Answer:
Step-by-step explanation:
Given that Student scores on exams given by a certain instruc-tor have mean 74 and standard deviation 14.
Group I X Group II Y
Sample mean 74 74
n 25 64
Std error (14/sqrtn) 2.8 1.75
a) P(X>80) =[tex]1-0.9839\\= 0.0161[/tex]
b) P(Y>80) = [tex]1-0.9997\\=0.0003[/tex]
c) X-Y is Normal with mean = 0 and std deviation = [tex]\sqrt{2.8^2+1.75^2} \\=3.302[/tex]
P(X-Y>2.2) = [tex]1-0.8411\\=0.1589[/tex]
d) [tex]P(\bar x -\bar Y>2.2) = 0.1589[/tex]