mb0040 statics for management
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STATICS FOR MANAGEMENT
Name RATI BHAN
Roll No. 511022630
Program MBA
Subject STATICS FOR
MANAGEMENT [Set 1]
Code MB040
Learning Centre
IICM KINGSWAY CAMP
RATI BHAN, MBA (1ST SEM), SUBJECT CODE-MB043, SET-1 Page 1 4/11/2023
STATICS FOR MANAGEMENT
Q.1 Elucidate the functions of statistics.?
Ans. Function of Statistics
Statistics is used for various purposes. It is used to simplify mass data and to make comparisons easier. It is also used to bring out trends and tendencies in the data as well as the hidden relations between variables. All this helps to make decision making much easier. Let us look at each function of Statistics in detail.
1. Statistics simplifies mass data
The use of statistical concepts helps in simplification of complex data. Using statistical concepts, the managers can make decisions more easily. The statistical methods help in reducing the complexity of the data and consequently in the understanding of any huge mass of data.
2. Statistics makes comparison easier
Without using statistical methods and concepts, collection of data and comparison cannot be done easily. Statistics helps us to compare data collected from different sources. Grand totals, measures of central tendency, measures of dispersion, graphs and diagrams, coefficient of correlation all provide ample scopes for comparison.
Hence, visual representation of numerical data helps you to compare the data with less effort and can make effective decisions. The graphical curve represented in figure 1.7 and figure 1.8 shows the profits of CBA Company and ZYX Company respectively, for ten years from 1998 to 2008. The profits are plotted on the Y-Axis and the timeline in years on X-Axis. From the graphs, we can compare the profits of two companies and derive to a conclusion that profits of CBA Company in the year 2008 are higher than that of ZYX Company. The graphical curve in case of figure 1.7 shows that the profits for CBA Company are increasing, whereas the profits curve in figure 1.8 is constant for ZYX Company from middle of the decade (1998-2008). 3. Statistics brings out trends and tendencies in the data After data is collected, it is easy to analyse the trend and tendencies in the data by using the various concepts of Statistics.
4. Statistics brings out the hidden relations between variables
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STATICS FOR MANAGEMENT
Statistical analysis helps in drawing inferences on data. Statistical analysis brings out the hidden relations between variables. 5. Decision making power becomes easier
With the proper application of Statistics and statistical software packages on the collected data, managers can take effective decisions, which can increase the profits in a business.
Q2. What are the methods of statistical survey? Explain briefly. ?Ans :-A Statistical survey is a scientific process of collection and analysis of numerical data. Statistical surveys are used to collect numerical information about units in a population. Surveys involve asking questions to individuals. Surveys of human populations are common in government, health, social science and marketing sectors.
2.2 Stages of Statistical Survey
Statistical surveys are categorized into two stages – planning and execution.
STATISTICAL SURVEY
Primary Execution
Planning a Statistical Survey
The relevance and accuracy of data obtained in a survey depends upon the care exercised in planning. A properly planned investigation can lead to best results with least cost and time.
Methods of Statistical Survey
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There are several ways of administering a survey, including:
Telephone
use of interviewers encourages sample persons to respond, leading to higher response rates.[1]
interviewers can increase comprehension of questions by answering respondents' questions.
fairly cost efficient, depending on local call charge structure
good for large national (or international) sampling frames
some potential for interviewer bias (e.g. some people may be more willing to discuss a sensitive issue with a female interviewer than with a male one)
cannot be used for non-audio information (graphics, demonstrations, taste/smell samples)
unreliable for consumer surveys in rural areas where telephone penetration is low[2]
three types:
o traditional telephone interviews
o computer assisted telephone dialing
o computer assisted telephone interviewing (CATI)
the questionnaire may be handed to the respondents or mailed to them, but in all cases they are returned to the researcher via mail.
