In: Statistics and Probability
Suppose that the IQs of university A's students can be described by a normal model with mean 140 and standard deviation 8 points. Also suppose that IQs of students from university B can be described by a normal model with mean 120 and standard deviation 11.
a) Select a student at random from university A. Find the probability that the student's IQ is at least 130 points. The probability is nothing. (Round to three decimal places as needed.)
b) Select a student at random from each school. Find the probability that the university A student's IQ is at least 10 points higher than the university B student's IQ. The probability is nothing. (Round to three decimal places as needed.)
c) Select 3 university B students at random. Find the probability that this group's average IQ is at least 125 points. The probability is nothing. (Round to three decimal places as needed.)
d) Also select 3 university A students at random. What's the probability that their average IQ is at least 10 points higher than the average for the 3 university B students? The probability is nothing. (Round to three decimal places as needed.)
We are given the distributions for Student A and student B here as:
a) The probability that the student A's IQ is at least 130 is
computed here as:
P(A >= 130)
Converting it to a standard normal variable, we have here:
Getting it from the standard normal tables, we have here:
Therefore 0.8944 is the required probability here.
b) As both A and B are normal variables, therefore any linear combination of the two scores would also be a normal variable. The distribution of A - B is obtained here as:
The probability that the university A student's IQ is at least 10 points higher than the university B student's IQ is computed here as:
P(A - B >= 10)
Converting it to a standard normal variable, we have here:
Getting it from the standard normal tables, we have here:
Therefore 0.7689 is the required probability here.
c) Probability that the average IQ score of the three B students selected is at least 125 points is computed here as:
Converting it to a standard normal variable, we have here:
Getting it from the standard normal tables, we have here:
Therefore 0.2156 is the required probability here.
d) The distribution of the sample means for A and B here is given as:
The probability required here is:
Converting it to a standard normal variable, we have here:
Getting it from the standard normal tables, we have here:
Therefore 0.8986 is the required probability here.