Question 1194410
Population Mean mu=4.91
Population Standard Deviation sigma=0.72
Sample size n=30
We have to find the probability that they will have an average GPA between 4.75 and 4.95.
For a normal distribution z=(x-mu)/(sigma/sqrt(n)) where n is the sample size.
x1=4.75
z(lower)=(x1-mu)/(sigma/sqrt(n))=(4.75-4.91)/(0.72/sqrt(30))​=−1.22
x2=4.95
z(upper)=(x2-mu)/(sigma/sqrt(n))=(4.95-4.91)/(0.72/sqrt(30))​=0.3
The required probability=P(z(lower<z<z(upper))=P(-1.22<z<0.3)=P(z<=0.3)P(z<-1.22)
=0.6195-0.1118=0.5078
Answer:
The required probability=0.5078