Question

5. A large freight elevator can transport a maximum of 9800 pounds. Suppose a load of cargo containing 49 boxes must be transported via the elevator. Experience has shown that the weight of boxes of this type of cargo follows a distribution with mean u = 205 pounds and standard deviation o = 15 pounds. Based on this information, what is the probability that all 49 boxes can be safely loaded onto the freight elevator and transported? A study involving stress is done on a college campus among the students. The stress scores follow a uniform distribution with the lowest stress score equal to 1 and the highest equal to 5. Using a sample of 75 students, find: a. The probability that the total of 75 stress scores is less than 220. b. The 90th percentile for the total stress score for the 75 students. 7. A study involving stress is done on a college campus among the students. The stress scores follow a uniform distribution with the lowest stress score equal to 1 and the highest equal to 5. Using a sample of 75 students, find: a. The probability that the average stress score for the 75 students is less than 2.8 b. The 90th percentile for the average stress score for the 75 students. 8. The number of hours per week that high school seniors spend on computers is normally distributed, with a mean of 5 hours and a standard deviation of 2 hours. 70 students are chosen at random. Let X represent the mean number of hours spent on the computer for this group. Find the probability that X is between 5.1 and 5.7.

Answer 1

5)

µx = n*µ=49*205 = 10045                                      
σx = √n * σ = √49 * 15 = 105                                      
  
X =   9800                                      
                                          
Z =   (X - µx )/(σx) = (   9800   -   10045.00   ) / (   105.000   ) =   -2.33  
                                          
P(X ≤   9800   ) = P(Z ≤   -2.333   ) =   0.0098                       (answer)

8)

µ =    5                                      
σ =    2                                      
n=   70                                      
we need to calculate probability for ,                                          
5.1   ≤ X ≤    5.7                                  
X1 =    5.1   ,    X2 =   5.7                          
                                          
Z1 =   (X1 - µ )/(σ/√n) = (   5.1   -   5   ) / (   2   / √   70   ) =   0.42  
Z2 =   (X2 - µ )/(σ/√n) = (   5.7   -   5   ) / (   2   / √   70   ) =   2.93  
                                          
P (   5.1   < X <    5.7   ) =    P (    0.42   < Z <    2.93   )       
                                          
= P ( Z <    2.93   ) - P ( Z <   0.42   ) =    0.998   -    0.662   =    0.3361       (answer)