# Actuarial. Calculate N using a normal approximation with the continuity correction.

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.

Facts:

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 :

a)$23$

b)$25$

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

d)$33$

e)$35$

## 1 Answer

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 $$\Pr\left[Z > \frac{N - 18.5}{\sqrt{10}}\right] < 0.1.$$

• 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? – Electro82 Oct 10 '15 at 6:28