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.?

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:

The standard deviation of the binomial distribution is:

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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 , .

In this problem, we have that:

So

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

Using continuity correction, this is . 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

has a pvalue of 0.9484

X = 49.5

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.

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step-by-step explanation:

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overpriced

step-by-step explanation: