a. P(492 < x-bar < 512)
z = (492-502)/100/√90
z = -0.95 is 0.1711
z = (512-502)/100/√90
z = 0.95 is 0.8289
b. P(505 < x-bar < 525)
z = (505-515)/100/ √90
z = -0.95
z = (525-515)/100/ √90
z = 0.95
P(-0.95< z < 0.95) = 0.6578
P(-0.95< z < 0.95) = 0.6578
c. P(484 < x-bar < 504)
z = (484-494)/100/√100
z = -1 is 0.1587
z = (504-494)/100/√100
z = 1 is 0.8413
P(-1< z <1) = 0.6826
The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009):
Assume that the population standard deviation on each part of the test is = 100.
a. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test (to 4 decimals)?
b. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test (to 4 decimals)?
c. What is the probability a sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test (to 4 decimals)?
3 answers
a. P(492 < x-bar < 512)
z = (492-502)/100/√90
z = -0.95 is 0.1711
z = (512-502)/100/√90
z = 0.95 is 0.8289
P(-0.95< z < 0.95) = 0.6578
b. P(505 < x-bar < 525)
z = (505-515)/100/ √90
z = -0.95
z = (525-515)/100/ √90
z = 0.95
P(-0.95< z < 0.95) = 0.6578
c. P(484 < x-bar < 504)
z = (484-494)/100/√100
z = -1 is 0.1587
z = (504-494)/100/√100
z = 1 is 0.8413
P(-1< z <1) = 0.6826
z = (492-502)/100/√90
z = -0.95 is 0.1711
z = (512-502)/100/√90
z = 0.95 is 0.8289
P(-0.95< z < 0.95) = 0.6578
b. P(505 < x-bar < 525)
z = (505-515)/100/ √90
z = -0.95
z = (525-515)/100/ √90
z = 0.95
P(-0.95< z < 0.95) = 0.6578
c. P(484 < x-bar < 504)
z = (484-494)/100/√100
z = -1 is 0.1587
z = (504-494)/100/√100
z = 1 is 0.8413
P(-1< z <1) = 0.6826
Yall are wrong