1)Permutation----nPr = n!
----
(n-r)!
2)Combination----nCr = nPr = n!
----- ------- n r! r! (n-r)! 3)Summation-----∑ X i
i =1
n
4)Product--------Л Xi
i=1

5)Age specific fertility rate(Asfr)=No of live birth at specific age ------------------------------------- No of women to specific age group Asfr=∑

6)Total fertility rate(TFR)= ∑ Asfr *group of year

7)Gross Reproductive Rate( GRR) =TFR*(100)sex ratio
205 f 8)Net Reproduction Rate (NRR)= {Σ( Asfr * nSx)} * group of year

9)Life Table-
nLx= Persons year live in the Interval
lx = No.of person live at particular age
nqx = Proportion of person live at particular age nSx = nLx
l0 * group of year , where l0 =5,00,000

10)Crude Death Rate (CDR)= Σ (Pi)*(asdri) →Direct Standardisation Σ Pi Where Pi= Population age wise,
asdri = Age specific death rate in particular age

At population level, s
Standard Death rate = Σ Pi*asdri
s s Σ Pi Where ,P =Standard Population ,i=age group

_
11)Mean(arithmetic mean)( X )=
_ For Ungroup data Mean ( X )= Σ Xx=Character observe, ----- n=the no. of observation n

__ For Group data Mean ( X)= ΣfiXifi=frequency,
----- , xi=Mid point of the
Σfi interval Σfi( n)=the no.of observation 12)Median=
For Ungroup data Median=Average of the middle two value =a+b
-----
2
>First arrange all data in Ascending order-small value
Comes first & larger in the last.
>Find out middle data

For Group data Median= L +(N/2-F) * C L=the lower limit ------------- of the median
fclass
N=the total
no.of observation F= The no. of observations up to the Median class(cumulative frequency/observation) f=frequency/observation in the median class C=the interval of the median class

13)Population(P) Change=
P2009-P2008
P2009=P2008 + [Births-Deaths] + [Immigrants – Emmigrants] 14) “Table” is used to summerise the data.Information derived from it is a frequency or percentage. Diagramatic representation of data is—

1) Bar diagram- for discrete data & continuous data
For this no. of observation/frequency is always taken onY-axis ....

...
INTRODUCTION TO NORMAL DISTRIBUTIONS
The normal distribution is the most important and most widely used distribution in statistics. It is sometimes called the "bell curve," although the tonal qualities of such a bell would be less than pleasing. It is also called the "Gaussian curve" after the mathematician Karl Friedrich Gauss. As you will see in the section on the history of the normal distribution, although Gauss played an important role in its history, Abraham de Moivre first discovered the normal distribution.
Strictly speaking, it is not correct to talk about "the normal distribution" since there are many normal distributions. Normal distributions can differ in their means and in their standard deviations. Figure 1 shows three normal distributions. The green (left-most) distribution has a mean of -3 and a standard deviation of 0.5, the distribution in red (the middle distribution) has a mean of 0 and a standard deviation of 1, and the distribution in black (right-most) has a mean of 2 and a standard deviation of 3. These as well as all other normal distributions are symmetric with relatively more values at the center of the distribution and relatively few in the tails.
Figure 1. Normal distributions differing in mean and standard deviation.
The density of the normal distribution (the height for a given value on the x axis) is shown below. The parameters μ and σ are the mean and standard deviation, respectively, and define the normal distribution. The...

