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)

$P(x0) + P(x1) + P(x2) + P(x3) + P(x4) + P(x5) = 1$

$P(x) = \frac{1}{10}$ for x = 0

$P(0) = \frac{1}{10}$

$P(x) = \frac{kx}{10}$ for x = 1, 2, 3

$P(1<= x <= 3) = \frac{k}{10} + \frac{2k}{10} + \frac{3k}{10} = \frac{6k}{10}$

$P(x) = k(6-x)$ for x = 4, 5

$P(4<= x <= 5) = k(6-4) + k(6-5) = 3k$

$1 = \frac{1}{10} + \frac{6k}{10} + 3k$

$1 - \frac{1}{10} = \frac{9}{10} = \frac{6k + 30k}{10} = \frac{36k}{10}$

$(9 * 10) = (36k * 10)$

Answer: 

$K = \frac{90}{360} = 0.25$

Solution b)

Part i) 3 visits in 1 hour

$P(x=3) = \frac{0.25x}{10} = \frac{0.25* 3}{10} = 0.075$

Answer: 0.075

Part ii) less than 2 visits in 1 hour

$P((x = 0 \, or \,x=1) = \frac{1}{10} + \frac{0.25}{10} = 0.125$

Answer: 0.125

Part iii) more than 6 visits in 2 hours

Let x be the visits in the first hour and y be visits in the second hour

$Max(x+y) = 10$

$Min(x+y) = 7$

$P(x + y > 6) = P(x + y = 7 \, or \,x + y = 8 \\\, or \,x + y = 9\, or \, x + y = 10 \, or \,x + y = 11\, or \,x + y = 12)$

$P(x + y = 7) = P(x = 5 \,and \, y = 2) + P(x = 4 \, and \, y = 3) + P(x = 3 \,and \, y = 4) + P(x = 2 \,and \,y = 5)$

$= (0.25(6-5) * \frac{0.25 * 2}{10} + 0.25(6-4) * \frac{0.25 * 3}{10} + \frac{0.25 * 3}{10} * 0.25(6-4) + \frac{0.25 * 2}{10} * (0.25(6-5)$

= 0.1

$P(x + y = 8) = P(x = 5 \, and \,y = 3) + P(x = 4 \,and \,y = 4) + P(x = 3 \,and \,y = 5)$

$= (0.25(6-5) * \frac{0.25 * 3}{10} + 0.25(6-4) * 0.25(6-4) + \frac{0.25 * 3}{10} * (0.25(6-5)$

= 0.2875

$P(x + y = 9) = P(x = 5 \, and \,y = 4)+ P(x = 4 \, and \,y = 5)$

$= (0.25(6-5) * (0.25(6-4) + (0.25(6-4) * (0.25(6-5)$

= 0.125

$P(x + y = 10) = P(x = 5 \, and \, y = 5)$

$= (0.25(6-5) * (0.25(6-5)$

= 0.0625

$P(x + y = 11) = 0$

$P(x + y = 12) = 0$

$P(x + y > 6) = 0.1 + 0.2875 + 0.125 + 0.0625$

Answer: 0.575

Part c) Probability of having less than 253 visits in 100 hours

Average visits per hour (x) = $\frac{253}{100}$ = 2.53

Expectation of x = $\mu = \sum{x p(x)}$

X            px                      xpx                        p(x) * (x-\mu)^2

0             $\frac{1}{10}$                       0                        1.296

1            $\frac{0.25}{10}$                   0.025                     0.169

2            $\frac{0.5}{10}$                     0.1                         0.128

3             $\frac{0.75}{10}$                  0.225                     0.027

4             0.5                        2                         0.08

5             0.25                   1.25                       0.49

$\mu = 0 + 0.025 + 0.1 + 0.225 + 2 + 1.25 = 3.6$

$Var(x) = \sum{ p(x) * (x-\mu)^2} = 2.433$

$\sigma = \sqrt{2.433} = 1.5598$

$Z value = \frac{x- \mu}{sd} = \frac{2.53- 3.6}{1.5598} = -0.686$

At z value $= -0.686$ ,$p = 0.2464$

Answer: 0.2464