Question

An exam consists of 41 multiple-choice questions. Each question has a choice of five answers, only one of which is correct. For each correct answer, a candidate gets 1 mark, and no penalty is applied for getting an incorrect answer. A particular candidate answers each question purely by guess-work. Using Normal approximation to Binomial distribution with continuity correction, what is the estimated probability this student obtains a score greater than or equal to 10? Please use R to obtain probabilities and keep at least 6 decimal places in intermediate steps.

Answer 1

Answer:

There is a 24.20% probability this student obtains a score greater than or equal to 10.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aaproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X), \sigma = \sqrt{V(X)}$

In this problem, we have that:

There are 41 question, so $n = 41$.

There are 5 options for each answer, one of which is correct. The student guesses each one. So $p = \frac{1}{5} = 0.2$.

This means that:

$\mu = E(X) = 41*0.2 = 8.2$

$\sigma = \sqrt{V(X)} = \sqrt{41*0.2*0.8} = 2.56$

What is the estimated probability this student obtains a score greater than or equal to 10?

This probability is 1 subtracted by the pvalue of Z when $X = 10$.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10 - 8.2}{2.56}$

$Z = 0.70$

$Z = 0.70$ has a pvalue of 0.7580.

This means that there is a 1-0.7580 = 0.2420 = 24.20% probability that this student obtains a score greater than or equal to 10.