...A. Genetics
Ribosomal 5S RNA can be presented as a sequence of 120 nucleotides. Each nucleotide can be represented by one of four characters: A (Adenine), G (Guanine), C (Cytosine), or U (Uracil). The characters occur with different probabilities for each person. We wish to test if a new sequence is the same as ribosomal 5S RNA. For this purpose, we replicate the new sequence 100 times and find that there are 60 A’s in the20th position. Use a 0.05 level of significance.
1. If the probability of an A in the 20th position is 0.79 in ribosomal 5S RNA, then test the hypothesis that the new sequence is the same as the ribosomal 5S RNA using critical method.
2. Report a p- value corresponding to your result in problem num 1
1A.
* Ho= there is no difference in the new sequence as the ribosomal 5S RNA using critical method
Ha= there is difference in the new sequence of ribosomal 5S RNA using critical method
* X= 60, level of significance= 0.05
* Reject null hypothesis
GENETICS | |
| |
Data |
Null Hypothesis = | 0.79 |
Level of Significance | 0.05 |
Population Standard Deviation | 20 |
Sample Size | 100 |
Sample Mean | 60 |
| |
Intermediate Calculations |
Standard Error of the Mean | 2 |
Z Test Statistic | 29.605 |
| |
Two-Tail Test | |
Lower Critical Value | -1.959963985 |
Upper Critical Value | 1.959963985 |
p-Value | 0 |
Reject the null hypothesis | |
* Conclusion: The new...

...Graded Assignment
Unit Test, Part 2: Polynomials and Power Functions
Answer the questions and show your work. When you are finished, submit this assignment to your teacher through the appropriate dropbox basket.
(3 pts)
1.) Factor
100x^2 – 49
to factor, use the difference of squares formula, because both the terms are perfect squares
the difference of squares formula is a^2 – b^2 = (a-b)(a+b)
therefore
100x^2 – 49 = (10x)^2 – 7^2 = (10x – 7)(10x +7)
(5 pts)
2.) Solve
x^2 – 11x = -30
(x^2 – 11x) + 30 = -30 +30
x^2 – 11x +30 = 0
Reorder the terms:
30 + 11x + x2 = 0
Solving
30 + 11x + x2 = 0
Factor a trinomial.
(6 + x)(5 + x) = 0
Subproblem 1
Set the factor '(6 + x)' equal to zero and attempt to solve:
Simplifying
6 + x = 0
Solving
6 + x = 0
Move all terms containing x to the left, all other terms to the right.
Add '-6' to each side of the equation.
6 + -6 + x = 0 + -6
Combine like terms: 6 + -6 = 0
0 + x = 0 + -6
x = 0 + -6
Combine like terms: 0 + -6 = -6
x = -6
Simplifying
x = -6
Subproblem 2
Set the factor '(5 + x)' equal to zero and attempt to solve:
Simplifying
5 + x = 0
Solving
5 + x = 0
Move all terms containing x to the left, all other terms to the right.
Add '-5' to each side of the equation.
5 + -5 + x = 0 + -5
Combine like terms: 5 + -5 = 0
0 + x = 0 + -5
x = 0 + -5
Combine like terms: 0 + -5 = -5
x = -5 ...

...BIOSTATISTICS
* Effect modification: effect of the main exposure on the outcome is affected by another variable. It is NOT a bias!
* Case-control study: also known as retrospective study. Divided into “cases” and “controls.” If disease is rare, the odds ratio will approximate the relative risk (following subjects over time.)
* Cohort study (Retrospective or prospective): divided into “exposed” and “non-exposed.” Study subjects are free of the outcome at the time the study begins.
* Cross-sectional study: exposure and outcome are studied at one point in time. (Think: snapshot study)
* Randomized control trial: gold standard for studying the efficacy of treatment or procedure. Subjects are randomized into experimental or control. Less bias and strong causal relationship.
* Cross-over study: group of participants are randomized to one treatment for a certain period of time and the other with the alternative for that same certain period of time. Then after period ends, the groups switched treatments for the duration of the trial.
* Parallel study: think “drug group vs. placebo group.” No other variables are measured
* Hazard ratio: the higher the ratio, that higher risk for hazardous events. If ratio is 1 (or value if closer to 1), then there’s little difference between the two entities
* Factorial design studies: randomization of different interventions with additional study of 2 or more variables
* Cluster...

