Wednesday 18 December 2013

University of the punjab old paper of computer organization and assembly language (IT) Course code IT-203 and IT-21402


University of the punjab Lahore past papers of BS-4 years program (ITthird semester BS-information technology,
computer organization and assembly language (IT) Course code IT-203
pu course code IT-21402
computer organization and assemmbly language (IT) Course code IT-203 and IT-21402

Punjab university past paper of discrete math (IT) Course code MATH-231 and IT-21404

Punjab university Lahore old papers of BS-4 years program (ITthird semester BS-information technology,
discrete math (IT) Course code MATH-231
pu course code IT-21404
discrete math (IT) Course code MATH-231 and IT-21404

Punjab university past paper of communication skills (IT) Course code ENG-231 and ENG-21452

Punjab university Lahore old papers of BS-4 years program (ITthird semester BS-information technology,
communication skills (IT) Course code ENG-231
pu course code ENG-21452
communication skills (IT) Course code ENG-231 and ENG-21452


Tuesday 17 December 2013

Punjab university old paper of object oriented programming (OOP) Course code IT-201 and IT-21400

Punjab university Lahore past papers of BS-4 years program (IT) third semester BS-information technology,
object oriented programming (OOP) Course code IT-201
pu course code it-21400
object oriented programming (OOP) Course code IT-201 and it-21400
Front side of paper

object oriented programming (OOP) Course code IT-201 and it-21400
Back side of paper

Thursday 12 December 2013

Find the value of k in given a standard normal distribution such that

The problem of finding the value of k in a given standard normal distribution has been solved in very easy steps. You only may need to have the table for z score or table of area under normal curve.
Some examples are given that how to find the value of k in a given standard normal distribution.

Given a standard normal distribution find the value of k such that
(a) P( Z < k ) = 0.0427
(b) P ( Z > k ) = 0.2946
(c) P(-0.93 < Z < k ) = 0.7235

Solution. (a)
P( Z < k ) = 0.0427
table(k) = 0.0427
from area table
k = -1.72

Solution (b)
P( Z > k ) = 0.2946
P( Z > k ) = 1-P( Z <= k ) = 0.2946
P( Z <= k ) = 1-0.2946
table(k)=0.7054
from area table
k = 0.54

Solution(c) 
P(-0.93 < Z < k ) = 0.7235
table(k) - table(-0.93) = 0.7235
table(k) - 0.1762 = 0.7235
table(k) = 0.7235 + 0.1762
table(k) = 0.8997
from area table
k = 1.28

normal-distribution-find-value-of-k
Table for area under normal curve