The Pew Research Center estimates that as of January 2014, 89% of 18-29 year olds in the United States use social networking sites.
1. Calculate the probability that at least 95% of 500 randomly sampled 18-29 year olds use social networking sites.

0% probability that at least 95% of 500 randomly sampled 18-29 year olds use social networking sites.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this problem.

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:

1. Calculate the probability that at least 95% of 500 randomly sampled 18-29 year olds use social networking sites.

This probability is 1 subtracted by the pvalue of Z when X = 0.95*500 = 475. So

has a pvalue of 1

1 - 1 = 0

0% probability that at least 95% of 500 randomly sampled 18-29 year olds use social networking sites.

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