PB0003 Statistics for Management
(3 credits)Assignment 1
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1.Briefly explain the functions and limitations of Statistics in
your own words.
Ans.The Functions and limitations of Statistics are:
Important functions of Statistics are:
1.It simplifies complexity of the data: Complex numerical data
are simplified by the application of statistical methods. For
instance, complex data regarding varying costs and prices of
commodities of daily use can be reduced to the form of cost of
living index number. This can be understood easily.2.It reduces the
bulk of the data: Voluminous data could be reduced to a few figures
making them easily understandable.3.It adds precision to thinking:
Statistics sharpens ones thinking.4.It helps in comparing different
sets of figures: The imports and exports of a country may be
compared among themselves or they may be compared with those of
another country.
5.It guides in the formulation of policies and helps in
planning: Planning and policy making by the government is based on
statistics of production, demand, etc6.It indicates trends and
tendencies: Knowledge of trend and tendencies helps future
planning.
7.It helps in studying relationship between different factors:
Statistical methods may be used for studying the relation between
production and price of commodities.
The Limitations of Statistics
Major limitations of Statistics are:
1.Statistics does not deal with qualitative data. It deals only
with quantitative data: Statistical methods can be applied only to
numerically expressed data. Qualitative characteristics can be
studied only if an alternative method of numerical measurement is
introduced.2. Statistics does not deal with individual fact:
Statistical methods can be applied only to aggregate of facts.
Single fact cannot be statistically studied.3.Statistical can be
misused: Increasing misuse of Statistics has led to increasing
distrust in Statistics.
4.Statistical inferences are not exact: Statistical inferences
are true only on an average. They are probabilistic statements.5.
Common men cannot handle Statistics properly: Only statisticians
can handle statistics properly. An illogical analysis of
statistical data leads to statistical
fallacies.____________________________________________________________________________________________________________________________________________
2.A Survey of 128 smokers revealed the following frequency
distribution of daily expenditure on smoking of these smokers. Find
the mean daily expenditure.Expenditure
(Rs.)10-2020-3030-4040-5050-6060-7070-80
No. of smokers23443512932
Solution:
Expenditure (Rs.)Frequency (f)Mid-value (x)fx
10-20
20-30
30-40
40-50
50-60
60-70
70-802344
35
12
9
3
21525
35
45
55
65
7534511001225540495195150
Total128-4050
The mean is
X=fx= 4050= Rs. 31.64
N 128
The mean daily expenditure is Rs.
31.64.____________________________________________________________________________________________________________________________________________
3.What do you mean by Marginal Probilities under statistical
dependence?
Ans.A marginal or unconditional probability is the simple of the
occurrence of an event. In a fair coin toss, P(H))=0.5, and
P(T)=0.5; that is, the probability of heads equals 0.5 and the
probability of tails equals 0.5. This is true for every toss, no
matter how many tosses have been made or what their outcomes have
been. Every toss stands along and is in no way connected with any
other loss. Thus, the outcome of each toss of a fair coin is an
event that is statistically independent of the outcomes of every
other toss of the coin.
Imagine that we have a biased or unfair coin that has been
altered in such a way that heads occurs 0.90 of the time and tails
0.10 of the time. On each individual toss, P(H)=0.90, and
P(T)=0.10. The outcome of any particular toss is completely
unrelated to the outcomes of the tosses that may precede or follow
it. The outcomes of several tosses of this coin are statistically
independent events too, even though the coin is biased.Marginal
probabilities under statistical dependence are computed by summing
up the probabilities of all the joint events in which the simple
event occurs. In the example above, we can compute the marginal
probability of the event colored by summing the probabilities of
the two joint events in which colored occurred:P(C) =P(CD) + P(CS)
= 0.3+0.1= 0.4
Similarly, the marginal probability of the event gray can be
computed by summing the probabilities of the 2 joint events in
which gray occurred:
P(G) = P(GD) + P(GS) = 0.2+.04 =0.6In like manner, we can
compute the marginal probability of the event dotted by summing the
probabilities of the two joints events in which dotted
occurred:
P(D) = P(CD) + P(GD) = 0.3+0.2= 0.5And, finally, the marginal
probability of the event striped can be computed by summing the
probabilities of the 2 joint events in which gray occurred:P (CS) =
P(CS) + P(GS) = 0.1+ 0.4 =0.5
____________________________________________________________________________________________________________________________________________4.The
probabilities of three events A,B and C occurring are P(A)=0.35,
P(B)= 0.45 and P(C) =0.2. Assuming that A,B and C has occurred, the
probabilities of another event X occurring are P(X/A) =0.8, P(X/B)
= 0.65 and P(X/C) = 0.3. Find P(A/X), P(B/X) and P(C/X).Ans.
