Answer: 0.0228
Explanation:
Given : In a hypothetical study looking at the scores of 100 people in an entrance exam, it was determined that the average score was 95 with a standard deviation of 10.
i.e.

Let x denotes the scores.
We assume that the scores are normally distributed.
Then, the proportion of individuals in this population have a score greater than 115 will be :-
![P(x>115)=1-P(x\leq115)\\\\=1-P((x-\mu)/(\sigma)<(115-95)/(10))\\\\=1-P(z\leq2)\ \ [\because\ z=(x-\mu)/(\sigma)]\\\\=1-0.9772\ \ [\text{By z-table}]\\\\=0.0228](https://img.qammunity.org/2020/formulas/mathematics/high-school/k4vsd1947cmaeiphnz0ffmow9rnb73wqie.png)
Hence , the proportion of individuals in this population have a score greater than 115 = 0.0228