Chapter 5: Problem 12

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 they obtained on a typical day. Researchers found that visually impaired students averaged 9.04 hours of sleep, with a standard deviation of 2.67 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.

(a) What is the probability that a visually impaired student gets less than 6.9 hours of sleep?

(b) What is the probability that a visually impaired student gets between 6.2 and 8.43 hours of sleep?

(c) Twenty percent of students get less than how many hours of sleep on a typical day?

Answer: hours

This question relates to topics within probability and normal distribution in mathematics. Probabilities of certain sleep durations for visually impaired students are calculated through z-score determination, given the mean and standard deviation for the students' sleep times.

Explanation:

The question is related to probability and normal distribution. In this situation, we are dealing with the mean (μ) of 9.04 hours of sleep and the standard deviation (σ) of 2.67 hours for those visually impaired students' sleep times.

(a) To find the probability that a visually impaired student gets less than 6.9 hours of sleep, you first need to calculate the z-score: Z = (X - μ) / σ = (6.9 - 9.04) / 2.67 = -0.80. Looking this z-score up in a z-table gives a probability of approximately 0.2119, or 21.19%.

(b) For the probability that a student sleeps between 6.2 and 8.43 hours, find the z-scores for both values then subtract the smaller cumulative probability from the larger one.

(c) To find the number of hours of sleep that less than 20% of students attain, you would look up the z-score that corresponds with 20% (which is approximately -0.84) in the z-table and then use that z-score in the equation X = μ + Zσ to solve for X.

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