = 0.20$, $0.10$, and $0.05$, respectively.

If we are not given answer choices, then the solution is to explicitly calculate the bounds: That is to say, suppose $$\Pr\left[Z > \frac{N-19.5}{\sqrt{10}}\right] = 0.1 = \Pr[Z > z_{0.90}] \approx \Pr[Z > 1.282].$$ Therefore, $$\frac{N - 19.5}{\sqrt{10}} \approx 1.282,$$ or $N = 23.554$. Since we require the probability to exceed $0.1$, this means $N$ must be an integer smaller than $23.554$, so the next lower value is $N = 23$. Then we check that this also satisfies $$\Pr\left[Z > \frac{N - 18.5}{\sqrt{10}}\right] < 0.1.$$

  • $\begingroup$ Thanks for your answer. I was wondering what will be the procedure If I have no the 5 answer.How to calculate the value of N. Not knowing the answer? $\endgroup$ – Electro82 Oct 10 '15 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.