Answer: A. 12%
Step-by-step explanation:-
Given : In an exam , Mean score : [tex]\mu=78\text{ points}[/tex]
Standard deviation : [tex]6\text{ points}[/tex]
Let X be a random variable that represents the scores of students.
We assume that the points are normally distributed.
Z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = 85, we have
[tex]z=\dfrac{85-78}{6}\approx1.17[/tex]
Then using standard normal distribution table, the probability that the students received more than 85 is given by :-
[tex]P(x>85)=P(z>1.17)=1-P(z<1.17)\\\\=1-0.8789995=0.1210005\approx0.12=12\%[/tex]
Hence, the percentage of students received more than 85 =12%
