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CS501 Assignment No. 4

Wednesday, June 15, 2011 Posted In Edit This
Assignment No. 04
Semester Spring 2011
Advance Computer Architecture - CS501


Total Marks: 15
Due Date: 19/06/2011


Objective:
To learn and understand basic concepts of I/O subsystems, and DMA
Instructions:
Please read the following instructions carefully before solving & submitting assignment:
Assignment should be in your own wordings not copied from net, handouts or books.
It should be clear that your assignment will not get any credit (zero marks) if:

  • The assignment is submitted after due date.
  • The submitted assignment does not open or file corrupt.
  • The assignment is copied (from other student or copy from handouts or internet).
  • Student ID is not mentioned in the assignment File or name of file is other than student ID.
Note: You have to provide solution with all the calculations and formulations involved, else no marks will be awarded.

For any query about the assignment, contact at cs501@vu.edu.pk

Q No. 1  Marks 10
An I/O device transfers data at a rate of 5MB/s over a 100MB/s bus. The data is transferred in 2KB blocks. If the processor operates at 500MHz, and it takes a total of 5000 cycles to handle each DMA request, find the fraction of CPU time handling the data transfer with and without DMA.

Q. No. 2 Marks 5
A 4 K byte block of data is read from a disk drive. What is the overall data transmission rate if the disk drive has a latency of 7 ms, and a burst bandwidth of 2 MB per second? 

CS408 - Human Computer Interaction QUIZ No.03 File

Wednesday, June 15, 2011 Posted In Edit This
CS408 - Human Computer Interaction QUIZ No.03 File

Question # 1 of 10 ( Start time: 04:26:12 PM ) Total Marks: 1
_______ and _________ are the same things
Select correct option:
Excise and Navigation
Excise and Testing
Excise and Evaluation
All of the given

Quiz Start Time: 04:26 PM Time Left 89
sec(s)

Question # 2 of 10 ( Start time: 04:27:25 PM ) Total Marks: 1
Scenario content and context are derived from information gathered during the
____________ phase and analyzed during the ______________ phase (respectively)
Select correct option:
Research, modeling
Modeling, implementation
Research, implementation
Modeling,Research

Question # 3 of 10 ( Start time: 04:28:21 PM ) Total Marks: 1
Usability testing works for _____________.
Select correct option:
Software products
Hardware products
All products
None of the given

Question # 4 of 10 ( Start time: 04:29:09 PM ) Total Marks: 1
What do you enjoy most about your job (or lifestyle) is an example of _________.
Select correct option:
Avoidance
Motivation
Exceptions
Attitude-oriented questions

Question # 5 of 10 ( Start time: 04:29:46 PM ) Total Marks: 1
________ represents the Early-phase of ethnographic interviews.
Select correct option:
Clarify user roles and behaviors.
Confirm patterns of use.
Clarifying questions.
Focused on domain knowledge.

Question # 6 of 10 ( Start time: 04:31:11 PM ) Total Marks: 1
_____________ are perhaps the least-documented patterns, but they are nonetheless in
widespread use.
Select correct option:
Postural
Structural
Behavioral
Mnemonic

Question # 7 of 10 ( Start time: 04:31:56 PM ) Total Marks: 1
There can only be one _________ persona per interface for a product
Select correct option:
Primary
Secondary
Supplemental
Customer

Question # 8 of 10 ( Start time: 04:32:40 PM ) Total Marks: 1
____________ capture the non-verbal dialogue between artifact and user over time.
Select correct option:
Persona
Scenario
Dialogue
Design model

Question # 9 of 10 ( Start time: 04:33:20 PM ) Total Marks: 1
Exploring how children talk together in order to see if an innovative new groupware
product would help them to be more engaged would probably be better informed by a
______________.
Select correct option:
Usability testing
Field study
Predictive evaluation
DECIDE framework

