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anonymous

  • one year ago

Suppose that the probability distribution of a random variable x can be described by the formula p(x)= x/15 for each of the values x=1,2,3,4,5. For example, p(x=2)=2/15 a. Write out the probability distribution of x in tabular form. Keep p(x) in Fraction b. Show that the probability distribution of x satisfies the properties of a discrete probability distribution. c. Calculate the mean and standard deviation of x d. Compute the interval (u-2(sigma),u+sigma) e. Find the actual probability that x falls within the interval of part d

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  1. anonymous
    • one year ago
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    are you need a documented answer ?

  2. anonymous
    • one year ago
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    Yes, I think so

  3. anonymous
    • one year ago
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    I believe I solved a and b, but and stuck on c right now

  4. anonymous
    • one year ago
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    @ganeshie8

  5. kropot72
    • one year ago
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    The expected value is given by \[\large E(X)=\sum_{}^{}xp(x)=\frac{1}{15}+\frac{4}{15}+\frac{9}{15}+\frac{16}{15}+\frac{25}{15}\]

  6. anonymous
    • one year ago
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    Thank you kropot, I got 55/15 for the mean

  7. anonymous
    • one year ago
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    The formula to figure out standard deviation seems a lot longer and I'm not sure how to format it here

  8. kropot72
    • one year ago
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    The standard deviation is found by taking the square root of the variance. \[\large Var(X)=E(X^{2})-[E(X)]^{2}\] \[=1^{2}\times\frac{1}{15}+2^{2}\times\frac{2}{15}+3^{2}\times\frac{3}{15}+4^{2}\times\frac{4}{15}+5^{2}\times\frac{5}{15}-(\frac{55}{15})^{2}\]

  9. anonymous
    • one year ago
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    The answer I got was 1.556 for standard deviation. Is that correct?

  10. anonymous
    • one year ago
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    σ = Square Root [ ∑ (X- µ) 2 P(X) ] this was the formula I used

  11. kropot72
    • one year ago
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    I get 1.556 for the variance. What do you need to do to get the standard variation?

  12. anonymous
    • one year ago
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    you need to square root it?

  13. kropot72
    • one year ago
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    Correct.

  14. anonymous
    • one year ago
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    Ok so it is 1.247?

  15. kropot72
    • one year ago
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    That is the correct S.D.

  16. anonymous
    • one year ago
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    Thank you. Now I'm lost on d and e. I feel confused on what to do

  17. anonymous
    • one year ago
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    for d we just plug in (55/15 - 2(1.247), 55/15 + 1.247) ?

  18. kropot72
    • one year ago
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    Yes, I believe you are correct.

  19. anonymous
    • one year ago
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    so, the answer for d is (1.173, 4.914) ?

  20. kropot72
    • one year ago
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    Yes, you are correct for part d.

  21. anonymous
    • one year ago
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    Thank you. How do we solve part e? Do we use chebyshev's theorem?

  22. kropot72
    • one year ago
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    This is a discrete probability distribution. Therefore the values that x can take between 1.173 and 4.914 are 2, 3 and 4. So just add the probabilities for x = 2, x =3 and x = 4.

  23. anonymous
    • one year ago
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    I got 9/15. I'm just curious where you got the numbers 2,3, and 4? Thank you

  24. kropot72
    • one year ago
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    The numbers 2, 3 and 4 are the values that x can take that lie between 1.173 and 4.914.

  25. anonymous
    • one year ago
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    Ah, I see. I just noticed that now. Thank you

  26. kropot72
    • one year ago
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    You're welcome :)

  27. anonymous
    • one year ago
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    I will be closing this question, thanks so much for your help. I have another question that I might post later.

  28. kropot72
    • one year ago
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    Cool. If I am logged in I'll take a look.

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