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.
Z = (score-mean)/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.
Z-scores: (9.5-8)/1=1.5 and (5.5-8)/1=-2.5. Corresponding probabilities: 94.41%
and 0.62%, so the answer is P=94.41%-0.62%=93.79%