Respuesta :
Answer:
a) P(X>825)
b) This low value of probability of the sample outcome (as 825 voters actually did vote) suggests that the 60% proportion may not be the true population proportion of eligible voters that actually did vote.
Step-by-step explanation:
We know a priori that 60% of the eligible voters did vote.
From this proportion and a sample size n=1309, we can construct a normal distribution probabilty, that is the approximation of the binomial distribution for large samples.
Its mean and standard deviation are:
[tex]\mu=n\cdot p=1309\cdot 0.6=785.4\\\\\sigma =\sqrt{np(1-p)}=\sqrt{1309\cdot 0.6\cdot 0.4}=\sqrt{314.16}=17.7[/tex]
Now, we have to calculate the probabilty that, in the sample of 1309 voters, at least 825 actually did vote. This is P(X>825).
This can be calculated using the z-score for X=825 for the sampling distribution we calculated prerviously:
[tex]z=\dfrac{X-\mu}{\sigma}=\dfrac{825-785.4}{17.7}=\dfrac{39.6}{17.7}=2.24\\\\\\P(X>825)=P(z>2.24)=0.0126[/tex]
This low value of probability of the sample outcome (as 825 voters actually did vote) suggests that the 60% proportion may not be the true population proportion of eligible voters that actually did vote.
