Begin2.DVI

I11-19. (i)
∑n

x=0

x^2 f(x) =

∑n

x=0

[x(x−1) +x]f(x)

=

∑n

x=0

x(x−1)f(x) +

∑n

x=0

xf(x)

=

∑n

x=0

x(x−1)f(x) +np

Usex(x−1)

(

n

x

)

=x(x−1)

n(n−1)(n−2)!

x(x−1)(x−2)!(n−x)!=n(n−1)

(

n− 1

x− 2

)

and write

∑n

x=0

x^2 f(x) =

∑n

x=2

n(n−1)

(

n− 2

x− 2

)

pxqn−x+np=n(n−1)p^2

∑n

x=2

(

n− 2

x− 2

)

px−^2 qn−x+np

Note that by shifting the summation index

n∑− 2

X=0

(

n− 2

X

)

pXqn−^2 −X=

n∑− 2

X=0

(

n− 2

n− 2 −X

)

pXqn−^2 −X= (p+q)n−^2 = 1

so that

∑n

x=0

x^2 f(x) =n(n−1)p^2 +np

σ^2 =

∑n

x=0

(x−μ)^2 f(x) =

∑n

x=0

(x^2 − 2 xμ+μ^2 )f(x)

=

∑n

x=0

x^2 f(x)− 2 μ

∑n

x=0

xf(x) +μ^2

∑n

x=0

f(x)

=

∑n

x=0

x^2 f(x)−μ^2 =E[(x^2 )]−(E[x])^2

=n(n−1)p^2 +np−n^2 p^2

=np(1−p) =npq

I11-21. (a)f(x) =

( 3

x

)( 9

5 −x

)

( 12

5

)

∑ 5

x=1f(x) = 1

f(x) = 7/ 44 , f(1) = 21/ 44 , f(2) = 7/ 22 , f(3) = 1/ 12 , f(4) =f(5) = 0

(b) The probability thatn= 6items are selected with zero nondefectives is

f(x) =

( 3

x

)( 9

6 −x

)

( 12

6

) evaluated atx= 0orf(0) = 1/ 11. The probability that 6 items are

selected and there is 1 defective isf(1) = 9/ 22. We are not interested inf(x)for x

Solutions Chapter 11