Question

A multiple choice exam has 40 questions, each with 5 possible answers. A well prepared student believes she has a probability of 0.5 of getting any particular question correct. She uses the normal approximation to determine the probability of getting less than 25 correct. In her first attempt she forgets to use the continuity correction and then corrects herself. Determine the probability both with and without the continuity correction.

Answer 1

Without continuity correction, an 89.71% probability of getting less than 25 correct.

With continuity correction, 92.22% probability of getting less than 25 correct.

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:

`√(V(X)) = √(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 = (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 = √(V(X))`.

In this problem, we have that:

`p = 0.5, n = 40`

So

`\mu = E(X) = np = 40*0.5 = 20`

`\sigma = √(V(X)) = √(np(1-p)) = √(40*0.5*0.5) = 3.1622`

Less than 25 correct

Without cc

`P(X \leq 24)`, which is the pvalue of Z when X = 24. So

`Z = (X - \mu)/(\sigma)`

`Z = (24 - 20)/(3.1622)`

`Z = 1.265`

`Z = 1.265` has a pvalue of 0.8971

Without continuity correction, an 89.71% probability of getting less than 25 correct.

With cc

`P(X \leq 24 + 0.5) = P(X \leq 24.5)`, which is the pvalue of Z when X = 24.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (24.5 - 20)/(3.1622)`

`Z = 1.42`

`Z = 1.42` has a pvalue of 0.9222

With continuity correction, 92.22% probability of getting less than 25 correct.