Answer:
a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.
b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.
Step-by-step explanation:
This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.
The parameters for the normal distribution will be
![\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353](https://tex.z-dn.net/?f=%5Cmu%3Dp%3D0.5%5C%5C%5C%5C%5Csigma%3D%5Csqrt%7Bp%281-p%29%2Fn%7D%20%3D%5Csqrt%7B0.5%2A0.5%2F200%7D%3D%200.0353)
We can calculate the z values for x1=0.47 and x2=0.51:
![z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28](https://tex.z-dn.net/?f=z_1%3D%5Cfrac%7Bx_1-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B0.47-0.5%7D%7B0.0353%7D%3D-0.85%5C%5C%5C%5Cz_2%3D%5Cfrac%7Bx_2-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B0.51-0.5%7D%7B0.0353%7D%3D0.28)
We can now calculate the probabilities:
![P(0.47](https://tex.z-dn.net/?f=P%280.47%3Cp%3C0.51%29%3DP%28-0.85%3Cz%3C0.28%29%3DP%28z%3C0.28%29-P%28z%3C-0.85%29%5C%5C%5C%5CP%28z%3C0.28%29-P%28z%3C-0.85%29%3D0.61026-0.19766%3D0.41260)
If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.
b) If the sample size change, the standard deviation of the normal distribution changes:
![\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05](https://tex.z-dn.net/?f=%5Cmu%3Dp%3D0.5%5C%5C%5C%5C%5Csigma%3D%5Csqrt%7Bp%281-p%29%2Fn%7D%20%3D%5Csqrt%7B0.5%2A0.5%2F100%7D%3D%200.05)
We can calculate the z values for x1=0.47 and x2=0.51:
![z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2](https://tex.z-dn.net/?f=z_1%3D%5Cfrac%7Bx_1-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B0.47-0.5%7D%7B0.05%7D%3D-0.6%5C%5C%5C%5Cz_2%3D%5Cfrac%7Bx_2-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B0.51-0.5%7D%7B0.05%7D%3D0.2)
We can now calculate the probabilities:
![P(0.47](https://tex.z-dn.net/?f=P%280.47%3Cp%3C0.51%29%3DP%28-0.6%3Cz%3C0.2%29%3DP%28z%3C0.2%29-P%28z%3C-0.6%29%5C%5C%5C%5CP%28z%3C0.2%29-P%28z%3C-0.6%29%3D0.57926-0.27425%3D0.30501)
If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.