Answer: 0.0228
Step-by-step 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. [tex]\mu=95\ \ \&\ \sigma=10[/tex]
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 :-
[tex]P(x>115)=1-P(x\leq115)\\\\=1-P(\dfrac{x-\mu}{\sigma}<\dfrac{115-95}{10})\\\\=1-P(z\leq2)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.9772\ \ [\text{By z-table}]\\\\=0.0228[/tex]
Hence , the proportion of individuals in this population have a score greater than 115 = 0.0228
