Question

Let πx denote the proportion of households in some county with x cars. Suppose π1 = 0.32, π2 = 0.38, π3 = 0.1, π4 = 0.02, π5 = 0.01 and all the other households in the county have 0 cars.

(a) What is the average number of cars per household in this county? What is the corresponding variance?

(b) If a household is selected at random and X is its number of cars, draw the pmf of X.

(c) What is the expectation and what is the standard deviation of X?

(d) Compute the skewness of X.

Answer 1

sum of all probability must be 1

so, π0 = 1 - 0.32 - 0.38 - 0.1 - 0.02-0.01 =0.17

a)

X	P(X)	X*P(X)	X² * P(X)
1	0.32	0.32	0.32
2	0.38	0.76	1.52
3	0.1	0.3	0.9
4	0.02	0.08	0.32
5	0.01	0.05	0.25
0	0.17	0	0

	P(X)	X*P(X)	X² * P(X)
total sum =	1	1.51	3.31

mean = E[X] = Σx*P(X) =            1.51
          
E [ X² ] = ΣX² * P(X) =            3.31
          
variance = E[ X² ] - (E[ X ])² =            1.0299

the average number of cars per household in this county = 1.51

b)

X	P(X)
1	0.32
2	0.38
3	0.1
4	0.02
5	0.01
0	0.17

c)

mean = E[X] = Σx*P(X) =            1.51
          
E [ X² ] = ΣX² * P(X) =            3.31
          
variance = E[ X² ] - (E[ X ])² =            1.0299
          
std dev = √(variance) =            1.014839889

so, expectation = 1.51

std dev = 1.0148

d)

X	P(X)			(x-mean)³P(x)
1	0.32			-0.04245
2	0.38			3.04
3	0.1			2.7
4	0.02			1.28
5	0.01			1.25
0	0.17			0

µ3 = Σ(x-mean)³P(x) = 8.2276

skewness=µ3/σ³ = 8.2276/1.01484³ = 7.8719

since,skewness is >0,so distribution is skewed to left