Question

Use normal approximation to estimate the probability of passing a true/false test of 50 questions if the minimum passing grade is 60% and all responses are random guesses.

Answer 1

Answer:

7.93% probability of passing the test.

Step-by-step explanation:

We use the binomial approximation to the normal 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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

Guessed true/false question.

Each have two options, one of which is correct, so $p = \frac{1}{2} = 0.5$

50 questions

This means that $n = 50$

So

$\mu = E(X) = np = 50*0.5 = 25$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.5*0.5} = 3.5356$

Probability of passing a true/false test of 50 questions if the minimum passing grade is 60% and all responses are random guesses.

This probability is 1 subtracted by the pvalue of Z when X = 0.6*50 = 30. So

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

$Z = \frac{30 - 25}{3.5356}$

$Z = 1.41$

$Z = 1.41$ has a pvalue of 0.9207

1 - 0.9207 = 0.0793

7.93% probability of passing the test.