Question # 10 of 10 ( Start time: 04:34:00 PM ) Total Marks: 1
____________ is particularly useful early in design. It is excellent technique to use with
the prototype, because it provides a wealth of diagnostic information.
Select correct option:
Co-discovery
Active intervention
Splendid research
None of the given
Quiz Start Time: 04:35 PM Time Left 89
sec(s)

Question # 1 of 10 ( Start time: 04:35:42 PM ) Total Marks: 1
Number of keystrokes is the type of ____________ work.
Select correct option:
Logical
Mnemonic
Physical
Structural

Question # 2 of 10 ( Start time: 04:36:46 PM ) Total Marks: 1
What are the most common things you do with the product is a type of ___________.
Select correct option:
Goal-oriented question.
System-oriented question.
Workflow-oriented question.
Attitude-oriented question.

Question # 3 of 10 ( Start time: 04:37:24 PM ) Total Marks: 1
Desktop applications fit into _____________ categories of posture.
Select correct option:
Two
Four
Five
Three

Question # 4 of 10 ( Start time: 04:37:58 PM ) Total Marks: 1
Evaluations done during design to check that product continues to meet users’ needs are
known as _____________ evaluation.
Select correct option:
Formative
Summative
Relative
None of the given

Question # 5 of 10 ( Start time: 04:38:32 PM ) Total Marks: 1
Goal-oriented context scenarios are _____________ task-oriented than key path scenario
Select correct option:
Less
Alike
More
None of the given

Question # 6 of 10 ( Start time: 04:39:55 PM ) Total Marks: 1
____________ patterns can be applied at the conceptual level.
Select correct option:
Postural
Structural
Behavioral
Mnemonic

Question # 7 of 10 ( Start time: 04:40:14 PM ) Total Marks: 1
____________ represent the user’s expectations of the tangible outcomes of using a
specific product.
Select correct option:
Non-user goals
End goals
Experience goals
Life goals

Question # 8 of 10 ( Start time: 04:40:47 PM ) Total Marks: 1
Usability testing works for _____________.
Select correct option:
Software products
Hardware products
All products
None of the given

Question # 9 of 10 ( Start time: 04:41:20 PM ) Total Marks: 1
There can only be one _________ persona per interface for a product
Select correct option:
Primary
Secondary
Supplemental
Customer

Question # 10 of 10 ( Start time: 04:41:49 PM ) Total Marks: 1
If the user requires access via a navigational portal relatively infrequently, the appropriate
posture is ______________.
Select correct option:
Sovereign
Transient
Temporary
None of the given
------------------------------------------------- --------
Question # 1 of 10 ( Start time: 05:10:07 PM ) Total Marks: 1
What are the most common things you do with the product is a type of ___________.
Select correct option:
Goal-oriented question.
System-oriented question.
Workflow-oriented question.
Attitude-oriented question.
Quiz Start Time: 05:10 PM
Time Left 85
sec(s)

Question # 2 of 10 ( Start time: 05:11:37 PM ) Total Marks: 1
____________ is particularly useful early in design. It is excellent technique to use with the
prototype, because it provides a wealth of diagnostic information.
Select correct option:
Co-discovery
Active intervention
Splendid research
None of the given

Question # 3 of 10 ( Start time: 05:12:25 PM ) Total Marks: 1
User personas that are not primary or secondary are ____________ personas.
Select correct option:
Served
Negative
Customer
Supplemental

Question # 4 of 10 ( Start time: 05:13:09 PM ) Total Marks: 1
The _____________ phase sets the stage for the core of the design effort
Select correct option:
Requirement definition
Modeling
Implementation
None of the given

Question # 5 of 10 ( Start time: 05:14:12 PM ) Total Marks: 1
If the product is new then _____________ time is usually invested in market research.
Select correct option:
Less
Less
More
None of the given (verify yourself )

