Suppose heights of men are normally distributed with a mean 69.0 inches and standard deviation 2.8 inches. What percent of men are over 6 feet tall? Round to the nearest tenth of a percent.
I understand finding the z score but I'm not sure how to find the percent part. Help?
2 answers
Are you using tables, charts, or software?
I assume charts. It's an online class and I'm basically teaching myself.
z = (x -mean)/ sd
z = (72-68)/2
z= 4/2= 2
1-.9772 = .0228
2.28%
I saw this example (I found the same question with different numbers) and it made sense to me until the 1-.9772 = .0228
Where is that coming from?
z = (x -mean)/ sd
z = (72-68)/2
z= 4/2= 2
1-.9772 = .0228
2.28%
I saw this example (I found the same question with different numbers) and it made sense to me until the 1-.9772 = .0228
Where is that coming from?
1 answer