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.
Answer: . a. 36.08 to 78.80
Step-by-step explanation:
The confidence interval for population standard deviation is given by :-
![\sqrt{\dfrac{(n-1)s^2}{\chi_{\alpha/2}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7B%28n-1%29s%5E2%7D%7B%5Cchi_%7B%5Calpha%2F2%7D%7D%7D%3C%5Csigma%5Csqrt%7B%5Cdfrac%7B%28n-1%29s%5E2%7D%7B%5Cchi_%7B1-%5Calpha%2F2%7D%7D%7D)
Given : Sample size = ![n=24](https://tex.z-dn.net/?f=n%3D24)
Significance level : ![\alpha:1-0.99=0.01](https://tex.z-dn.net/?f=%5Calpha%3A1-0.99%3D0.01)
Using the chi-square distribution table , the critical values would be :-
![\chi_{n-1,\alpha/2}=\chi_{23, 0.005}=44.18127525](https://tex.z-dn.net/?f=%5Cchi_%7Bn-1%2C%5Calpha%2F2%7D%3D%5Cchi_%7B23%2C%200.005%7D%3D44.18127525)
![\chi_{n-1,1-\alpha/2}=\chi_{23, 0.995}=9.26042478](https://tex.z-dn.net/?f=%5Cchi_%7Bn-1%2C1-%5Calpha%2F2%7D%3D%5Cchi_%7B23%2C%200.995%7D%3D9.26042478)
Then , a 99% confidence interval for the population standard deviation will be :-
![\sqrt{\dfrac{(23)(50)^2}{44.18127525}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7B%2823%29%2850%29%5E2%7D%7B44.18127525%7D%7D%3C%5Csigma%5Csqrt%7B%5Cdfrac%7B%2823%29%2850%29%5E2%7D%7B9.26042478%7D%7D%5C%5C%5C%5C%3D36.075702658%3C%5Csigma%3C78.7985939401%5C%5C%5C%5C%5Capprox36.08%3C%5Csigma%3C78.80)
Answer:
wheres the table?
Step-by-step explanation:
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Answer: It‘s already written in standard form. f(x) = x^2 - 8x
Step-by-step explanation: