The given information is:
- The average test score is 85 points
- The standard deviation is 9
- It is a normal distribution
We need to find the best estimate for the proportion of students earned a 70 points or higher.
First, we need to find the z-score for 70, by applying the following formula:
![z=(x-\mu)/(\sigma),\text{ where }\mu=average,\sigma=standard\text{ }deviation,x=value](https://img.qammunity.org/2023/formulas/mathematics/college/f05g2gwkpohywthbu5w832qeo42j08xaqm.png)
Then, as x=70, we replace the given values and find z:
![z=(70-85)/(9)=(-15)/(9)=-1.67](https://img.qammunity.org/2023/formulas/mathematics/college/242p5su5q3labbl6dhpcgx5s37iqmgq0oa.png)
Then, we can find the estimate as follows:
![P(X\ge70)=P(z\ge-1.67)=1-P(z<-1.67)](https://img.qammunity.org/2023/formulas/mathematics/college/pl36qobnhtt90ge7njka7m7ov4gl1uyvwd.png)
Now, in a standard normal table, we search the cumulative probability of z<-1.67, and it is 0.0475:
Thus, replace this value to find P(X>=70):
![P(X\ge70)=1-0.0475=0.9525](https://img.qammunity.org/2023/formulas/mathematics/college/8v3ewyv61ags76o7l59fzyd8d1ops9sjqe.png)
The probability is 0.9525, if we multiply it by 100%, we obtain: 95.25%
The best estimate is 95%.