in a certain region, the ages of licensed drivers is normally distributed with a mean of 44.5 years and a standard deviation of 9,1 years. Find the percentage that a randomly selected driver is younger than 25
How to make 44.5 years and a standard deviation of 9,1 years into a standard normal? Do you have the formula?
I dont know the formula...that is the problem. I have all these notes and i am so lost..
Okay z = (x - mean)/ standard deviation I would start a new page of notes and keep this formula in a safe place.
i tried that but that doesnt give me the right answer
the answers are a 1.6% b 2.7% c 4.9% d 88.4%
z = (25 - 44.5)/9 = -2.17 What do we do next?
I have no idea :/
Hmm. Are you okay with the formula? Did you get the same answer?
yes, well i got - 2.14 i did 25-44.5 seperate then divided by 9.1
Ohh, 9,1 meant 9.1?
Do you have your z-table with all of those numbers in it?
i never got a z table..i dont even know what that means
i was absent 3 days i had to go to Ohio for an emergency
Ohh, hope everything is okay. Google has a z-table if you missed school on that day https://www.google.com/search?q=z-table&hl=en&prmd=imvns&tbm=isch&tbo=u&source=univ&sa=X&ei=PCoCT9zSPKaciQLswMDQDg&sqi=2&ved=0CCkQsAQ&biw=1024&bih=613
thank you..i dont understand it at all..but thanks anyways
did you look at one of the charts? We need to find 2.14 using the outside of the chart?
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