A census of persons recovering from lowerextremity fractures

A census of persons recovering from lower-extremity fractures find that during the first six months of recovery they spend a mean of 8 hours per week working, with a standard deviation of 2 hours per week. Assume that the numver of weekly work hours is normally distributed.

(A) What proportion of persons in the first 6 months of recovery from lower-extremity fractures spend at least 10.5 hours per week working?

(B) If We were to Select a random sample of 40 persons in the first 6 months of recovery from lower-extremity fractures what is the probability that their sample mean weekly time spent working will be less than 6.5?

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    10.5      
u = mean =    8      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    1.25      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.25   ) =    0.105649774 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    6.5      
u = mean =    8      
n = sample size =    40      
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -4.74341649      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -4.74341649   ) =    0.00000105072 [ANSWER]