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

- anonymous

- schrodinger

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- anonymous

are you need a documented answer ?

- anonymous

Yes, I think so

- anonymous

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

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## More answers

- anonymous

- 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

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

- anonymous

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

- 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

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

- anonymous

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

- kropot72

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

- anonymous

you need to square root it?

- kropot72

Correct.

- anonymous

Ok so it is 1.247?

- kropot72

That is the correct S.D.

- anonymous

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

- anonymous

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

- kropot72

Yes, I believe you are correct.

- anonymous

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

- kropot72

Yes, you are correct for part d.

- anonymous

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

- 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

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

- kropot72

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

- anonymous

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

- kropot72

You're welcome :)

- anonymous

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

- kropot72

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

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