Managing Ashland Multicomm Services
As stated, the data follows a normal distribution. Thus, z-scores together with the normal table can be applied to determine the probabilities. The provided data reveals that the mean is 1.005, while the standard deviation is 0.10.
- (a). We denote the random variable upload speed with X. Thus, the probability that upload speed will be less than 1.0 will be given by P(X<1).
The Z-scores for 1.0 upload speed will be Z=(x-µ)/δ
X=1.0, µ=1.005, δ=0.10
Z=(1.0-1.005)/0.10= -0.05
From the normal table, the corresponding probability for z=-0.05 is o.4801
Therefore, P(X<1.0) is 0.4801
- The probability that upload speed is between 0.95 and 1.0 will be given by P(0.95<X<1.0)
We find the Z-scores for the two values of X: P(x>0.95) and P(x<1.0)
Z1= (0.95-1.005)/0.10= -0.55
Z2=(1.0-1.005)/0.10= -0.05
P(-0.55<z<-0.05)= P(z<-0.05)-P(z>-0.55)= 0.4801-0.29116= 0.18894
Thus, P(0.95<X<1.0)= 0.189
- The probability that upload speed will be between 1.0 and 1.05 will be denoted by P(1.0<X<1.05)
The Z-score for the probabilities will be given by P(x>1) and P(x<1.05)
Z1= (1-1.005)/0.1= -0.05
Z2= (1.05-1.005)/0.1= 0.45
Thus, P(-0.05<z<0.45)= P(z<0.45)-P(z>-0.05)= 0.67364-0.4801= 0.19354
The probability that upload speed is between 1.0 and 1.05 is 0.194
- To find the probability that the upload speed is below 0.95 or above 1.05, we use the formula; P(X<0.95)+P(X>1.05)
Z-scores will be calculated for P(X<0.95) as (0.95-1.005)/0.1= -0.55
The Z-score for P(x>1.05) will be given by (1.05-1.005)/0.1= 0.45
From the normal table, P(z<-0.55)=0.29116
From the normal table, P(z>0.45)= 1-0.67364=0.32636
The probability will be 0.29116+0.32636= 0.61752
- The team aims to reduce the probability that the upload speed will be below 1.0, which is denoted as P(x<1). The team has two options that they want to compare and find the one that results in the smallest probability. The options are to increase the mean from 1.005 to 1.05 or reduce the standard deviation from 0.1 to 0.075.
To determine which options result in the smallest probability, we calculate the Z-scores for each option and find the corresponding probabilities.
The Z-score for the increased mean will be (1-1.05)/0.1= -0.5
The Z-score for the reduced standard deviation is (1-1.005)/0.075= -0.67
We find the probabilities for P(z<-0.5) and P(z<-0.67).
The results from the table show that P(z<-0.5)=0.30854 and P(z<-0.67)=0.25143
The option to improve processes to reduce standard deviation results is the smallest probability than improving the process to increase the mean upload time. From the results, it is recommended that the team should choose a process improvement that reduces the standard deviation.