Question # 6 of 10 ( Start time: 05:15:28 PM ) Total Marks: 1
In DECIDE frame work _______________ comes just after choosing the evaluation paradigm
and techniques.
Select correct option:
Deciding about ethical issues
Exploring the questions
Identifying the practical issues
None of the given (verify yourself )

Question # 7 of 10 ( Start time: 05:16:55 PM ) Total Marks: 1
Evaluations done during design to check that product continues to meet users’ needs are known
as _____________ evaluation.
Select correct option:
Formative
Summative
Relative
None of the given (verify yourself )
Quiz Start Time: 05:10 PM
Time Left 88
sec(s)

Question # 8 of 10 ( Start time: 05:17:50 PM ) Total Marks: 1
Which of the following is least likely to be revealed by a paper prototype?
Select correct option:
Your users don’t know the term algorithm
Toolbar buttons are too small to press.
The Help menu isn’t in the right place.
None of the given
Quiz Start Time: 05:10 PM
Time Left 88
sec(s)

Question # 9 of 10 ( Start time: 05:19:06 PM ) Total Marks: 1
_______ and _________ are the same things
Select correct option:
Excise and Navigation
Excise and Testing
Excise and Evaluation
All of the given (verify yourself )
Quiz Start Time: 05:10 PM
Time Left 87
sec(s)

Question # 10 of 10 ( Start time: 05:20:31 PM ) Total Marks: 1
_____________ is needed to check that users can use the product and like it.
Select correct option:
Coding
Evaluation
Guideline
None of the given

Mgmt630 Assignment No. 2

Wednesday, June 15, 2011 Posted In Edit This

Semester “Spring 2011”
“Knowledge Management (MGMT630)”
Assignment No. 02 Marks: 15

“Assignment Title”
U“ABC (pvt) limited, a manufacturing organization, wants to check its return on knowledge of its new CEO. It takes 250 hours for the new CEO to learn the processes with 60 hours spent on learning the manufacturing process, 110 hours spent on learning the packaging process and 80 hours spent on learning the marketing process. The total cost of manufacturing, packaging and marketing of each product is Rs. 200/-. The cost of manufacturing is Rs. 50; the cost of packaging is Rs. 50 and the cost of marketing is Rs. 100.

Based on the information given above, calculate return on knowledge (ROK).”

Important Tips
1. UThis Assignment can be best attempted from the knowledge acquired after watching video lecture no. 1 to lecture no 28 and reading handouts as well as recommended text book).

2. Video lectures can be downloaded for free from Online VU Lectures

Schedule
Opening Date and Time June 09 , 2011 At 12:01 A.M. (Mid-Night)
Due Date and Time June 15 , 2011 At 11:59 P.M. (Mid-Night)

Note: Only in the case of Assignment, 24 Hrs extra / grace period after the above mentioned due date is usually available to overcome uploading difficulties which may be faced by the students on last date. This extra time should only be used to meet the emergencies and above mentioned due dates should always be treated as final to avoid any inconvenience.

Important Instructions:
Please read the following instructions carefully before attempting the assignment solution.

Deadline:
• Make sure that you upload the solution file before the due date. No
assignment will be accepted through e-mail once the solution has been
uploaded by the instructor.

Formatting guidelines:
• Use the font style “Times New Roman” and font size “12”.
• It is advised to compose your document in MS-Word 2003.
• Use black and blue font colors only.

Solution guidelines:
• Use APA style for referencing and citation. For guidance search “APA
reference style” in Google and read various website containing
information for better understanding or visit
http://linguistics.byu.edu/faculty/henrichsenl/apa/APA01.html


Solution:

Manufacturing Process 60 = 60/250 0.24
Packaging Process 110 = 10/250 0.44
Marketing Process 80 = 80/250 0.32
Total 250

Manufacturing cost 50 = 50/200 0.25
Packaging cost 50 = 50/200 0.25
Marketing cost 100 = 100/200 0.50
Total 200

ROK on Manufacturing Process =0.24/0.25*100 96%
ROK on Packaging Process =0.44/0.25*100 176%
ROK on Marketing Process =0.32/0.50*100 64%

CS601 Assignment No. 4 solution

Wednesday, June 15, 2011 Posted In Edit This
Question # 1 
A telephone line normally has bandwidth of 5000 Hz, the signal to noise ratio is usually 25dB.Calculate thecapacity of the channel? 


