Question

A state end-of-grade exam in American History is a multiple-choice test that has 50 questions with 4 answer choices for each question. A student must get at least 25 correct to pass the test, and the questions are very difficult. Question 1. If a student guesses on every question, what is the probability the student will pass

Answer 1

Answer:

0.004% probability the student will pass

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

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:

$n = 50, p = \frac{1}{4} = 0.25$

So

$\mu = E(X) = np = 50*0.25 = 12.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.25*0.75} = 3.06$

If a student guesses on every question, what is the probability the student will pass

Using continuity correction, this is $P(X \geq 25 - 0.5) = P(X \geq 24.5)$, which is 1 subtracted by the pvalue of Z when X = 24.5. So

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

$Z = \frac{24.5 - 12.5}{3.06}$

$Z = 3.92$

$Z = 3.92$ has a pvalue of 0.99996

1 - 0.99996 = 0.00004

0.004% probability the student will pass