Posted by lindsay on .
A survey was conducted to measure the number of hours per week adults spend on home
computers. In the survey, the number of hours was normally distributed, with a mean of 8
hours and a standard deviation of 1 hour. A survey participant is randomly selected. Find the
probability that the hours spent on the home computer by the participant are between 5.5 and
9.5 hours per week.

statistics 
PsyDAG,
Z = (scoremean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z scores. 
statistics 
Anonymous,
0.3258

statistics 
Anonymous,
Zscores: (9.58)/1=1.5 and (5.58)/1=2.5. Corresponding probabilities: 94.41%
and 0.62%, so the answer is P=94.41%0.62%=93.79%