Answer:
a) P(X≥143)=0
b) This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).
Explanation:
If we use the normal approximation to the binomial distribution we have the following parameters (mean and standard deviation):
![\mu=np=0.68*206=140.1\\\\ \sigma=√(np(1-p))=√(206*0.68*0.32)=√(44.8256)=6.7](https://img.qammunity.org/2021/formulas/mathematics/college/1n5uugnquj8p8kpnv4frbb4tvs296lpl44.png)
Then, we can calculate the probability of X being equal or more than 143 using the z-score:
![z=(X-\mu)/(\sigma/√(n))=(143-140.1)/(6.7/√(206))=(2.9)/(0.4668)=6.2124\\\\\\P(x\geq143)=P(z>6.2124)=0](https://img.qammunity.org/2021/formulas/mathematics/college/fgs3v21rfy9dmr53rgd6349ayyl67114mq.png)
This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).