...Application to the new Swedish Institute’s leadership programme Young Connectors of the Future (YCF) 2013 for selected countries in South Asia
Make sure you answer all the questions below. All information should be submitted in English.
General Information
LAST NAME
Nadir
FIRST NAME
Shoaib
DATE OF BIRTH (YEAR-MONTH-DAY)
1989-1-31
FEMALE MALE
CITY OF RESIDENCE
ISLAMABAD
E-MAIL
Shoaibnadir@gmail.com
PHONE
00922277246
MOBILE PHONE
00923335441313
COUNTRY OF CITIZENSHIP
AFGHANISTAN
BANGLADESH
INDIA
PAKISTAN
Are you a leader with a vision? Do you want to make a change? Then apply to the leadership programme Young Connectors of the Future!
This is a new leadership programme for young active change makers from selected countries in South Asia, which aims to build a foundation for dialogue, mutual understanding and knowledge sharing. The programme combines intercultural leadership training with theory and practice in the fields of transparency, democracy and human rights, connecting people from different sectors of society – from civil society to political and entrepreneurial sectors. The programme has the ambition of building a dynamic network of young leaders and connectors of the future.
In the questions below, please motivate your participation in the programme. Don’t repeat information already contained in your CV, but feel free to explain and elaborate on the driving forces in your life and what...

...I<ONICA MINOLTA
January 30, 2013 Purchase Order Renewal Notice
DHHS
MARK MCGEE
DBA FOOD AND DRUG ADMINISTRATION HFA 720
12345 PARKLAWN DR
ROCKVILLE MD 20857
Dear Valued Customer:
Please be advised that your Purchase Order CREDIT CARD with Konica Minolta Business Solutions will expire on 09/30/2012. In order to avoid an interruption in the provision of service and supplies, please provide
a renewal purchase order prior to expiration. Your Renewal Purchase Order should reflect an expiration
date of 09/30/2013, as well as reference the data below. If your purchase order includes units that are not referenced below, please include them on your renewal. The pricing below does not include tax, if applicable:
Model Serial#
Meter
Type
Meter
Freq.
Freq Annual Allowable Ovg. Base Copies Rate Rate
Base
Freq
**BIZHUB C353 PRINTERIC A02E014000343 Black
Color
MON MON
4000 0.01100 528.96
No minimum 0.08460 0.00
MON MON
In the event your Purchase Order lapses, all service and supplies will be charged at then current Konica
Minolta rates and will require a credit card for processing.
To avoid a lapse in coverage, please complete this form and return to the address, fax, or Email listed below.
If you are renewing, please check one of the following and include your Purchase Order, if required:
1 PO for maintenance coverage...

...99003 99699/ 0 97419 00225 / 080-32552008 Email : onlineies.com@gmail.com Site: www.onlineIES.com Google+: http://bit.ly/gplus_iesgate FB: www.facebook.com/onlineies
Institute Of Engineering Studies (IES,Bangalore)
Formulae Sheet in ECE/TCE Department
Drift velocity of e Poisson equation =
≤1 =
cm/sec ⇒ =E=
Transistor : = + = –α =–α +
→ Active region (1- e / ) β= . = - 0.25 mv / C
Common Emitter : = (1+ β) +β
= → Collector current when base open → Collector current when = 0 > or → - 2.5 mv / C ; → Large signal Current gain β = D.C current gain β (β =h = =h >
C R
) ≈ β when
Small signal current gain β = Over drive factor =
→
= h =
(
)
h C
∵
=β
Conversion formula :CC ↔ CE h =h ; h =1; CB ↔ CE h =
h = - (1+ h ) ;
h
=h
; h =
-h
;h =
; h
=
CE parameters in terms of CB can be obtained by interchanging B & E . Specifications of An amplifier : = = =h +h =h = =
.
= =
.
=
.
.
.
Choice of Transistor Configuration : For intermediate stages CC can’t be used as 10 R
For thermal stability [
- 2 (R + R )] [ 0.07
. S] < 1/
;
<
Hybrid –pi(π)- Model :g =| |/
r = h /g r =h -r r =r /h g = h - (1+ h ) g For CE : = =h
(
)
= =
(
)
;
=
C = C + C (1 + g R )
= S.C current gain Bandwidth product = Upper cutoff frequency For CC : = ≈ = =
( )
For CB: =
(
)
= (1 + h )
= (1 + β)
22...

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