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

LILCOMBO 2TIME · Accepted Answer

Final Answer : B, Explanation :  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 = 1200Standard deviation = 90/√100 = 9For university Y:Mean = 1215Standard deviation = 110/√100 = 11Thus, we can write:Z = (X̄ - μ) / (σ/√n)For university X:Z = (1190 - 1200) / 9 = -1.11For university Y:Z = (1190 - 1215) / 11 = -2.27Now we need to find the probability that   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) = 0.1335P(Z  < -2.27) = 0.0116Therefore, the probability that  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(  < 1190) = P(Z  < -1.11) + P(Z  < -2.27) = 0.1335 + 0.0116 = 0.1451, which is closest to option B (0.1335).

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