Consider the probability that no less than 95 out of 152 registered voters will vote in the presidential election. Assume the probability that a given registered voter will vote in the presidential election is 61%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

Step-by-step explanation:

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 , .

Assume the probability that a given registered voter will vote in the presidential election is 61%.

This means that

152 registed voters:

This means that

Mean and Standard deviation:

Probability that no less than 95 out of 152 registered voters will vote in the presidential election.

This is, using continuity correction, , which is 1 subtracted by the pvalue of Z when X = 94.5. So

has a pvalue of 0.6179

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.