1)
Let X denote IQ of an individual. Then
a)
i.e. 0.57%
b)
i.e. 0.043%
Required number of students = 0.00043(12000) = 5.16 ~ 5
c)
i.e. 2.275%
d)
We want to find ‘x’ such that
From Normal probability tables, we get
which is the required IQ.
Q2.
a) z = (3 – 5.2)/1.3
z = – 0.733
P = 0.2327
b) z = (2 – 5.2)/1.3
z = – 2.46
P = 0.006947
Get back = 150000 * 0.006947
Get back = about 1042
c) z score for lowest 10% = – 1.28
(x – 5.2)/1.3 = – 1.28
x = 3.536