The Journal of Visual Impairment & Blindness (May-June 1997) published a study of the lifestyles of visually impaired students. Using diaries, the students kept track of several variables, including the number of hours of sleep obtained in a typical day. These visually impaired students had a mean of 9.06 hours and a standard deviation of 2.11 hours. Assume that the distribution of the number of hours of sleep for this group of students is approximately normal. a. Find the probability that a visually impaired student obtains more than 10 hours of sleep on a typical day. (5 points) b. Find the probability that a visually impaired student gets between 7 and 8 hours of sleep on a typical night. (5 points)
Solution :
Given that ,
mean =
= 9.06 hours
standard deviation =
= 2.11 hours
a) P(x > 10) = 1 - p( x< 10)
=1- p P[(x -
) /
< (10 - 9.06) / 2.11]
=1- P(z < 0.45)
Using z table,
= 1 - 0.6736
= 0.3264
b) P( 7 < x < 8) = P[(7 - 9.06 ) / 2.11) < (x -
) /
<
(8 - 9.06) / 2.11 ) ]
= P( -0.98 < z < -0.50)
= P(z < -0.50) - P(z < -0.98)
Using z table,
= 0.3085 - 0.1635
= 0.1450
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