Question

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

Answer 1

are you need a documented answer ?

Answer 2

Yes, I think so

Answer 3

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

Answer 4

@ganeshie8

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

you need to square root it?

Answer 13

Correct.

Answer 14

Ok so it is 1.247?

Answer 15

That is the correct S.D.

Answer 16

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

Answer 17

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

Answer 18

Yes, I believe you are correct.

Answer 19

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

Answer 20

Yes, you are correct for part d.

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

You're welcome :)

Answer 27

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

Answer 28

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