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:
![\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](https://tex.z-dn.net/?f=%5Cmu%3Dn%5Ccdot%20p%3D1309%5Ccdot%200.6%3D785.4%5C%5C%5C%5C%5Csigma%20%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B1309%5Ccdot%200.6%5Ccdot%200.4%7D%3D%5Csqrt%7B314.16%7D%3D17.7)
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:
![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](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3D%5Cdfrac%7B825-785.4%7D%7B17.7%7D%3D%5Cdfrac%7B39.6%7D%7B17.7%7D%3D2.24%5C%5C%5C%5C%5C%5CP%28X%3E825%29%3DP%28z%3E2.24%29%3D0.0126)
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.
3x-9=2x-28
3x-2x=9-28
x=-19
Answer:
3/10. 1/2 and 7/10
Step-by-step explanation:
For P(x) = x/10
When x = 3;
Then;
P(3) = 3/10
When x = 5
Then;
P(5) = 5/10
= 1/2
When, x = 7
Then;
P(7) = 7/10
Answer:
160
Step-by-step explanation:
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