Posted by Amy on Saturday, July 11, 2009 at 2:54pm.
Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 9.36 hours of sleep, with a standard deviation of 2.28 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.
What is the probability that a visually impaired student gets between 6.4 and 9.41 hours of sleep?
**I took (9.419.36)/2.28 = .0219 and (6.419.36)/2.28 = 1.2982. I found their corresponding values in the normal distribution table, and I subtracted the 6.4 value from the 9.41 value. It's not right, and I don't know why.
Forty percent of students get less than how many hours of sleep on a typical day?
**I'm not really sure how to set this one up. If there's some sort of formula that I could use, that'd be great.
Thanks :)

Probability/Statistics  Reiny, Saturday, July 11, 2009 at 3:18pm
your zscores are correct
using the zscore of .0219 you should have found a value of .5087 in your tables, telling you that 50.87% would have less than 9.41 hours of sleep
your zscore of 1.2982 should have given you a value of .09711 telling you that 9.71% would have less than 6.4 hours of sleep.
so the "betwee" percentage would be
50.87%  9.71% = 41.16%
Instead of a table I use the following website
http://davidmlane.com/hyperstat/z_table.html
set the output for "between" and just enter the values, you don't even have to find the zscores.
Of course it will also work if you use the zscores, but make sure you set the mean to 0 and your SD to 1
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