Respuesta :
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
153 voters:
This means that [tex]n = 153[/tex]
Assume the probability that a given registered voter will vote in the presidential election is 63%.
This means that [tex]p = 0.63[/tex]
Mean and standard deviation:
[tex]\mu = E(X) = np = 153(0.63) = 96.39[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.63*0.37} = 5.97[/tex]
Consider the probability that at least 88 out of 153 registered voters will vote in the presidential election.
Using continuity correction, this is: [tex]P(X \geq 88 - 0.5) = P(X \geq 87.5)[/tex], which is 1 subtracted by the p-value of Z when X = 87.5.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{87.5 - 96.39}{5.97}[/tex]
[tex]Z = -1.49[/tex]
[tex]Z = -1.49[/tex] 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.
