Question

The College Board reported the following mean scores for the three parts of the SAT (The World Almanac, 2009):

Critical reading	502
Mathematics	515
Writing	494

Assume that the standard deviation on each part of the test is 100

a) What is the probability that a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical reading part of the test?

b) What is the probability that a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test? Compare this probability to the value computed in part (a).

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	502
std deviation   =σ=	100.000
sample size       =n=	90
std error=σx̅=σ/√n=	10.541

probability that a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical reading part of the test:

probability =P(492<X<512)=P((492-502)/10.541)<Z<(512-502)/10.541)=P(-0.95<Z<0.95)=0.8289-0.1711=0.6578

b)

probability =P(505<X<525)=P((505-515)/10.541)<Z<(525-515)/10.541)=P(-0.95<Z<0.95)=0.8289-0.1711=0.6578

(both values are same)