Answer:
50% of students have a GPA higher than 3.1.
Step-by-step explanation:
We are given that the mean GPA of students in a neighboring school is 3.1 with a standard deviation of 0.3.
Assuming that the data follows normal distribution.
Let X = GPA of students in a neighboring school
So, X ~ Normal([tex]\mu=3.1,\sigma^{2} =0.3^{2}[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean GPA = 3.1
[tex]\sigma[/tex] = standard deviation = 0.3
Now, the probability that the students have a GPA higher than 3.1 is given by = P(X > 3.1)
P(X > 3.1) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{3.1-3.1}{0.3}[/tex] ) = P(Z > 0) = 0.50
The above probability is calculated by looking at the value of x = 0 in the z table which has an area of 0.50.
Therefore, 50% of students have a GPA higher than 3.1.