EventP(Event)P(X/Event)P(X and Event)P(Event/X)= P(Event &
X)/P(X)
A0.350.8=0.3*0.8= 0.28000.2800/0.6325= 0.4427
B0.450.65=0.45*0.65 = 0.29250.2925/0.6325= 0.4625
C0.20.3=0.2*0.3= 0.06000.0600/0.6325= 0.0949
P (X) = P(XA)+P(XB)+P(XC)0.6325
Therefore P(A/X) = 0.4427,P(B/X) =0.4625and P(C/X)= 0.0949
____________________________________________________________________________________________________________________________________________
5.Write short notes on Bernoulli distribution.Ans. One widely
used probability distribution of a discrete random variable is the
binomial distribution. It describes a variety of processes of
interest to mangers. The describes a variety of processes of
interest to managers. The binomial distribution describes discrete,
not continuous, data, resulting from an experiment known as
Bernoulli process, after the seventeenth-century Swiss
mathematician Jacob Bernoulli. The tossing of a fair coin a fixed
number of times is a Bernoulli process, and the outcomes of such
tosses can be represented by the binomial probability distribution.
The success of failure of interviewees on a aptitude test may also
be described by a Bernoulli process. On the other hand, the
frequency distribution of the lives of fluorescent lights in a
factory would be measured on a continuous scale of hours and would
not qualify as binomial distribution.Use of the Bernoulli Process
We can use the outcomes of a fixed number of tosses of a fair coin
as an example of a Bernoulli process. We can describe this process
as follows:1. Each trial (each toss, in this case) has only two
possible outcomes; heads or tails, yes or no, success or failure.2.
The probability of the outcome of any trial (toss) remains fixed
overtime. With a fair coin, the probability of heads remains 0.5
for each toss regardless of the number of times that coin is
tossed.
3. The trials are statistically independent; that is, the
outcome of one toss does not affect the outcome of any other
toss.The probability of r successes in n trials is given as : nCr
Pr Qn-r= n!__Pr Qn-r r! (n-r)!The mean of a Binomial distribution
is given as =np andThe standard deviation of a binomial
distribution as =npq.Using of Bernoulli processOne of the
requirements for using as Bernoulli process is that that
probability of the outcome must be fixed over time. This is a very
difficult condition to meet in practice. Even a fully automatic
machine making parts will experience some wear as the number of
parts increases and this will affect the probability of producing
acceptable parts. Still another condition for its uses that the
trials (manufacture of parts in our machine example) be
independent. This too is a condition that is hard to meet. If our
machine produces a long series of bad parts, this could affect the
position (or sharpness) of the metal-cutting tool in the machine.
Here, as in every other situation, going from the textbook to the
real world is often difficult, and smart managers use their
experience and intuition to known when a Bernoulli process is
appropriate.____________________________________________________________________________________________________________________________________________
6.What do you mean by Stratified Sampling?
Ans.In stratified sampling, we divide the population into
relatively homogeneous groups, called strata. Then we use one of
two approaches. Either we select at random from each stratum a
specified number of elements corresponding to the proportion of
that stratum in the population as a whole or we draw an equal
number of elements from each stratum and give weight to the results
according, stratified sampling guarantees that every element in the
population has a chance of being selected.
Uses of Stratified Sampling
Stratified Sampling is appropriate when the population is
already divided into groups of different sizes and we wish to
acknowledge this fact. Suppose that a Doctors patients are divided
into four groups say, according to age. The doctor wants to find
out how many hours his patients sleep. To obtain an estimate of
this characteristic of the population, he could take a random
sample from each of the four age groups and give weight to the
samples according to the percentage of patients in that group. This
would be an example of a stratified sample.
