Homework Help: Probability/Statistics

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 (9.41-9.36)/2.28 = .0219 and (6.41-9.36)/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 z-scores are correct
using the z-score 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 z-score 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 z-scores.
Of course it will also work if you use the z-scores, but make sure you set the mean to 0 and your SD to 1

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