Assume that 1,000 scores are normally distributed with a mean of 72 and a standard deviation of 4. How many scores can we expect to be between 75 and 82?

Question

anonymous · Answer

z-score for 75 = (75-72)/4 = 3/4 = .75
z-score for 82 = (82-72)/4 = 10/4 = 2.5

anonymous · Answer

use a table such as http://training.ce.washington.edu/pgi/Modules/08_specifications_qa/Images/main_pictures/normal_table.gif to find the area between them

anonymous · Answer

actually that table sucks

anonymous · Answer

http://www.math.umt.edu/elias/NormalDistributionTable.jpg

anonymous · Answer

P( .75 < Z < 2.5) = P(Z < 2.5) - P(Z < .75) = .9938 - .7734 = .2204

anonymous · Answer

thansk man