(1 pt) Using diaries for many weeks, a study on the lifestyles of visually impai
ID: 3252229 • Letter: #
Question
(1 pt) 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 8.33 hours of sleep, with a standard deviation of 1.9 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.(b) What is the probability that a visually impaired student gets between 6.7 and 8.38 hours of sleep?
answer: so is it using the formula x(bar)-mean(u)/Standaded deviation, then looking on z-chart for each and subtracting the Z-scores? (1 pt) 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 8.33 hours of sleep, with a standard deviation of 1.9 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.
(b) What is the probability that a visually impaired student gets between 6.7 and 8.38 hours of sleep?
answer: so is it using the formula x(bar)-mean(u)/Standaded deviation, then looking on z-chart for each and subtracting the Z-scores? (1 pt) 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 8.33 hours of sleep, with a standard deviation of 1.9 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.
(b) What is the probability that a visually impaired student gets between 6.7 and 8.38 hours of sleep?
answer: so is it using the formula x(bar)-mean(u)/Standaded deviation, then looking on z-chart for each and subtracting the Z-scores?
Explanation / Answer
P(6.7 < x < 8.38) = ?
P(6.7 < x < 8.38) = P(z < 8.38) - P(z < 6.7)
= P(Z < 8.38 - 8.33 / 1.9) - P( 6.7- 8.33/1.9)
= P(z < 0.0263) - P(z < -0.8579)
= 0.5105 - (1-0.80453)
= 0.31503
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