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

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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
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