Residences of U. S. Citizens According to the U. S. Census, 67.5% of the U. S. population were born in their state of residence. In a random sample of 203 Americans, what is the probability that fewer than 126 were born in their state of residence? Round the final answer to at least four decimal places and intermediate -value calculations to two decimal places. P ( X < 126)=?

P(X < 126) = 0.0418

Step-by-step explanation:

To solve this question, we 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:

The standard deviation of the binomial distribution is:

Normal probability distribution

Problems of normally distributed distributions 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 , .

Sample of 203, 67.5% of the U.S. population were born in their state of residence.

This means that

So, for the approximation:

What is the probability that fewer than 126 were born in their state of residence?

Using continuity correction, this is P(X < 126 - 0.5) = P(X < 125.5), which is the pvalue of Z when X = 125.5. So

has a pvalue of 0.0418

So

P(X < 126) = 0.0418

step-by-step explanation:

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