Please show work A journal published a study of the lifestyl

Please show work

A journal published a study of the lifestyles of visually impaired students. Using diaries, the students kept track of several variables, including number of hours of sleep obtained in a typical day. These visually impaired students had a mean of 8.29 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. Complete parts a and b below. a. Find the probability that a visually impaired student obtains less than 6 hours of sleep on a typical day. P(x

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    6      
u = mean =    8.29      
n = sample size =    1      
s = standard deviation =    2.11      
          
Thus,          
          
z =    -1.085308057      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -1.085308057   ) =    0.1389 [ANSWER]

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We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    8      
x2 = upper bound =    10      
u = mean =    8.29      
n = sample size =    1      
s = standard deviation =    2.11      
          
Thus, the two z scores are          
          
z1 = lower z score =    -0.137440758      
z2 = upper z score =    0.81042654      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.445341208      
P(z < z2) =    0.791152465      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.3458   [ANSWER]