Solution Of Assignment No.2 (Course STA301) Fall 2012
Question 1:
A)The fourth mean moment of a symmetric distribution is 243. What would be the value of the standard deviation in order that the distribution may be meso-kurtic?
SOLUTION:-
Standard is equal to 2nd moment We are that m4 = 243 . Therefore
m4 = 243
m4/m2 = b3 = 3 i.e = 243/m2^2 = 3
or 243/3 = 81 or m2 = 9
Standard deviation is equal to m 2nd moment
B)Two candidates X and Y at the B.A (Hons.) Examination obtained the following data.
Candidate X Candidate Y
= 43.2 = 20.5
S.D = 24.25 S.D = 15.6
Which of the candidate showed a more consistent performance?
SOLUTION:-
By calculating the co-efficient of variation in each case.
Candidate A
C V = s/x * 100C V = 24.25/43.2 * 100
C V = 56.13%
Candidate B
C V = s/y * 100C V = 15.6/20.5*100
C V = 76.097%
We see that coefficient of variation for the candidate A smaller than that for the
candidate B.Hence candidate A is more consistent than candidate B
Question 2:
A)If coefficient of skewness = 0, then what would you say about the shape of the distribution?
SOLUTION:-
For symmetric distributions or curves, the coefficient is zero. The shape is symmetric
X = Number of years of service
Y = weekly wage rate
n = 23; ∑X = 2433; ∑Y = 4254, ∑ = 281019; = 841786 and = 482788
If the correlation coefficient indicates that there does exist a relationship between X and Y, compute the least squares line of regression Y = a + b X.
SOLUTION:-
Where bxy = n (XY)-(X)(Y) / n (X^2)-(X)^2
bxy = 23(482788) (2433)(4254) / 23(281019) (2433)^2
bxy = 11104124-10349982 / 6463437-5919489
bxy = 754142/543948
b = 1.3864
a = y-bx
y = ￥y/n =
4254/23 =
184.96
x =
￥x/n =
2433/23 = 105.78
a = 184.96-1.39(105.78)
a = 184.96-147.0342
a = 37.92
Hence
y = 37.92+1.39X
Is estimated regression equation appropriate for predicting the no
of years (x) given weekly wage rate