Question

First consider the following probability distribution for the total number of devices that connect to a home router.

X             p(x)

1             6%

2             4%         

3             2%

4             3%

5             4%

6             4%         

7             4%

8             5%

9             8%

10           10%

11           9%

12           8%

13           8%

14           6%

15           5%

16           4%

17           3%

18           3%

19           2%

20           2%

• Generate plots of the PDF and CDF for this distribution.
  • • Calculate the following o The expectation and the variance of the number of devices.
  • o The probability that a customer has 10 or fewer devices connecting.
  • o The probability that a customer has 14 or more devices connecting.
  • o The 50th percentile of the distribution.
  • • Now assume that the amount of data transferred per day is normally distributed with a mean of 15 GB and a standard deviation of 5 GB. Assume the data transferred each day is independent of other days. Calculate the following o The probability a customer transfers more than 18 GB in a day.
  • o The probability a customer transfers more than 18 GB for 3 days in a row.
  • o The 90th percentile of the daily transfer amount.

Answer 1

Answer -->

1. Plot of PDF and CDF are as below
2. The expectation and the variance of the number of devices.

X	P(x)	x*P(x)	x^2*P(x)	F(x)
1	6.0%	0.06	0.06	6.0%
2	4.0%	0.08	0.16	10.0%
3	2.0%	0.06	0.18	12.0%
4	3.0%	0.12	0.48	15.0%
5	4.0%	0.2	1	19.0%
6	4.0%	0.24	1.44	23.0%
7	4.0%	0.28	1.96	27.0%
8	5.0%	0.4	3.2	32.0%
9	8.0%	0.72	6.48	40.0%
10	10.0%	1	10	50.0%
11	9.0%	0.99	10.89	59.0%
12	8.0%	0.96	11.52	67.0%
13	8.0%	1.04	13.52	75.0%
14	6.0%	0.84	11.76	81.0%
15	5.0%	0.75	11.25	86.0%
16	4.0%	0.64	10.24	90.0%
17	3.0%	0.51	8.67	93.0%
18	3.0%	0.54	9.72	96.0%
19	2.0%	0.38	7.22	98.0%
20	2.0%	0.4	8	100.0%

expected value of X and Variance of X is

Expected Value = E(X) = Sum(X*P(x) )

= 10.21

Variance (x) = E(x^2) - E(x) ^2

= sum (X^2 * p(x) ) - 10.21 ^ 2

=127.75 -104.2441

= 23.5059

3. The probability that a customer has 10 or fewer devices connecting.

P(x<=10) = P(x=1) + P(x=2) +......+ P(x=10) = F(x=10)

= 0.5

4. The probability that a customer has 14 or more devices connecting

P(x >= 14) = 1- P(X<14) = 1-0.75 = 0.25

5. The 50th percentile of the distribution, we need to find x such that P(x<=X) >= 0.50

from the CDF we can observe 50th Percentile of X is 10

6. probability a customer transfers more than 18 GB in a day

Z score =x – μ /σ

=18 – 15/ 5

=0.6

P-value from Z-Table:

P(x<18) = 0.72575

P(x>18) = 1 - P(x<18) = 0.27425

7. probability a customer transfers more than 18 GB for 3 days in a row

Since data transferred each day is independent of other days

probability a customer transfers more than 18 GB for 3 days in a row = probability a customer transfers more than 18 GB for 1st day * probability a customer transfers more than 18 GB for 2nd day * probability a customer transfers more than 18 GB for 3rd day

= 0.27425 *0.27425*0.27425

=0.27425^3

=0.020627

8. The 90th percentile of the daily transfer amount, we need to find X such that P(x<=X) >=0.90

P (x< = X ) = 0.9

X = 20.2077

90th percentile of the daily transfer amount is 20.4077 GB.