STA 301 _Assignment 1_Spring_2012

Question # 1
Frequency Distribution:

Class Limit    Frequency ( f )    Class Boundaries    Class-Mark(Midpoint) X      
20 – 24    1    19.5 – 24.5    22      
25 – 29    4    24.5 – 29.5    27      
30 – 34    9    29.5 – 34.5    32      
35 – 39    10    34.5 – 39.5    37      
40 – 44    15    39.5 – 44.5    42      
45 – 49    9    44.5 – 49.5    47      
50 – 54    2    49.5 – 54.5    52      
Total    50           

Question # 2 (i)
     _
Mean ( X )

Classes    Frequency ( f )    Class Mark (Midpoint) X    fX      
0.7312 – 0.7313    10    0.73125    7.3125      
0.7314 – 0.7315    15    0.73145    10.97175      
0.7316 – 0.7317    20    0.73165    14.633      
0.7318 – 0.7319    25    0.73185    18.29625      
0.7320 – 0.7321    30    0.73205    21.9615      
0.7322 – 0.7323    8    0.73225    5.858      
0.7324 – 0.7325    2    0.73245    1.4649      
Total    110        80.4979   

Applying Formula   
    _      ∑fx
    X = -------
            ∑f
        = 80.4979 / 110
        = 0.7317991
Question 2
Part ii     ^
Mode ( X )

     Classes    Frequency ( f )    Class Boundaries ( C.B)      
0.7312 – 0.7313    10    0.73115 – 0.73135      
0.7314 – 0.7315    15    0.73135 – 0.73155      
0.7316 – 0.7317    20    0.73155 – 0.73175      
0.7318 – 0.7319    25    0.73175 – 0.73195      
0.7320 – 0.7321    30    0.73195 – 0.73215      
0.7322 – 0.7323    8    0.73215 – 0.73235      
0.7324 – 0.7325    2    0.73235 – 0.73255      
Total    110       
Applying Formula
       ^        ( fm – fi )
Mode =  X = L + -------------------------- * h
           ( fm – fi ) + ( fm – f2)

                            (30 - 25)
            = 0.73195 +--------------------------- * 0.0002
              (30 – 25) + (30 – 8)
                5
         = 0.73195 + ----------------- *0.0002
                  (5) + (22)
                  5
         = 0.73195 + ----------- * 0.0002
                  27
        = 0.73195 + 0.1851852 * 0.0002
        = 0.73195 + 0.000037
        = 0.731987

Working:
Class Mark is obtained by adding the sum of two limits of the and dividing by 2.
e.g. 0.7312 + 0.7313 /2 = 0.73125
fX is obtained X multiplying with (Frequency) f.
e.g. 10*0.73125 =7.3125
The Mode lies somewhere between 0.73195 and 0.73215.
By applying formula we noted that
fm = 30, fi = 25, f2 = 8

CS601 _ Assignment 1 Spring 2012 Solution

Q: 1
Solution:

STAR would be more appropriate as it is good with multiple computers connected in a big network plus it can also perform troubleshooting and new devices can be easily added in this network at any time.
According to me STAR would be more suitable as it is good with many computers connected in a network plus can also perform troubleshooting and new devices can be easily added in the network.
Q.2
Solution:

The physical describe of devices on a network, each and every LAN has a topology or the mode that the devices on a network are arranged and how they communicate with each other. They want that the workstations are linked to the network through the real cables that transmit data. The physical structure of the network is called the physical topology. The logical topology, is contrast, is the way that the signals act on the network media. On the way that the data passes through the network from one device to the another without regard to the physical interconnection of the devices.

      Because multipoint topologies share an ordinary channel, each device needs a way to identify itself and the device to which it wants to send information. The method used to categorize senders and receivers is called addressing four types of physical topologies are often used in computer networking:

    1.Star Topology
    2.Bus Topology
    3.Ring Topology
    4.Hybrid Topology
Q.3
Solution:
Following devices can be shared on network.

Scanner
CD-ROM drives
Light pen
Printer