Question

Based on the analysis of a company's records, a material control manager estimates that the average and standard deviation of the "delivery time" required when buying a small valve are 8 and 1.5 days, respectively. He does not know the distribution of delivery time, but he is willing to assume that the estimates of the mean and the standard deviation are correct. 
Calculate:

1-How many units of standard deviations would be separated from the mean if x =11?

2-How many days below the average should you wait for the probability of the first valve to arrive?

3-Determine a time interval such that the probability of being received during that time is at least 8/9.

Answer 1

    Given    μ = 8                          
   Let    σ = 1.5                          
                                  
1)     X = 11                               
   => it is (11-8 = 3) days away from the mean                              
   3 is 2 times 1.5 that is 2 times the standard deviation                              
   Answer :                              
   X=11 is separated   2   units of standard deviation from the mean                              
                                  
                                  
2)     On an average the valves arrive at 8 days delivery time                              
   P(Valve arrives before 8 days)                              
   = P(X < 8)                              
   Since 8 is the mean, P(X < 8) = 0.5                              
                                  
                                  
                                  
3)     Let X1 and X2 be the limits of the required interval                              
   Then P(X1 < X < X2) ≥ 8/9 = 0.8889                              
   P(X1 < X < X2) = P(X < X2) - P(X < X1) ≥ 0.8889                              
   There can be many such time intervals                              
   From Empirical Rule we know 95% of the values lie between 2 standard deviations about the mean                              
   2 standard deviations = 2*1.5 = 3                              
   8-3 = 5                              
   8 + 3 =11                              
   P(5 < X < 11) = 0.95 ≥ 8/9                              
   Interval is 5 to 11 days