Question

1. A mail-order computer business has six telephone lines. X denotes the number The probability mass function of X is given of the lines in use at a specified time. in the accompanying table. 0 2 20 3 25 20 4 5 06 6 04 P(Xx) 10 .15 Find P(1< X <4) and P(X-2) b. E(X), Var(X) c. In a random sample of 10 randomly selected times, let Y be the number of the times that exactly two lines are in use. ldentify the probability distribution of Y and find P(Ys 2) d. Find the exact probability that in a random sample of 64 selected times, there will be 12 times that two lines are in use. Now approximate the probability using Normal distribution

Answer 1

a)

P(1<X<4) =P(X=2))+P(X=3)=0.2+0.25=0.45

P(X=2)=0.2

b)

x	P(x)	xP(x)	x2P(x)
0	0.1	0.000	0.000
1	0.15	0.150	0.150
2	0.2	0.400	0.800
3	0.25	0.750	2.250
4	0.2	0.800	3.200
5	0.06	0.300	1.500
6	0.04	0.240	1.440
	total	2.640	9.340
	E(x) =μ=	ΣxP(x) =	2.6400
	E(x2) =	Σx2P(x) =	9.3400
	Var(x)=σ2 =	E(x2)-(E(x))2=	2.370

from above E(X)=2.64

Var(X)=2.370

c)

Y follows binomial distribution with parameter n=10 and p=0.2

hence P(Y<=2) =P(Y=0)+P(Y=1)+P(Y=2)

=¹⁰C₀(0.2)⁰(0.8)¹⁰+¹⁰C₁(0.2)¹(0.8)⁹+¹⁰C₂(0.2)²(0.8)⁸=0.6778

d)

n=	64	p=	0.2000
here mean of distribution=μ=np=		12.8
and standard deviation σ=sqrt(np(1-p))=		3.2000
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

probability =

P(11.5<X<12.5)

=

P(-0.41<Z<-0.09)=

0.4641-0.3409=

0.1232