Let X represent the SAT total score of an entering freshman at University X.The random variable X is known to have a N(1200,90)distribution.Let Y represent the SAT total score of an entering freshman at University Y.The random variable Y is known to have a N(1215,110)distribution.A random sample of 100 freshmen is obtained from each university.Let = the sample mean of the 100 scores from University X,and = the sample mean of the 100 scores from University Y.What is the probability that will be less than 1190?
A)0.0116
B)0.1335
C)0.4090
D)0.4562

Question

Let X represent the SAT total score of an entering freshman at University X.The random variable X is known to have a N(1200,90)distribution.Let Y represent the SAT total score of an entering freshman at University Y.The random variable Y is known to have a N(1215,110)distribution.A random sample of 100 freshmen is obtained from each university.Let  = the sample mean of the 100 scores from University X,and  = the sample mean of the 100 scores from University Y.What is the probability that  will be less than 1190?
A)0.0116 
B)0.1335 
C)0.4090 
D)0.4562

Quizplus · Accepted Answer

We use the central limit theorem to approximate the sample mean distributions by normal distributions. The mean and standard deviation of the sample mean distribution are:
For university X:
Mean = 1200
Standard deviation = 90/√100 = 9
For university Y:
Mean = 1215
Standard deviation = 110/√100 = 11
Thus, we can write:
Z = (X̄ - μ) / (σ/√n)
For university X:
Z = (1190 - 1200) / 9 = -1.11
For university Y:
Z = (1190 - 1215) / 11 = -2.27
Now we need to find the probability that 11eab14b_c866_43be_9166_25aac393529a_TB5065_11 will be less than 1190, which means we need to find P(Z < -1.11) and P(Z < -2.27). Using a standard normal distribution table or calculator, we find that:
P(Z < -1.11) = 0.1335
P(Z < -2.27) = 0.0116
Therefore, the probability that 11eab14b_c866_43be_9166_25aac393529a_TB5065_11 will be less than 1190 is the probability that the sample mean from university X will be less than 1190 or the sample mean from university Y will be less than 1190, which is equal to the sum of the probabilities we just calculated:
P(11eab14b_c866_43be_9166_25aac393529a_TB5065_11 < 1190) = P(Z < -1.11) + P(Z < -2.27) = 0.1335 + 0.0116 = 0.1451, which is closest to option B (0.1335).

Let X Represent the SAT Total Score of an Entering