Answer:
3.04
Explanation:
Mean = 3.34
Standard deviation= 0.4
n = 163
X ~ N(3.3, 0.4)
The cut off gpa = j
P(x>j) = 121/163
P(x<=j) = 42/163 = 0.26
O.26 represents the 26th percentile
We solve further by using Microsoft Excel. We use the function NORMINV on excel to get GPA
X =. NORMINV(0.26, 3.3, 0.4)
X = 3.04 approximately
We conclude that the required gpa is 3.04