Question

# 7.1 Regional College increased their admission standards so that they only admit students who have scored...

7.1 Regional College increased their admission standards so that they only admit students who have scored in the 80th percentile or higher (top 20%) of all scores on the SAT. What SAT cutoff will the school use this year? (Recall that for the SAT μ = 500, σ = 100, and the scores are normally distributed)

7.2 Even with the increase in standards, the Registrar at Regional College notices that people who don’t meet the minimum continue to apply for admission. This year, the college randomly sampled 12 of the applicants; they had a mean SAT score of 525. What is the probability that a group of 12 applicants would have a mean SAT score of 525 or lower (assuming those applicants were randomly selected from all SAT takers)? SHOW WORK

7.3 What is the probability that a group of 12 applicants (randomly selected from the distribution of SAT sample scores) would have a mean SAT score that is greater than 525 but below the current admission standard for Regional College? SHOW WORK

7.4 Out of all possible SAT sample means, what is the likelihood of obtaining a sample mean SAT score of 550 or higher given a random sample of 20 SAT takers?

7.5 Dr. Spock has given the same grammar test in his classes for the last 15 years. From it he has created a population of his students’ individual grammar test scores (the scores are normally distributed). After a catastrophic computer failure, he no longer has a record of the grammar test μ or σ. What Dr. Spock does know is how his current class scored on the grammar test. This semester 20 students took the test; they had a mean score of 66 (z = -0.38), and SEM of 2.8. Use this information to figure out what μ and σ are for his population of test scores. SHOW WORK

7.6 Given the sample size 20 and SEM 2.8 from the previous problem, on average, how far away (in points) would we expect the current sample mean to be from Dr. Spock’s population mean?

mean = 500 , s 100

1)
z value at 80% = 0.8416

z = (x -mean)/s
0.8416 = ( x -500)/100
x = 584.1621234

2)
n =12

P(x < 525)

z = (x -mean)/(s/sqrt(n))
= ( 525 -500)/(100/sqrt(12))
= 0.8660

P(x < 525) = P(z <0.8660) = 0.8068

3)

n =12

P(x > 525)

z = (x -mean)/(s/sqrt(n))
= ( 525 -500)/(100/sqrt(12))
= 0.8660

P(x > 525) = P(z > 0.8660) = 0.1932

4)

n = 20
P(x> 550)

z = (x -mean)/(s/sqrt(n))
= ( 550 -500)/(100/sqrt(20))
=2.2361

P(x > 550) = P(z > 2.2361) = 0.0127

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