Problem 2 The scores of students on the ACT college entrance examination in a re
ID: 3340647 • Letter: P
Question
Problem 2 The scores of students on the ACT college entrance examination in a recent year had the Normal distribution with mean 18.6 and standard deviation 5.9. 1. What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher? 2. Consider 20 randomly selected students who took the test. What is the distribution of the average score of 20 students? What are parameter values (mean and standard deviation) for this distribution? 3. What is the probability that the average score is 21 or higher?Explanation / Answer
Problem 2
(a) Mean scores of students = 18.6
Standard deviation of scores = 5.9
(1) Let say X is the score of a random student
Pr(X >= 21) = Pr(X >= 21; 18.6 ; 5.9) = 1 - (Z)
where is the standard normal cumulative distribution.
And, Z = (21 - 18.6)/ 5.9 = 0.4068
Pr(X >= 21) = Pr(X >= 21; 18.6 ; 5.9) = (Z) = 1 - (0.4068) = 1- 0.6579 = 0.3421
(2) sample size = 20
the distribution of average score of 20 students is approximate normal distribution.
Mean of this distribution = 18.6
standard deviation of this distrbiution = 5.9 / sqrt(20) = 1.32
(3) Let say X is the sample mean of sample size 20
Pr(x >= 21) = Pr(x >= 21; 18.6 ; 1.32) = 1 - (Z)
where is the standard normal cumulative distribution.
And, Z = (21 - 18.6)/ 1.32 = 1.82
Pr(X >= 21) = Pr(X >= 21; 18.6 ; 1.32) = 1 -(Z) = 1- (1.82) = 1 - 0.9655 = 0.0345
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