Question

Scotland has been a member of the United Kingdom since the 1707 Act of Union. In a recent poll, 38% of Scots would vote in favor of leaving the U.K. A random sample of 135 Scots was selected. What is the probability that between 50 and 60 of them were in favor of leaving the U.K.?

Answer 1

Answer:

57.39% probability that between 50 and 60 of them were in favor of leaving the U.K.

Step-by-step explanation:

I am going to use the normal approximmation 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:

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

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:

$n = 135, p = 0.38$

So

$\mu = E(X) = np = 135*0.38 = 51.3$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{135*0.38*0.62} = 5.6397$

What is the probability that between 50 and 60 of them were in favor of leaving the U.K.?

Using continuity correction, this is $P(50-0.5 \leq X \leq 60 + 0.5) = P(49.5 \leq X \leq 60.5)$. So this is the pvalue of Z when X = 60.5 subtracted by the pvalue of Z when X = 49.5. So

X = 60.5

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

$Z = \frac{60.5 - 51.3}{5.6397}$

$Z = 1.63$

$Z = 1.63$ has a pvalue of 0.9484

X = 49.5

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

$Z = \frac{49.5 - 51.3}{5.6397}$

$Z = -0.32$

$Z = -0.32$ has a pvalue of 0.3745

0.9484 - 0.3745 = 0.5739

57.39% probability that between 50 and 60 of them were in favor of leaving the U.K.