Complete Question
The complete question is shown on the first uploaded image
Answer:
The mean of p is ![\mu = 0.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.5)
The standard deviation is ![\sigma = 0.025](https://tex.z-dn.net/?f=%5Csigma%20%3D%200.025)
The probability is ![P(p \ge 0.52) = 0.2119](https://tex.z-dn.net/?f=P%28p%20%5Cge%200.52%29%20%3D%200.2119)
Step-by-step explanation:
From the question
The proportion of those with IQ over 100 is p
The selected sample is ![n =400](https://tex.z-dn.net/?f=n%20%3D400)
From the question the mean is of p is
![\mu = 0.50](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.50)
The standard deviation is
![\sigma = \sqrt{\frac{p(1-p)}{n} }](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D)
![= \sqrt{\frac{0.5 *0.5}{400} }](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7B%5Cfrac%7B0.5%20%2A0.5%7D%7B400%7D%20%7D)
![= 0.025](https://tex.z-dn.net/?f=%3D%200.025)
We are asked to determine
![P(\r p \ge 0.52)](https://tex.z-dn.net/?f=P%28%5Cr%20p%20%5Cge%200.52%29)
By applying normal approximation
→
Where ![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{0.52-0.5}{0.025}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.52-0.5%7D%7B0.025%7D)
= 0.6
So the the probability is
![P(Z \ge 0.8)](https://tex.z-dn.net/?f=P%28Z%20%5Cge%200.8%29)
We can also look at this probability in this manner
![P(Z \ge 0.8) = 1- P(Z < 0.8)](https://tex.z-dn.net/?f=P%28Z%20%5Cge%200.8%29%20%3D%201-%20P%28Z%20%3C%200.8%29)
From the z table z value of 0.8 is 0.78814
![P(Z \ge 0.6)=1- 0.78814](https://tex.z-dn.net/?f=P%28Z%20%5Cge%200.6%29%3D1-%200.78814)
![= 0.2119](https://tex.z-dn.net/?f=%3D%200.2119)