When given a bandwidth and a S/N ratio, you need to use the Shannon equation to determine the capacity of the channel in bits/second. Add 1 to your S/N ratio (25+1=26) and take the base 2 logarithm (Log2 of 26=4.7, that is, 2^4.7 = 26) and then multiply this result by the bandwidth (5000). In this case you'll get a capacity of 5000x4.7 = 23,500 bits/second.


You can solve Q No 1 with the help of this question
We can calculate the theoretical highest bit rate of a regular telephone line. A 
telephone line normally  has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 


3162. For this channel the capacity is calculated as 
BitRate = 2 × 3000 × log2 2 = 6000 bps 
BitRate = 2 × 3000 × log2 4 = 12,000 bps 
Capacity = Bandwidth × log2 (1 + SNR) 
C = B log2 (1 + SNR) = B log2 (1 + 0) = B log2 (1) = B × 0 = 0 
C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163) 
C = 3000 × 11.62 = 34,860 bps


Question 2: [10] 
Why Digital transmission is preferred over analog transmission justify your answer with solid reasons? 


Analog vs. Digital Transmission
Compare at two levels: 
1. Data continuous (audio) vs. discrete (text) 
2. Signaling continuously varying electromagnetic wave vs. sequence of voltage pulses. 
Also Transmission transmit without regard to signal content vs. being concerned with signal content. Dierence in how attenuation is handled, but not focus on this.


Improving digital technology 
 Data integrity. Repeaters take out cumulative problems in transmission. Can thus transmit longer distances. 
 Easier to multiplex large channel capacities with digital Easy to apply encryption to digital data 
 Better integration if all signals are in one form. Can integrate voice, video and digital Data. 


Solid Reasons:
1. Analog circuits require appliers, and each ampler adds distortion and noise to t signal. 


2. In contrast, digital appliers regenerate an exact signal, eliminating cumulative errors. An incoming (analog) signal is sampled, its value is determined, and the nod then generates a new signal from the bit value; the incoming signal is discarded. With analog circuits, intermediate nodes amplify the incoming signal, noise and al 


3. Voice, data, video, etc. can all by carried by digital circuits. What about carrying digital signals over analog circuit? The modem example shows the deputies in carrying digital over analog. 
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Solution:



Q#2) Why Digital transmission is preferred over analog transmission justify your answer with solid reasons?
There are a wide variety of analog and digital services in which we can compare these two like in ordinary things analog clock and digital clock we compare these two. But if we are talking about services we will talk about telephone services and the internet services.


Analog Services:
In early days our telephone systems are started with the analog system the subscribers were used the analog services and the analog network is called PSTN (Public Switched Telephone Network) and when we connect the modem to our computers the analog signals are digitized for faster speed. Let’s have a look on this aspect:


1) The speed of this service was low
2) There can be more chances of attenuation
3) There can be occurrence of noise while conversation.
4) The speed of the service depends on the bandwidth if carries.
5) The analog services are not more reliable.


Digital Service:
Nowadays digital services are taking place of analog services because in this IT world every thing is becoming more faster and saying Bye to the slower services.


THESE ARE THE POINTS OF WHICH WE PREFFERED DIGITAL TRANSMISSION:
SPEED:


1) The speed of this service is very high.


SMOOTH PERFORMANCE:


2) There can be less chances of attenuation.


NOISE REDUCTION:


3) There can be less noise


BIT RATES:


4) The speed depends upon the bit rates


RELIABILITY:.


5) The digital services are more reliable
Hence that time is not far away when every service and task will be done digitally.

