anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
are you need a documented answer ?
anonymous
  • anonymous
Yes, I think so
anonymous
  • anonymous
I believe I solved a and b, but and stuck on c right now

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
kropot72
  • kropot72
The expected value is given by \[\large E(X)=\sum_{}^{}xp(x)=\frac{1}{15}+\frac{4}{15}+\frac{9}{15}+\frac{16}{15}+\frac{25}{15}\]
anonymous
  • anonymous
Thank you kropot, I got 55/15 for the mean
anonymous
  • anonymous
The formula to figure out standard deviation seems a lot longer and I'm not sure how to format it here
kropot72
  • kropot72
The standard deviation is found by taking the square root of the variance. \[\large Var(X)=E(X^{2})-[E(X)]^{2}\] \[=1^{2}\times\frac{1}{15}+2^{2}\times\frac{2}{15}+3^{2}\times\frac{3}{15}+4^{2}\times\frac{4}{15}+5^{2}\times\frac{5}{15}-(\frac{55}{15})^{2}\]
anonymous
  • anonymous
The answer I got was 1.556 for standard deviation. Is that correct?
anonymous
  • anonymous
σ = Square Root [ ∑ (X- µ) 2 P(X) ] this was the formula I used
kropot72
  • kropot72
I get 1.556 for the variance. What do you need to do to get the standard variation?
anonymous
  • anonymous
you need to square root it?
kropot72
  • kropot72
Correct.
anonymous
  • anonymous
Ok so it is 1.247?
kropot72
  • kropot72
That is the correct S.D.
anonymous
  • anonymous
Thank you. Now I'm lost on d and e. I feel confused on what to do
anonymous
  • anonymous
for d we just plug in (55/15 - 2(1.247), 55/15 + 1.247) ?
kropot72
  • kropot72
Yes, I believe you are correct.
anonymous
  • anonymous
so, the answer for d is (1.173, 4.914) ?
kropot72
  • kropot72
Yes, you are correct for part d.
anonymous
  • anonymous
Thank you. How do we solve part e? Do we use chebyshev's theorem?
kropot72
  • kropot72
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.
anonymous
  • anonymous
I got 9/15. I'm just curious where you got the numbers 2,3, and 4? Thank you
kropot72
  • kropot72
The numbers 2, 3 and 4 are the values that x can take that lie between 1.173 and 4.914.
anonymous
  • anonymous
Ah, I see. I just noticed that now. Thank you
kropot72
  • kropot72
You're welcome :)
anonymous
  • anonymous
I will be closing this question, thanks so much for your help. I have another question that I might post later.
kropot72
  • kropot72
Cool. If I am logged in I'll take a look.

Looking for something else?

Not the answer you are looking for? Search for more explanations.