cost is very low, since bulk postage is cheap in most countries
long time delays, often several months, before the surveys are returned and statistical analysis can begin
not suitable for issues that may require clarification
respondents can answer at their own convenience (allowing them to break up long surveys; also useful if they need to check records to answer a question)
no interviewer bias introduced
large amount of information can be obtained: some mail surveys are as long as 50 pages
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response rates can be improved by using mail panels
o members of the panel have agreed to participate
o panels can be used in longitudinal designs where the same respondents are surveyed several
Online surveys
can use web or e-mail web is preferred over e-mail because interactive HTML forms can be used
often inexpensive to administer
very fast results
easy to modify
response rates can be improved by using Online panels - members of the panel have agreed to participate
if not password-protected, easy to manipulate by completing multiple times to skew results
data creation, manipulation and reporting can be automated and/or easily exported into a format which can be read by PSPP, DAP or other statistical analysis software
data sets created in real time
some are incentive based (such as Survey Vault or YouGov)
may skew sample towards a younger demographic compared with CATI
often difficult to determine/control selection probabilities, hindering quantitative analysis of data
use in large scale industries.
Personal in-home survey
respondents are interviewed in person, in their homes (or at the front door) very high cost
suitable when graphic representations, smells, or demonstrations are involved
often suitable for long surveys (but some respondents object to allowing strangers into their home for extended periods)
suitable for locations where telephone or mail are not developed
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skilled interviewers can persuade respondents to cooperate, improving response rates
potential for interviewer bias
Personal mall intercept survey
shoppers at malls are intercepted - they are either interviewed on the spot, taken to a room and interviewed, or taken to a room and given a self-administered questionnaire
socially acceptable - people feel that a mall is a more appropriate place to do research than their home
potential for interviewer bias
fast
easy to manipulate by completing multiple times to skew results
Q3.Tabulate the following data: Age: 20-40; 40-60;60-above Departments: English, Hindi, Political science, History, sociology Degree level: Graduates, Post graduates; PhD, Total students in age group and in degree level. Ans.
Department
Age TOTAL20-40 40-60 60 ABOVE
G P.G. PH.D. G P.G. PH.D. G P.G. PH.D.EnglishHindiPolitical ScienceHistorySociologyTotal
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Q4. The data given below is the distribution of employees of a business according to their efficiency. Find the mean deviation and coefficient of mean deviation from Mean and Median:
Efficiency Index
22 - 26 26 - 30 30 - 34 34 – 38 38 – 42
Employees 25 35 15 5 2
Ans 4 :-Efficiency Index
Freq.(F)
Mid Value (X)
D = X-32 4
FD F [X-28.29]
CF [X-MED]
F [X-MED]
22 – 26 25 24 - 2 - 50
107.25 25 3.82 95.5
26 – 30 35 28 - 1 - 35
962.15 60 0.18 6.3
30 – 34 15 32 0 0 392.35 75 4.18 62.734 – 38 5 36 1 5 105.45 80 8.18 40.938 – 42 2 40 2 4 16.58 82 12.18 24.36
-76 1583.78 229.76
Mean deviation:- _(X) = A + £FD × I £F
= 32 + (- 76) × 4 82
= 32 – 3.707
= 28.29 _(X) = A + £FD × I £F
Mediation => Nth value = 82 => 41
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2 2
Median class is 26 – 30
Median = 26 + 41 – 25 × 4 35
= 26 + 16 × 4= 26 + 1.82 = 27.82 35
MD of median = 229.76 82 = 2.80
Coefficient of M.D. :- _ _ Mean (X) = 19.31 or [MD (X)]
28.29 X = 0.6825
Median = 2.8027.82 = 0.10064
Q.5 What is conditional probability? Explain with an example.
Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the (conditional) probability of A, given B" or "the probability of A under the condition B". When in a random experiment the event B is known to have occurred, the possible outcomes of the experiment are reduced to B, and hence the probability of the occurrence of A is changed from the unconditional probability into the conditional probability given B.
Joint probability is the probability of two events in conjunction. That is, it is the probability of both events together. The joint probability of A and B is written
or
Marginal probability is then the unconditional probability P(A) of the event A; that is, the probability of A, regardless of whether event B did or did not occur. If B can be thought of as the event of a random variable X having a given outcome, the marginal probability of A can be obtained by summing (or integrating, more generally) the joint probabilities over all outcomes for X. For example, if there are two possible outcomes for X with corresponding events B and B', this means that . This is called marginalization.