STA301 Assignment No. 3 solution

Wednesday, June 15, 2011 Posted In Edit This

Assignment No.3 (Course STA301)

Spring 2011 (Total Marks 30)

Deadline

Your Assignment must be uploaded/ submitted before or on 15th June, 2011

STUDENTS ARE STRICTLY DIRECTED TO SUBMIT THEIR ASSIGNMENT BEFORE OR BY DUE DATE. NO ASSIGNMNENT AFTER DUE DATE WILL BE ACCEPTED VIA E.MAIL).

Rules for Marking

It should be clear that your Assignment will not get any credit IF:
  • The Assignment submitted, via email, after due date.
  • The submitted Assignment is not found as MS Word document file.
  • There will be unnecessary, extra or irrelevant material.
  • The Statistical notations/symbols are not well-written i.e., without using MathType software.
  • The Assignment will be copied from handouts, internet or from any other student’s file. Copied material (from handouts, any book or by any website) will be awarded ZERO MARKS. It is PLAGIARISM and an Academic Crime.
  • The medium of the course is English. Assignment in Urdu or Roman languages will not be accepted.
  • Assignment means Comprehensive yet precise accurate details about the given topic quoting different sources (books/articles/websites etc.). Do not rely only on handouts. You can take data/information from different authentic sources (like books, magazines, website etc) BUT express/organize all the collected material in YOUR OWN WORDS. Only then you will get good marks.

Objective(s) of this Assignment:


The assignment is being uploaded to strengthen
  •  The students’ BASIC concepts regarding the Probability and Probability Distributions.
  •  To learn the applications of the Probability Distributions.
  • To learn the use of Bivariate distribution/function for finding the probabilities and correlations.

Assignment 3 (Lessons 23-30)
Question 1:                                                                                   Marks:2+3+3+3+4=15
a) If Z is a standard normal random variable with mean 0 and variance 1, then find the lower quartile.
b) A random variable ‘x’ is said to be uniformly distributed such that  
(i)  Write down the density function of ‘x’
(ii) Prove that total area between the given limits is unity.
c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
d) A certain type of storage battery lasts on the average 3.0 years, with a standard deviation of 0.5 year. Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years. 
e) The incident of occupational disease is such that the workers have 20 percent chance of suffering from it. What is the probability that out of six more than 4 workers will come in contact of the disease?
Question 2:                                                                                          Marks:3+3+4+5=15
a)      The Probability distribution of X is given below:
Compute,
i)         P(x=2)
ii)                  E(X)

X
0                 1               2                3                4
P(X)
                                               

b) The p. d. f. of a random variable is given
            f(x) = (2x-1)                1<x < 2
0                   Elsewhere
Calculate P(X<1.5).
c) The given information is obtained from a joint distribution of X and Y. Calculate Correlation coefficient () between X and Y.
  
d) Find the value of A, such that the function f(x, y) is a density function.

          


Solution:



QUESTION 1; Part B


here is the Q:1 part B solution. Also include it in your file..
b) If Z is a standard normal random variable with mean 0 and variance 1, then find the Lower quartile.
Solution
Mean , µ = 0
Variance б2 = 1
Q1 = µ - 0.6745 б
Q1 = 0 - .6745 (1)
Q1 = -.6745


c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
Answer; -
First, find the probability that the first card is an Ace and the 2nd to 5th are not:
(4/52) * (48/51) * (47/50) * (46/49) * (45/48).
Next, find the probability that the 2nd card is an Ace and the first, 3rd, 4th and 5th are
not:
(48/52) * (4/51) * (47/50) * (46/49) * (45/48).
Next, the 3rd card is an Ace; the others are not:
(48/52) * (47/51) * (4/50) * (46/49) * (45/48).
Next, 4th card is an Ace; not the others:
(48/52) * (47/51) * (46/50) * (4/49) * (45/48).
Lastly, the 5th card is an Ace; not the others:
(48/52) * (47/51) * (46/50) * (45/49) * (4/48).
Adding these, the answer is:
0.29947... which is just under 30%.




