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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 (u2(sigma),u+sigma)
e. Find the actual probability that x falls within the interval of part d
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 (u2(sigma),u+sigma) e. Find the actual probability that x falls within the interval of part d

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you need a documented answer ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe I solved a and b, but and stuck on c right now

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2The expected value is given by \[\large E(X)=\sum_{}^{}xp(x)=\frac{1}{15}+\frac{4}{15}+\frac{9}{15}+\frac{16}{15}+\frac{25}{15}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you kropot, I got 55/15 for the mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The formula to figure out standard deviation seems a lot longer and I'm not sure how to format it here

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2The standard deviation is found by taking the square root of the variance. \[\large Var(X)=E(X^{2})[E(X)]^{2}\] \[=1^{2}\times\frac{1}{15}+2^{2}\times\frac{2}{15}+3^{2}\times\frac{3}{15}+4^{2}\times\frac{4}{15}+5^{2}\times\frac{5}{15}(\frac{55}{15})^{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer I got was 1.556 for standard deviation. Is that correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0σ = Square Root [ ∑ (X µ) 2 P(X) ] this was the formula I used

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2I get 1.556 for the variance. What do you need to do to get the standard variation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need to square root it?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2That is the correct S.D.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you. Now I'm lost on d and e. I feel confused on what to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for d we just plug in (55/15  2(1.247), 55/15 + 1.247) ?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2Yes, I believe you are correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, the answer for d is (1.173, 4.914) ?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2Yes, you are correct for part d.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you. How do we solve part e? Do we use chebyshev's theorem?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2This 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 9/15. I'm just curious where you got the numbers 2,3, and 4? Thank you

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2The numbers 2, 3 and 4 are the values that x can take that lie between 1.173 and 4.914.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, I see. I just noticed that now. Thank you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I will be closing this question, thanks so much for your help. I have another question that I might post later.

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2Cool. If I am logged in I'll take a look.
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