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In these definitions, note that there need not be a causal or temporal relation between A and B. A may precede B or vice versa or they may happen at the same time. A may cause B or vice versa or they may have no causal relation at all. Notice, however, that causal and temporal relations are informal notions, not belonging to the probabilistic framework. They may apply in some examples, depending on the interpretation given to events.
Conditioning of probabilities, i.e. updating them to take account of (possibly new) information, may be achieved through Bayes' theorem. In such conditioning, the probability of A given only initial information I, P(A|I), is known as the prior probability. The updated conditional probability of A, given I and the outcome of the event B, is known as the posterior probability, P(A|B,I).
Introduction
Consider the simple scenario of rolling two fair six-sided dice, labelled die 1 and die 2. Define the following three events (not assumed to occur simultaneously):
A: Die 1 lands on 3.
B: Die 2 lands on 1.
C: The dice sum to 8.
The prior probability of each event describes how likely the outcome is before the dice are rolled, without any knowledge of the roll's outcome. For example, die 1 is equally likely to fall on each of its 6 sides, so P(A) = 1/6. Similarly P(B) = 1/6. Likewise, of the 6 × 6 = 36 possible ways that a pair of dice can land, just 5 result in a sum of 8 (namely 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2), so P(C) = 5/36.
Some of these events can both occur at the same time; for example events A and C can happen at the same time, in the case where die 1 lands on 3 and die 2 lands on 5. This is the only one of the 36 outcomes where both A and C occur, so its probability is 1/36. The probability of both A and C occurring is called the joint probability of A and C and is written
, so . On the other hand, if die 2 lands on 1, the dice cannot
sum to 8, so .
Now suppose we roll the dice and cover up die 2, so we can only see die 1, and observe that die 1 landed on 3. Given this partial information, the probability that the dice sum to 8 is no longer 5/36; instead it is 1/6, since die 2 must land on 5 to achieve this result. This is called the conditional probability, because it is the probability of C under the condition that A is observed, and is written P(C | A), which is read "the probability of C given A." Similarly,
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P(C | B) = 0, since if we observe die 2 landed on 1, we already know the dice can't sum to 8, regardless of what the other die landed on.
On the other hand, if we roll the dice and cover up die 2, and observe die 1, this has no impact on the probability of event B, which only depends on die 2. We say events A and B are statistically independent or just independent and in this case
In other words, the probability of B occurring after observing that die 1 landed on 3 is the same as before we observed die 1.
Intersection events and conditional events are related by the formula:
In this example, we have:
As noted above, , so by this formula:
On multiplying across by P(A),
In other words, if two events are independent, their joint probability is the product of the prior probabilities of each event occurring by itself.
Definition
Given a probability space ( ,Ω F, P) and two events A, B ∈ F with P(B) > 0, the conditional probability of A given B is defined by
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If P(B) = 0 then P(A | B) is undefined (see Borel–Kolmogorov paradox for an explanation). However it is possible to define a conditional probability with respect to a -algebraσ of such events (such as those arising from a continuous random variable).
For example, if X and Y are non-degenerate and jointly continuous random variables with density ƒX,Y(x, y) then, if B has positive measure,
The case where B has zero measure can only be dealt with directly in the case that B={y0}, representing a single point, in which case
If A has measure zero then the conditional probability is zero. An indication of why the more general case of zero measure cannot be dealt with in a similar way can be seen by noting that that the limit, as all yδ i approach zero, of
depends on their relationship as they approach zero. See conditional expectation for more information.
Q6.The probability that a football player will play Eden garden is 0.6 and on Ambedkar Stadium is 0.4. The probability that he will get knee injury when playing in Eden is 0.07 and that in Ambedkar stadium is 0.04.What is the probability that he would get a knee injury if he played in Eden.
Ans: - Eaden Garden P (A) = 0.6
Ambedkar Stadium, P (B) = 0.4
After knee injury, P (A/D) = 0.07(Eaden Garden)
After knee injury, P (B/D) = 0.04(Ambedkar Stadium)
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So, P (D/B) = ____[ P(B) { P(B/D)} ] . P {P (A/D) + {P (B) P (B/D)}
= 0.4 * 0.04 . 0.6 * 0.07 + 0.4 * 0.04
= 0.016 . 0.042 +0.016
= 2.75
Therefore, the probability that he get a knee injury if he played in Eden garden is 2.75
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