QUESTION 1; Part C


100% correct but its not my subject.... Plz do make changes in it.
ALI Also include this one as well in your file. i have solved it for urself...
c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
Solution:

Since it is "at least" one ace, you have to account for drawing 1, 2, 3, or 4 aces. It is easier to calculate the following: Probability of drawing 0 aces. The probability of drawing
"at least" 1 ace is:


Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]


The probability of drawing 0 aces is:


C(48,5)/C(52,5)


The numerator is the ways we can draw 5-card hands out of 48 cards
(all the cards except for the aces). The denominator is the ways we
can draw any 5-card hands. So the ratio is the number of ways of
drawing no aces. If we subtract it from 1, we get the probability of
drawing "at least" 1 ace.
Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]
= 1 - (C(48,5)/C(52,5))
= 1 – (1712304/ 2598960)
= 1 - 0.658842
= 0.34 Ans
 ::::::::::::::::::::::::::::::::::::::::::::::::

DOWNLOAD another solution of STA301


::::::::::::::::::::::::::::::::::::::::::::::::


STA301 Assignment No. 3 solution
QUESTION 1; Part B


here is the Q:1 part B solution. Also include it in your file..
b) If Z is a standard normal random variable with mean 0 and variance 1, then find the Lower quartile.
Solution
Mean , µ = 0
Variance б2 = 1
Q1 = µ - 0.6745 б
Q1 = 0 - .6745 (1)
Q1 = -.6745

c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
Answer; -
First, find the probability that the first card is an Ace and the 2nd to 5th are not:
(4/52) * (48/51) * (47/50) * (46/49) * (45/48 ).
Next, find the probability that the 2nd card is an Ace and the first, 3rd, 4th and 5th are
not:
(48/52) * (4/51) * (47/50) * (46/49) * (45/48 ).
Next, the 3rd card is an Ace; the others are not:
(48/52) * (47/51) * (4/50) * (46/49) * (45/48 ).
Next, 4th card is an Ace; not the others:
(48/52) * (47/51) * (46/50) * (4/49) * (45/48 ).
Lastly, the 5th card is an Ace; not the others:
(48/52) * (47/51) * (46/50) * (45/49) * (4/48 ).
Adding these, the answer is:
0.29947... which is just under 30%.


QUESTION 1; Part C

100% correct but its not my subject.... Plz do make changes in it.
ALI Also include this one as well in your file. i have solved it for urself...
c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
Solution:

Since it is "at least" one ace, you have to account for drawing 1, 2, 3, or 4 aces. It is easier to calculate the following: Probability of drawing 0 aces. The probability of drawing
"at least" 1 ace is:

Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]

The probability of drawing 0 aces is:

C(48,5)/C(52,5)

The numerator is the ways we can draw 5-card hands out of 48 cards
(all the cards except for the aces). The denominator is the ways we
can draw any 5-card hands. So the ratio is the number of ways of
drawing no aces. If we subtract it from 1, we get the probability of
drawing "at least" 1 ace.
Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]
= 1 - (C(48,5)/C(52,5))
= 1 – (1712304/ 2598960)
= 1 - 0.658842
= 0.34 Ans
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::


Question 1: Marks:2+3+3+3+4=15
a) If Z is a standard normal random variable with mean 0 and variance 1, then find the lower quartile.
ANS
Mean,��= 0
Variance 62= 1
Q 1 =��-0.67456
Q1=0-.6745(1)
Q1= -.6745 ANS

b) A random variable ‘x’ is said to be uniformly distributed such that
(i) Write down the density function of ‘x’
(ii) Prove that total area between the given limits is unity.

