Answer:
0.9319 = 93.19% probability that at least 88 out of 153 registered voters will vote in the presidential election.
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x successes 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 ,
.
153 voters:
This means that
Assume the probability that a given registered voter will vote in the presidential election is 63%.
This means that
Mean and standard deviation:
Consider the probability that at least 88 out of 153 registered voters will vote in the presidential election.
Using continuity correction, this is: , which is 1 subtracted by the p-value of Z when X = 87.5.
has a p-value of 0.0681.
1 - 0.0681 = 0.9319
0.9319 = 93.19% probability that at least 88 out of 153 registered voters will vote in the presidential election.
