Comparison of Mean and Median

Statistics
Student Name –
Student ID –
Random Number Generator ( using MATLAB ) :
The numbers u51 , u52 and u53 are 0.0628, 0.7091 and 0.0728 respectively as obtained by
the MATLAB code.
Simulation Problem :
1. Formulation of the model
1.1 Definition of notation :
The various elements of the model have been denoted as follows :
A : Turn on the phone and dial a number ( 6 seconds )
B: Detect a busy signal ( 3 seconds )
C: wait for 5 rings to conclude than no one will answer ( 25 seconds )
D: End the call ( 1 second )
E: Dialed 4 times
F: Customer is using the line ( P = 0.2 )
G: Customer is unavailable ( P = 0.3 )
H: Customer is available and can answer the call within X seconds
1.2 Cumulative distribution function of X :
X is a continuous random variable with the mean value of 12 seconds and the
exponential distribution.
Mean = m = 12

Parameter l = 1/m = 1/12
Cumulative distribution function = 1- e –lx , x >=0
And 0 , else
Hence, CDF = 1- e –x/12 , x >=0
And 0 , else
Expression for inverse :
F (x) = 1- e –x/12 ,
Hence, x = F-1 (u), u = (0,1)
F (x) – 1 = - e –x/12 ,
- F (x) + 1 = e –x/12 ,
-x/12 = ln (- F (x) + 1 )
X = -12 ln (1 – F(u))
Hence , the expression for inverse has been derived.
1.3 Tree diagram for the calling process :
A
Busy Ring
B C
4 times Pick
D
4 times done
F G H
END

2. Data Collection :
An ad – hoc experiment is performed using the phone. The time measurements are as
follows for the various tasks.
TASK -> A B C D H
1 7 2 24 1 8
2 5 3 22 1 10
3 5 3 25 1 12
4 4 2 21 1 14
5 5 1 22 1 6
6 4 2 24 1 8
7 4 1 24 1 10
8 5 2 24 1 9
9 5 2 22 1 11
10 6 2 22 1 12
Average 5 2 23 1 10
The average values found for the elements A, B , C, D and H are 5, 2, 23, 1 and 10 seconds
respectively.
3. Monte – Carlo Simulation Algorithm :
In the given simulation problem, W represents the total time which the representative
spends on calling 1 customer.
There are 2 cases here:
Case 1 : Customer answers the call –
The value of W can be found as follows :

0 unsuccessful calls : W = ( A + B + H ) = ( 6 + 3 + 12(1- e –x/12 )) = (9+12(1- e –x/12 ) )
1 unsuccessful calls : W = ( A + B + C + D ) + ( A + B + H ) = ( 6 + 3 + 25 + 1 ) + ( 6
+ 3 + 12(1- e –x/12 )) = 35 + (9+12(1- e –x/12 ) )
2 unsuccessful calls : W = 2 * ( A + B + C + D ) + ( A + B + H ) = 2 * ( 6 + 3 + 25 +
1 ) + ( 6 + 3 + 12(1- e –x/12 )) = 70 + (9+12(1- e –x/12 ) )
3 unsuccessful calls : W = 3 * ( A + B + C + D ) + ( A + B + H ) = 3 * ( 6 + 3 + 25 +
1 ) + ( 6 + 3 + 12(1- e –x/12 )) = 105 + (9+12(1- e –x/12 ) )
Case 2 : Customer does not answer the call –
The value of W can be found as follows :
W = 4 * ( A + B + C + D ) = 4 * ( 6 + 3 + 25 + 1 ) = 4*35 = 140
4. Simulation :
The process is simulated 1000 times
5. Estimation :
i) Mean -
76.3694
ii) First quartile - 41.8217
Median - 79.9595

Third quartile - 123.3760
iii) Probabilities :
P ( W < = 15 ) = 0.08
P ( W < = 20 ) = 0.24
P ( W < = 30 ) = 0.24
P ( W > 40 ) = 0.76
P ( W > w5 ) = 0.68
P ( W > w6 ) = 0.52
P ( W > w7 ) = 0.48
The values of w5, w6 and w7 have been chosen as 50, 60 and 80 respectively.
6. Analysis :
6.1 Comparison of mean and median : The value of mean is less than the value of
median. It means the data distribution is skewed towards the left.
6.2 Sample space of W:
>> unique(val)
( 10.8422 13.0891 15.3316 16.9384 18.9147 19.3760 49.3036 49.8390
50.7848 51.2018 52.2632 52.5619 79.9595 82.4016 88.0897 89.5058
116.6544 121.5854 123.3224 123.5365 123.7335 140.0000 )
6.3 Graph :

0 5 10 15 20 25
0
20
40
60
80
100
120
140
The graph shows that :
X 15 20 30 40 50 60 80
Y 0.08 0.24 0.24 0.24 0.32 0.48 0.52

10 20 30 40 50 60 70 80
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
The graph shows that W can be an exponential random variable as per the graph.
7. Comments :
i) The step 3 was the most challenging and the step 2 was the least challenging.
ii) The step 1 was the most time consuming and the step 6 was the least time
consuming.
iii) The results of simulation do not contain errors as finally the exponential
distribution is obtained.