Question

3. A recent news article stated that only 17% of college students between the ages of 18 to 24 years old voted in the last presidential election. Assuming the voting rate stays the same, what is the probability that from a random sample of 500 college students from a local university, at least 20% will vote in the next presidential election

Answer 1

Answer:

3.67% probability that from a random sample of 500 college students from a local university, at least 20% will vote in the next presidential election

Step-by-step explanation:

I am going to 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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$p = 0.17, n = 500$.

So

$\mu = E(X) = np = 500*0.17 = 85$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{500*0.17*0.83} = 8.4$

Assuming the voting rate stays the same, what is the probability that from a random sample of 500 college students from a local university, at least 20% will vote in the next presidential election

This is 1 subtracted by the pvalue of Z when X = 500*0.2 = 100. So

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

$Z = \frac{100 - 85}{8.4}$

$Z = 1.79$

$Z = 1.79$ has a pvalue of 0.9633

1 - 0.9633 = 0.0367

3.67% probability that from a random sample of 500 college students from a local university, at least 20% will vote in the next presidential election