ANSWER
DEFINTION
THE PROBIBLITY DENSITY OF A CONTINOUS RANDOM VARIABLE IS A FUNCTION WHICH CAN BE INTEGRATED TO ONTAIN THE PROBIBLITY THAT THE RANDOM VARIABLE TAKE A VALUE IN A GIVEN INTERVAL.
PART B
PROVE THAT TOTAL AREA BETWEEEN THE GIVEN LIMIT IS UNITY
ANS
THE TOTAL PROBIBLITY FOR ALL POSSIBLE VALUE OF THE CONTINOUS RANDOM VARIABLE X IS 1:
∫F(x) dx= 1

b) What is the probability that a poker hand of 5 cards contain exactly 1 ace?

Answer

First find the probility that the first card is an Ace and the 2nd to 5th are not:
(4/25) *(48/51)*(47/50)*(46/49)*(45/48).
Next to find the probability that he second card is an Ace and the first ,3rd, 4th and 5th are.
(48/52)*(4/51)*(47/50)*(46/49)*(45/48).
Next , the 3rd card is an Ace; the other is not:
(48/52)*(47/51)*(4/50)*(46/49)*(45/48).
Next , 4rd card is an Ace ; not the other:
(45/52)*(47/51)*(46/50)*(4/49)*(45/48).
Lastly the 5th card is an Ace: not the others;
(48/52)*(47/51)*(46/50)*(45/49)*(4/48).
Adding this answer is,
0.29947…which is just under 30%

c) A certain type of storage battery lasts on the average 3.0 years, with a standard deviation of 0.5 year. Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years.
Answer
Z=2.3-3/0.5 = -1.4→p(x<2.3)=p( z < -1.4)=0.0808.

d) The incident of occupational disease is such that the workers have 20 percent chance of suffering from it. What is the probability that out of six more than 4 workers will come in contact of the disease?
Answer
P(4< x<6)=20/100= 0.2
Question 2: Marks:3+3+4+5=15
a) The Probability distribution of X is given below:
Compute,
i) P(x=2)
Answer
P(x=2)=35/100=0.35
ii) E(X)
Answer
E(x) =∑(x)p(x)=1.36 because


(x) p(x)
0

0.5

0.7

0.12
0.04
∑(x) p(x)=1.36

X 0 1 2 3 4
P(X)


b) The p. d. f. of a random variable is given
f(x) = (2x-1) 1<x < 2
0 Elsewhere
Calculate P(X<1.5).
Answer
⌠ (2x-1)dx=[2x-x]=1
c) The given information is obtained from a joint distribution of X and Y. Calculate Correlation coefficient ( ) between X and Y.


Answer
Pxy=Cov(X,Y)/√ver(x) ver(y)
Cov(x,y)=E(xy)-E(x)E(y)
=3.26-(1.4)(2.26)=3.26-3.164=0.096
0.096 NOW,
Var(x)=E(x)² -[E(x)]² =0.24
Var(y)=E(y)²-[E(y)]²=0.6
Pxy=0.096/√0.24 x 0.6= 0.25


d) Find the value of A, such that the function f(x, y) is a density function.

∫∫ ay (1-x)dxdy=1

a∫y(1-x)dx)dy=1

a∫(∫(y-yx)dx)dy=1


a∫[yx-y x²/2 dy]=1

after addining limit we get
A∫[y²/2-y²/4]=1
A[2-1/2]=1
A[1/4]=1
A=4


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Correction in Assignment no.3 Dated: Jun 20, 11

Dear Students,

It is to inform you that there was a mistake in the Question 2 (b), Assignment No. 3.

The given (incorrect) pdf was f(x) = (2x-1), 1<x < 2. This violates the basic property of a probability density function (pdf); that its total area under the curve MUST be equal to one.

The correct pdf is: f(x) = 2(x-1), 1<x < 2.

This ambiguity created difficulty for student in solving the problem. So, ALL the students will be awarded FULL marks in Q. No. 2 (b)

STA department sincerely apologize for the inconvenience!
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