A sociological investigation is concerned with the home enviroment in relatively

Question

A sociological investigation is concerned with the home enviroment in relatively large family. suppose the number of children of a family is denoted by an ordered pair(boys,girls). the ten family in the investigation pool have children as given by(0,5),(0,6),(1,3),(1,4),(1,5),(1,6),(2,3),(4,4),(3,3),(4,3). A family is selected randomly out of those 10 families and let Y be the random variable indicates the total number of children in the selected family. Find:
a) Possible values of Y
b) the probability function of Y
c)the distribution function of Y
d)the graphs of the probability function and the distribution function of Y
e)the probability that a selected family has more than 6 children
f)E(Y)
g)E(3Y-10)
h)E(Y^2)
i)V(Y)
j)V(3Y-10)

Explanation / Answer

There is quite a bit here, but I'll go through step by step.

a)
Just adding up the values we can find the possible values of Y:
4, 5,6,7, and 8

b)
Now we need to account for the number of times each value of Y comes up. Recall we have a total of 10 families.

Y=4, 1 time (so it has a probability p=1/10 or 0.1)
Y=5, 3 times (p=3/10=0.3)
Y=6, 3 times (p=0.3)
Y=7, 2 times (p=0.2)
Y=8, 1 time (p=0.1)

This will give us:

f(Y)= {0.1  Y=4
         0.3 Y=5
         0.3 Y=6
         0.2 Y=7
         0.1 Y=8
         0 else

c)
The CDF is the cumulative distribution function (or just the distribution function). This means we should add the probabilities as we go along:

F(Y)= {0.0 Y<4
          0.1 4Y<5
          0.4 5Y<6
          0.7 6Y<7
          0.9 7Y<8
          1.0 Y>8

d) These graphs can be easily obtained as step-wise functions from the answers detailed in b and c.

e)
This is basically asking for P(Y>6). This is the same as 1-P(Y6) which can be obtained from the CDF in part c. This gives our answer = 1-0.7

f) A discrete expectation is the sum of the Y values multiplied by their probabilities:
E(Y) = (0.1)(4)+(0.3)(5)+(0.3)(6)+(0.2)(7)+(0.1)(8)

g) Expectation is a linear operator, so we can manipulate and pull out constants, and don't forget that the expectation of a constant is that constant:
E(3Y-10) = 3E(Y)-E(10) = 3E(Y)-10
Just plug in the answer from part f as your expectation.

h) When you square an expectation, you obtain it the same way as you did in f, but the values of Y are squared.
E(Y2)= (0.1)(42)+(0.3)(52)+(0.3)(62)+(0.2)(72)+(0.1)(82)

i)Var(Y)=E(Y2)-E(Y)2
You have already determined E(Y) and E(Y2) in parts g and h.

j)Variance is similar to expectation as numbers can be pulled out, but they're squared and the variance of a constant is always 0.
Var(3Y-10) = 9Var(Y)
We just determined the value of Var(Y) in part i.

These answers aren't solved all the way, but pretty much almost. Good luck!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.