Answer:
3.67% probability that from a random sample of 500 college students from a local university, at least 20% will vote in the next presidential election
Step-by-step explanation:
I am going to 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:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
.
So
![\mu = E(X) = np = 500*0.17 = 85](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20500%2A0.17%20%3D%2085)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{500*0.17*0.83} = 8.4](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B500%2A0.17%2A0.83%7D%20%3D%208.4)
Assuming the voting rate stays the same, what is the probability that from a random sample of 500 college students from a local university, at least 20% will vote in the next presidential election
This is 1 subtracted by the pvalue of Z when X = 500*0.2 = 100. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{100 - 85}{8.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B100%20-%2085%7D%7B8.4%7D)
![Z = 1.79](https://tex.z-dn.net/?f=Z%20%3D%201.79)
has a pvalue of 0.9633
1 - 0.9633 = 0.0367
3.67% probability that from a random sample of 500 college students from a local university, at least 20% will vote in the next presidential election