Statistics

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).
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.)

  1. 👍 7
  2. 👎 1
  3. 👁 1,405
  1. 2880

    1. 👍 0
    2. 👎 5
  2. E[X]=450
    Var(X)=4500
    (1100<=P<=1200)=0.2996
    (but i'm not sure)

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

    1. 👍 1
    2. 👎 2
  4. 30*E_phones*E_minutes

    1. 👍 3
    2. 👎 0
  5. 3. 0.2712

    1. 👍 5
    2. 👎 0
  6. How did you guys find the variance?

    1. 👍 1
    2. 👎 0
  7. Variance, var(x) = 7200 ?

    1. 👍 0
    2. 👎 3
  8. var = 135000= 30^2*fmla

    1. 👍 0
    2. 👎 3
  9. Var[∑i=1NXi] = E[N]Var[X1]+(E[X1])2Var(N).
    =(5*3+25*5)*30=4500

    Does that look right?

    1. 👍 5
    2. 👎 1
  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

    1. 👍 0
    2. 👎 0
  11. @df .. how did you get 7200 as var?

    1. 👍 0
    2. 👎 0
  12. 7200 is wrong

    1. 👍 2
    2. 👎 0
  13. Var[∑i=1NXi] = E[N]Var[X1]+(E[X1])2Var(N).
    =(3*sqrt(5)+25*3) *30 --?

    1. 👍 0
    2. 👎 0
  14. I missed calculated (30*15+(15^2)* 30) = 7200 :(

    1. 👍 0
    2. 👎 1
  15. does anyone have final correct answers

    1. 👍 0
    2. 👎 0
  16. X1 is an exponential RV so the mean cant be equal to variance

    1. 👍 0
    2. 👎 0
  17. I got
    E[X]=450
    Var(X)=4500
    (1100<=P<=1200)=0.2712

    1. 👍 4
    2. 👎 0
  18. how do you get 0.2712?
    1100 - 1080 / sqrt(1080) < P < 1200 - 1080 / sqrt(1080) ?

    1. 👍 0
    2. 👎 0
  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

    1. 👍 1
    2. 👎 1
  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?

    1. 👍 0
    2. 👎 0

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