Question

The number of days, X, that it takes the post office to deliver a notarized letter cross- country between City A and City B has the probability distribution 4 f(x) .5 3.2 (a) Find the expected number of days and the standard deviation of the number of days. (b) A company in City A sends a notarized letter to a company in City B with a notarized return receipt request that is to be mailed immediately upon receiving the letter. Find the probability distribution of total number of days from the time the letter is mailed until the return receipt arrives back at the company in City A. Assume the two delivery times have the same distribution and are independent. (c) A single notarized letter is sent from City A on each of 100 different days. What is the approximate probability that more than 25 of the letters take 5 days to reach City B?

Answer 1

a) $E[X]=\sum x_if(x_i)=3(0.5)+4(0.3)+5(0.2)=3.7$

$E[X^2]=\sum x^2_if(x_i)=3^2(0.5)+4^2(0.3)+5^2(0.2)=14.3$

Standard deviation = $\sqrt{Var(x)}=\sqrt{E[X^2]-(E[X])^2}=\sqrt{14.3-(3.7)^2}=0.78$

c) Ther is a two way trip From A to B and then back B to A

If it took 3 days for the letter from A to B then it could take

3days or 4 or 5 days from B to A i.e (3,3),(3,4) or (3,5)

Similarly If it took 4 days for the letter from A to B then it could take

3days or 4 or 5 days from B to A i.e (4,3),(4,4) or (4,5)

Similarly If it took 5 days for the letter from A to B then it could take

3days or 4 or 5 days from B to A i.e (5,3),(5,4) or (5,5)

x(Total number of days)	6	7	8	9	10
f(x)	(0.5)(0.5)=0.25	(0.5)(0.3)+(0.3)(0.5)=0.3	(0.3)(0.3)+(0.5)(0.2)+(0.2)(0.5)=0.29	(0.3)(0.2)+(0.2)(0.3)=0.12	0.2(0.2)=0.04

c) This can be considered as binomial with probability of success (letter reasching at 5 days) p = 0.2

Total number of trials n=100

$P(X>25)=\sum_{x=26}^{100}b(x,0.2,100)=1-P(X\leq 25)=1-0.9125=0.0875$