Using diaries for many weeks, a study on the lifestyles of visu- ally impaired s

Question

Using diaries for many weeks, a study on the lifestyles of visu- ally 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 im- paired students averaged 9.63 hours of sleep, with a standard de- viation of 2.23 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 6.8 hours of sleep?

answer:

(b) What is the probability that a visually impaired student gets between 6.9 and 9.8 hours of sleep?

answer:

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

answer: _______ hours

Thank you.

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    6.8      
u = mean =    9.63      
          
s = standard deviation =    2.23      
          
Thus,          
          
z = (x - u) / s =    -1.269058296      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.269058296   ) =    0.102210136 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    6.9      
x2 = upper bound =    9.8      
u = mean =    9.63      
          
s = standard deviation =    2.23      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.224215247      
z2 = upper z score = (x2 - u) / s =    0.076233184      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.11043552      
P(z < z2) =    0.530383209      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.419947689   [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.2      
          
Then, using table or technology,          
          
z =    -0.841621234      
          
As x = u + z * s,          
          
where          
          
u = mean =    9.63      
z = the critical z score =    -0.841621234      
s = standard deviation =    2.23      
          
Then          
          
x = critical value =    7.753184649   [ANSWER]  
  

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