Question

The IQs of 600 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviation of 12. Note that IQs are recorded to the nearest integers.

a) If the college requires an IQ of at least 95, how many of these students will be rejected on this basis of IQ, regardless of their other qualifications?

b) What is the minimum IQ of the top 30 applicants?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	115
std deviation   =σ=	12

P(student has a IQ of 95 or less) :

probability =P(X<95)=(Z<(95-115)/12)=P(Z<-1.67)=0.0475

therefore expected rejection =np=600*0.0475 =28.5~ 29 rejections

b)

for top 30 applicants will have percentile =(600-30)*100/600 =95th percentile:

for 95th percentile critical value of z=		1.645
therefore corresponding value=mean+z*std deviation=			134.740 ~ 135