## 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

1. anonymous

are you need a documented answer ?

2. anonymous

Yes, I think so

3. anonymous

I believe I solved a and b, but and stuck on c right now

4. anonymous

@ganeshie8

5. kropot72

The expected value is given by $\large E(X)=\sum_{}^{}xp(x)=\frac{1}{15}+\frac{4}{15}+\frac{9}{15}+\frac{16}{15}+\frac{25}{15}$

6. anonymous

Thank you kropot, I got 55/15 for the mean

7. anonymous

The formula to figure out standard deviation seems a lot longer and I'm not sure how to format it here

8. kropot72

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

9. anonymous

The answer I got was 1.556 for standard deviation. Is that correct?

10. anonymous

σ = Square Root [ ∑ (X- µ) 2 P(X) ] this was the formula I used

11. kropot72

I get 1.556 for the variance. What do you need to do to get the standard variation?

12. anonymous

you need to square root it?

13. kropot72

Correct.

14. anonymous

Ok so it is 1.247?

15. kropot72

That is the correct S.D.

16. anonymous

Thank you. Now I'm lost on d and e. I feel confused on what to do

17. anonymous

for d we just plug in (55/15 - 2(1.247), 55/15 + 1.247) ?

18. kropot72

Yes, I believe you are correct.

19. anonymous

so, the answer for d is (1.173, 4.914) ?

20. kropot72

Yes, you are correct for part d.

21. anonymous

Thank you. How do we solve part e? Do we use chebyshev's theorem?

22. 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.

23. anonymous

I got 9/15. I'm just curious where you got the numbers 2,3, and 4? Thank you

24. kropot72

The numbers 2, 3 and 4 are the values that x can take that lie between 1.173 and 4.914.

25. anonymous

Ah, I see. I just noticed that now. Thank you

26. kropot72

You're welcome :)

27. anonymous

I will be closing this question, thanks so much for your help. I have another question that I might post later.

28. kropot72

Cool. If I am logged in I'll take a look.