Each phone call by Ali consumes an amount of time that follows an exponential distribution with mean 5 minutes. The number of different phone calls Ali makes on any given day has a Poisson distribution with mean 3. Assume that a single call always falls within a single day (no calls continue past midnight).

Question

Further, suppose that the number of phone calls that Ali makes on different days are independent, and that the lengths of the phone calls are also independent of each other. For simplicity, also assume that different phone calls never overlap and that there are 30 days in each given month.
Let 𝑋 be the total number of minutes Ali spends on the phone during one month.
Find 𝐄(𝑋) and 𝖵𝖺𝗋(𝑋).

Using the central limit theorem and a standard normal table or calculator, find the probability that the total number of phone calls Ali makes during an entire year (12 months of 30 days each) is between 1100 and 1200.
(Note that in this part of the question, you are asked about the number of phone calls, not the number of minutes.)

Answer 1

2880

Answer 2

E[X]=450
Var(X)=4500
(1100<=P<=1200)=0.2996
(but i'm not sure)

Answer 3

For 3 I calculated as 0.2741. 1 and 2 are correct.

Answer 4

30*E_phones*E_minutes

Answer 5

3. 0.2712

Answer 6

How did you guys find the variance?

Answer 7

Variance, var(x) = 7200 ?

Answer 8

var = 135000= 30^2*fmla

Answer 9

Var[∑i=1NXi] = E[N]Var[X1]+(E[X1])2Var(N).
=(5*3+25*5)*30=4500

Does that look right?

Answer 10

Exp distribution mean = 5
mean = 1/lambda
variance = 1/lamdba^2
although the sum of exponential random variables over a day should be
mean=k/lambda variance = k/lambda^2 where k = 3 here

Answer 11

@df .. how did you get 7200 as var?

Answer 12

7200 is wrong

Answer 13

Var[∑i=1NXi] = E[N]Var[X1]+(E[X1])2Var(N).
=(3*sqrt(5)+25*3) *30 --?

Answer 14

I missed calculated (30*15+(15^2)* 30) = 7200 :(

Answer 15

does anyone have final correct answers

Answer 16

X1 is an exponential RV so the mean cant be equal to variance

Answer 17

I got
E[X]=450
Var(X)=4500
(1100<=P<=1200)=0.2712

Answer 18

how do you get 0.2712?
1100 - 1080 / sqrt(1080) < P < 1200 - 1080 / sqrt(1080) ?

Answer 19

Var[∑i=1NXi] = E[N]Var[X1]+(E[X1])2Var(N).
=(3*25+25*3)*30=4500

X1 is exponential distribution, var does not equal E
N is poisson distribution, var = E

Answer 20

E[x]=450 minutes is OK for a month
var(X)=4500 minutes is the number you get from calculaltions, but does it make any sense as a variance?

Answer 21

I'm skeptical of var(x) = 4500. That's taking the variance for a day and multiplying it by 30, which is the naive approach. But the 30 days constant should go inside of the variance, so it get's effectively squared.

Answer 22

Agree with mean = 450

As the variance is concerned: we CAN'T consider 30*Xi and so 30^2*var(Xi) because it does not make sense to multiply by 30 the 'randomness' of a single day. Actually, is it more appropriate to think at X as x1+x2+...+x30.

Given that all the days are independent, Var(x1+...+x30) = var(x1) + ... + var(x30) = 30 * var(xi) = 4500

Answer 23

I think the answer others have gotten for part 3 is slightly off: P(1100 ≤ P ≤ 1200) = P((1100 - 1080)/sqrt(1080) ≤ Z ≤ (1200-1080)/sqrt(1080)) = P(20/sqrt(1080) ≤ Z ≤ 120/sqrt(1080)) = P(Z ≤ 3.65148...) - P(Z ≤ 0.060858...) = 0.99987 - 0.72907 = 0.2708, not 0.2712. But if you're only giving 2 decimal places, it doesn't matter.