Answer to Question #127771 in Statistics and Probability for Jack

Question

The number of visits to a website in one hour has the following probability mass function

{1/10 , 𝑥 = 0

𝑝(x)= { 𝑘𝑥/10 , 𝑥 = 1,2,3

{ 𝑘(6−𝑥)/15 , 𝑥 = 4,5

a) Determine the value of 𝑘.

With the value of 𝑘 obtained:

b) Find the probability of having

i) 3 visits in one hour.

ii) less than 2 visits in one hour.

iii) more than 6 visits in two hours.

c) By using central limit theorem, estimate the probability of having less than 253 visits in 100 hours.

Answer 1

Solution:

a)

The sum of all probability is 1.

So,

P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=1.

1/10+k/10+2k/10+3k/10+2k/15+k/15=1.

1/10+3k/5+k/5=1

4k/5=9/10

k=9/8.

So,

P(X=0)=1/10,

P(X=1)=9/80,

P(X=2)=9/40,

P(X=3)=27/80,

P(X=4)=3/20,

P(X=5)=3/40.

b)

i) P(X=3)=27/80,

The probability of  3 visits in one hour is equal to 27/80.

ii) P(X<2)=P(X=0)+P(X=1)=1/10+9/80=17/80,

The probability of  event that less than 2 visits in one hour is equal to 17/80.

iii) k₁+k₂>6:

(k₁,k₂)=(2,5),(3,4),(3,5),(4,3)(4,4),(4,5),(5,5)

P(X1+X2>6)=P(X1=2,X2=5)+P(X1=3,X2=4)+P(X1=3,X2=5)+P(X1=4,X2=3)+P(X1=4,X2=4)+P(X1=4,X2=5)+P(X1=5,X2=5)

P(X1+X2>6)=(9/40)(3/40)+(27/80)(3/20)+(27/80)(3/40)+(3/20)(27/80)+(3/20)(3/20)+(3/20)(3/40)+(3/40)(3/40)=117/640

The probability of  event that more than 6 visits in two housr is equal to 117/640.

c) $\mu=0\cdot P(X=0)+1\cdot P(X=1)+2\cdot P(X=2)+3\cdot P(X=3)+4\cdot P(X=4)+5\cdot P(X=5)=0\cdot 1/10+1\cdot 9/80+2\cdot 9/40+3\cdot 27/80+4\cdot 3/20+5\cdot 3/40=51/20.$

$\mu=51/20.$

$\sigma^2=E[X^2]-\mu^2$

$E[X^2]=0\cdot P(X=0)+1\cdot P(X=1)+4\cdot P(X=2)+9\cdot P(X=3)+16\cdot P(X=4)+25\cdot P(X=5)=0\cdot 1/10+1\cdot 9/80+4\cdot 9/40+9\cdot 27/80+16\cdot 3/20+25\cdot 3/40=333/40.$

$\sigma^2=333/40-(51/20)^2=729/400$

$\sigma=\sqrt{729/400}=27/20.$

$n=100$

$\beta=(253-n\cdot\mu)/(\sigma\cdot n^{1/2})$

$\beta=(253-100\cdot\mu)/(10\sigma)=-4/27$

$P(S_{n}<253)=P((S_{n}-n\cdot\mu)/(\sigma\cdot n^{1/2})<(253-n\cdot\mu)/(\sigma\cdot n^{1/2}))$

$P(S_{n}<253)=P((S_{n}-n\cdot\mu)/(\sigma\cdot n^{1/2})<-4/27)=\Phi(-4/27)$

$P(S_{100}<253)=\Phi(-4/27)$

$\Phi(-4/27)\approx0.4411$ (Here we use Wolfram Alpha)

The probability of having less than 253 visits in 100 hours is approximately equal to 0.4411.