Question

Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large university are absent on Mondays. If a random sample of 120 names is called on a Monday, what is the probability that less than twenty students are absent?

Answer 1

60.26% probability that less than twenty students are absent

Explanation:

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:

`n = 120, p = 0.15`.

So

`\mu = E(X) = np = 120*0.15 = 18`

`\sigma = √(V(X)) = √(np(1-p)) = √(120*0.15*0.85) = 3.91`

If a random sample of 120 names is called on a Monday, what is the probability that less than twenty students are absent?

This is the pvalue of Z when X = 20-1 = 19. So

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

`Z = (19 - 18)/(3.91)`

`Z = 0.26`

`Z = 0.26` has a pvalue of 0.6026.

60.26% probability that less than twenty students are absent