Answer:
81.86%
Step-by-step explanation:
We are given that the mean GPA of students in a course at UC Davis is 3.2 with a standard deviation of 0.3.
Assuming that the data follows normal distribution.
Let X = GPA of students in a course at UC Davis
So, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean GPA = 3.2
= standard deviation = 0.3
Now, the probability that the students in the course have a GPA between 2.9 and 3.8 is given by = P(2.9 < X < 3.8)
P(2.9 < X < 3.8) = P(X < 3.8) - P(X 2.9)
P(X < 3.8) = P( < ) = P(Z < 2) = 0.97725
P(X 2.9) = P( ) = P(Z -1) = 1 - P(Z < 1)
= 1 - 0.84134 = 0.15866
The above probability is calculated by looking at the value of x = 2 and x = 1 in the z table which has an area of 0.97725 and 0.84134 respectively.
Therefore, P(2.9 < X < 3.8) = 0.97725 - 0.15866 = 0.8186
Hence, 81.86% of students in the course have a GPA between 2.9 and 3.8.
