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

Answer:

60.26% probability that less than twenty students are absent

Step-by-step 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:

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

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:

$n = 120, p = 0.15$.

So

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

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.15*0.85} = 3.91$

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

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

$Z = \frac{19 - 18}{3.91}$

$Z = 0.26$

$Z = 0.26$ has a pvalue of 0.6026.

60.26% probability that less than twenty students are absent