CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
JAN2016
ASSESSMENT_CODE MCA4020_JAN2016
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
39123
QUESTION_TEXT
a. Write 5 properties of expectation of a single variable.
b. Prove that if X is a random variable, then V(aX+b)= a2 V(X), a and b
are constants.
SCHEME OF
EVALUATION
a.
1. if x and y are 2 random variables then E(x+y) = E(x)+ E(y).
2. if x and y are independent random variable then E(xy) = E(x).E(y).
3. If x is a random variable and a and b are constants then E(ax+b) =
aE(x) + b.
4. If x≥0, E(x) ≥ 0.
5. If x and y are 2 random variables such that y≤x then E(x) ≤ E(y).
(5 marks)
b.
Let T = ax+b then
E(T) = aE(x)+b
T–E(T) = ax+b–aE(x)–b
=a[x–E(x)]
Squaring both the sides
E[T–E(t)]2 = a2 E[x–E(x)]2
V(T) = a2 V(x).
(5 marks)
(total 10 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
39126
QUESTION_TEXT
Define the Time Series. Briefly explain the 4 components of Time Series.
SCHEME OF
EVALUATION
Time series is defined as the set of ordered pair of observation taken at
successive points of time. That is it is a series of values of variables whose
values varies with passage of time.
(2 marks)
Components are: (4×2=8 marks)
● Secular trend
● Seasonal variation
● Cyclical variation
● Irregular variation
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
72291
a.
What are the requisites of Ideal measure of Dispersion?
b.
List out the Absolute Measure of variation.
QUESTION_TEXT
a)
1.
It should be easy to understand and simple to calculate.
2.
It should be based on all values.
3.
It should be rigidly defined.
4.
It should not be affected by extreme values.
5.
It should not be affected by sampling fluctuations
SCHEME OF
EVALUATION
6. It should be capable of further algebraic
treatment
(6 Marks)
b)
1.
Range:
2.
Quartile Deviation:
3.
Mean Deviation
4.
Standard Deviation
(4 Marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
125367
QUESTION_TEXT
Explain the Merit & demerits of Harmonic Mean, Median &
Mode.
Merits of Harmonic Mean
1.
2.
3.
4.
It is based on all observations.
It is rigidly defined.
It is suitable in case of series having wide dispersion.
It is suitable for further mathematical treatment.
Demerits of Harmonic Mean
1. It is not easy to compute.
2. Cannot used when one of the item is zero.
It cannot represent distribution.
SCHEME OF
EVALUATION
Merits of Median
It can be easily understood and computed
It is not affected by extreme values
It can be determined graphically (Ogives)
Demerits of Median
It is not based on all values
It is not capable of further algebraic treatment
It is not based on all values
Merits of Mode
In many cases it can be found by inspection
It is not affected by extreme values
It can be calculated for distributions with open end classes
It can be located graphically
Demerits of Mode
It is not based on all values
It is not capable of further mathematical treatment
It is much affected by sampling fluctuations
QUESTION_T
DESCRIPTIVE_QUESTION
YPE
QUESTION_ID 125369
For the following data
i.
QUESTION_T
ii.
EXT
iii.
SCHEME OF
EVALUATION
Find the straight line trend
Find the trend values
Estimate the sale for the year 2012
(3 Marks)
a=
=
b=
=90
=
(1 Mark)
=2
(1 Mark)
i.
Hence, YC = a+b X = 90+2X
ii.
Computation of trend values :
1 Mark)
When X= 3, YC = 90+2X = 90+2(-3) = 84
(0.5 Mark)
X= -2, YC = 90+2X = 90+2(-2) = 86 (0.5 Mark)
iii.
QUESTION_TYPE
X= -1, YC = 90+2X = 90+2(-1) = 88
(0.5 Mark)
X= 0, YC = 90+2X = 90+2(0) = 90
(0.5 Mark)
X= 1, YC = 90+2X = 90+2(1) = 92
(0.5 Mark)
X= 2, YC = 90+2X = 90+2(2) = 94
(0.5 Mark)
X= 3, YC = 90+2X = 90+2(3) = 96
(0.5 Mark)
For the year 2012, X=5 so YC = 90+2(5) = 90+10 =100 (0.5 Mark)
DESCRIPTIVE_QUESTION
QUESTION_ID
125370
QUESTION_TEXT
Define covariance. State its 8 properties
Covariance is a measure of association between two random
variables. (2mark)
Let X and Y be two random variables. Then the covariance is
defined as
Cov(X,Y)=E[{X-E(X)}{Y-E(Y)}]
=E[XY-XE(Y)-YE(X)+E(X)E(Y)]
=E(XY)-E(Y)E(X)-E(X)E(Y)+E(X)E(Y)
=E(XY)-E(X)E(Y)
Properties : (1+1+1+1+1+1+1+1)
SCHEME OF
EVALUATION
1.
Cov(aX,bY)=abCov(XY)
2.
Cov(X+a,Y+b)=Cov(XY)
3.
Cov(
4.
Cov(aX+b, cY+ d)
5.
Cov(X+Y, Z)=Cov(XZ) + Cov(Y,Z)
6.
7.
V(
,
)=
= acCov(XY)
Cov(aX+bY, cX+dY)= ac
+(ad+bc)
Cov(X,Y)
+ ac
If X1, X2 … Xn be random variable, then
)=
+2
8.
If X1, X2 … Xn be independent random variable random
variable, then
V(
)=