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:
[tex]z=\frac{x-\mu}{\sigma},\text{ where }\mu=average,\sigma=standard\text{ }deviation,x=value[/tex]
Then, as x=70, we replace the given values and find z:
[tex]z=\frac{70-85}{9}=\frac{-15}{9}=-1.67[/tex]
Then, we can find the estimate as follows:
[tex]P(X\ge70)=P(z\ge-1.67)=1-P(z<-1.67)[/tex]
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):
[tex]P(X\ge70)=1-0.0475=0.9525[/tex]
The probability is 0.9525, if we multiply it by 100%, we obtain: 95.25%
The best estimate is 95%.
