Answer:
20.90% probability that more than 192 of the confidence intervals cover the true proportions
Step-by-step explanation:
I am going to use the normal approximation to the binomial 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:
A single 95% confidence interval has a 95% probability of covering the true proportions, so ![p = 0.95](https://tex.z-dn.net/?f=p%20%3D%200.95)
200 intervals, so ![n = 200](https://tex.z-dn.net/?f=n%20%3D%20200)
Then
![\mu = E(X) = np = 200*0.95 = 190](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20200%2A0.95%20%3D%20190)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.95*0.05} = 3.0822](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%7B200%2A0.95%2A0.05%7D%20%3D%203.0822)
What is the probability that more than 192 of the confidence intervals cover the true proportions
Using continuity correction, this is
, which is 1 subtracted by the pvalue of Z when X = 192.5. 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{192.5 - 190}{3.0822}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B192.5%20-%20190%7D%7B3.0822%7D)
![Z = 0.81](https://tex.z-dn.net/?f=Z%20%3D%200.81)
has a pvalue of 0.7910
1 - 0.7910 = 0.2090
20.90% probability that more than 192 of the confidence intervals cover the true proportions