Question

Given a test that is normally distributed with a mean of 100 and a standard deviation of 12, find: the probability that a single score drawn at random will be greater than 110

Answer 1

To do this, use your calculator to compute the normalcdf for these statistics. Your lower bound is 110. Your upper bound is 9999 (as there is no true upper bound, it's best to pick this number because of its largness). Your mean and standard deviation are provided. Your equation should look like [normalcdf(110, 9999, 100, 12)].

The answer is .202

Hope this helped.

Answer 2

the probability that a single score drawn at random will be greater than 110 =P(Z>110)
such that Z->N(100,12)
so T=(Z-100)/12->N(0,1) standard normal disstribution
so P(Z>110)=P(T>(110-100)/12)
=P(T>0.83)
= 1-P(T<0.83)
= 0.203