The advantage of stratified samples is that when they are
properly designed, they more accurately reflect characteristics of
the population from which they were chosen than do other kinds of
samples.
The advantage of stratified samples is that when they are
properly designed, they more accurately reflect characteristics of
the population from which they were chosen than do other do other
kinds of samples.
____________________________________________________________________________________________________________________________________________
PB0003 Statistics for Management
(3 credits)
Assignment
2____________________________________________________________________________________________________________________________________________1.A
bank calculates that its individual savings accounts are normally
distributed with a mean of Rs. 2,000 and a standard deviation of
Rs. 600. If the bank takes a random sample of 100 accounts, what is
the probability that the sample mean will lie between Rs. 1,900 and
Rs. 2,050.AnsStandard error of the mean =600
=60.
100
By using the Normal Table we can find the probability of the
sample mean lying between Rs. 1900 and Rs. 2050.Two z values
because: please note that the mean of the population is given as
2000 and the probability to be found are between the values 1900
and 2050 which is either side of the population mean.Z=x-
1900-2000-100=-1.67 and similarly
x60
602050-2000=50=0.83
60
60
Normal distribution table for a value of z=1.67 gives an area of
0.4525 and of a z value of 0.83 the area as 0.2967 hence adding
these two values we get 0.7492 as the total probability that the
mean of the sample will lie between Rs. 1900 and Rs, 2050.
____________________________________________________________________________________________________________________________________________
2.Briefly explain in your own words The Central Limit
Theorem
Ans.
____________________________________________________________________________________________________________________________________________
3.What is the criteria of a good estimator?
Ans.
____________________________________________________________________________________________________________________________________________4.From
a population of 540, a sample of 60 individuals is taken. From this
sample the mean is found to be 6.2 and the standard deviation SD to
be 1.368a)Find the estimated standard error of the mean.
b)Construct a 96% confidence interval of the mean.
Ans.A)
Given: = 1.368N= 540 n=60 x= 6.2Therefore Se=x=*N-nas n >
0.05
n
N-1 N
=x=1.368*540-60=0.167
60
540-1B)x2.05 Se= 6.2 2.05 (0.167)=5.86 and 6.54 and LCL and UCL
respectively.____________________________________________________________________________________________________________________________________________
5.In 120 throws of a single die, the following distribution of
faces was obtainedFaces123456Total
f302518102215120
Do these results constitute an example of the equal
probability?
Ans.Assuming a unbiased dice is used, we all know for such a die
the probability for getting any number is 1/6 and the theoretical
frequencies for each is will be 120/6= 20. Hence 20 is the expected
of frequency of Fe
Therefore using the formula x= (f0-fe)
Fe
X= (30-20) + (25-20) + (18-20) + (10-20) + (22-20) + (15-20)
20 20 20
20 20 s20
=1[100+25+4+100+4+25]
20
=258= 129=12.9
20 10Now degrees of freedom df is 6-1=5. At 5% level of
significance, the critical value of x from the tables given at the
end of this units is 11.070. Since the calculated value is far
greater that the critical value, the null hypothesis of equal
probability is rejected. Consequently, the die is
blased.____________________________________________________________________________________________________________________________________________
6.Calculate Karl Pearsons coefficient of correlation from the
following dataYear19851986198719881989199019911992
Index of Production10010210410710511210399
NuPBer of unemployed1512131112121926
Ans.Calculation of Karl Pearsons Correlation
CoefficientsYearIndex of Production XX-X=xxNo. of
unemployedY-Y=yyxy
19851986
1987
1988
1989
1990
1991
1992100102
104
107
105
112
103
99-4-2
0
+3+1+8-1-5164
0
9
1
64
J
251512
13
11
12
12
19
260-3
-2
-4
-3
-3
+4
+1109
4
16
9
9
16
1210+6
0
-12
-3
-24
-A
-55
X=832x=0x=120Y=120y=0y=184xy=-92
X=X/N = 832/8=104
Y=Y/N=120/8=15
r= xy =r= -92
=-0.619
x y
120*184
Therefore a Correlation between production and unemployed is
negative.____________________________________________________________________________________________________________________________________________
