In: Statistics and Probability
Full-time college students report spending a mean of 30 30 hours per week on academic activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 4 4 hours. Complete parts (a) through (d) below. b. If you select a random sample of 25 25 full-time college students, there is an 83 83% chance that the sample mean is less than how many hours per week?
µ =    30      
           
           
σ =    4      
           
           
n=   25          
           
       
proportion=   0.8300      
           
           
          
           
           
Z value at    0.83   =  
0.954   (excel formula =NORMSINV(  
0.83   ) )      
   
z=(x-µ)/(σ/√n)          
           
           
so, X=z * σ/√n +µ=   0.954   *  
4   / √    25   +  
30   =   30.76
If you select a random sample of 25 25 full-time collegestudents, there is an 83 83% chance that the sample mean is less than 30.76 hours per week