Check out some similar questions!

azbrowneyedgirl · May 27, 2010, 07:32 PM

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, then,
P (x=2) =p (2) =2/15

a) Write out the probability distribution of x.

b) Show that the probability distribution of X satisﬁes the properties of a discrete probability distribution.

c) Calculate the mean of x.

d) Calculate the variance, and the standard deviation,

Unknown008 · May 28, 2010, 09:18 AM

Note that we are not supposed to give you direct answers.

For your first part, can you complete this table?

$\begin{array}{|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5\\ <br /> \hline <br /> P(X=x) &... &\frac{2}{15} &... &... &...\\ \hline \end{array}$

This will be your answer to a)

b) For this one, you need to check your notes, there should be somewhere saying how to do that.

c) The mean of x ( $E(X)$ ) is given by $\Sigma xP(X)$ , or $(1\times P(X=1)) + (2 \times P(X = 2)) +... + (5\times P(X = 5))$

d) Variance ( $Var(X)$ ) is given by $E(X^2) - (E(X))^2$

You already have (E(X))^2 from c), to find E(X^2), you need to do something similar to that of E(X), but you square the value of x like this:

$(1^2\times P(X=1)) + (2^2 \times P(X = 2)) +... + (5^2\times P(X = 5))$

The standard deviation( $Sd(X)$ ) is given by $\sqrt{Var(X)}$

Post your answers! :)

duffer987 · Sep 18, 2010, 05:51 PM

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, and 5. For example, then, P(x = 2) = p(2) = 2=15.
a. Write out the probability distribution of x.
p(x) = x 15
p(1) = 1/15
p(2) = 2/15
p(3) = 3/15
p(4) = 4/15
p(5) = 5/15
b. Show that the probability distribution of x satisfies the properties of a discrete probability distribution.
p(1) = 1/15 = .067
p(2) = 2/15 = .133
p(3) = 3/15 = .20
p(4) = 4/15 = .267
p(5) = 5/15 = .333
c. Calculate the mean of x.
µx = ∑ xp(x)
µx = 1p(1) + 2p(2) + 3p(3) + 4p(4) + 5p(5)
µx = 1(.067) + 2(.133) + 3(.20) + 4(.267) + 5(.333)
µx = .067 + .266 + .60 + 1.068 + 1.665
µx = 3.666
d. Calculate the variance, s2x, and the standard deviation, sx.
s2x
o2 = ∑ (x - µx)2 p(x)
= (1 – 3.666)2 p(1) + (2 – 3.666)2 p(2) + (3 – 3.666)2 p(3) + (4 – 3.666)2 p(4) + (5 – 3.666)2 p(5)
= (-2.666)(.067) + (-1.666)(.133) + (-.666)(.20) + (.334)(.267) + (1.334)(.333)
= -.178 + -.222 + -.133 + .089 + .444
= 0
sx
o = √0
o = 0

Unknown008 · Sep 19, 2010, 06:11 AM

For b), you actually have to show that all the probabilities add up to 1. So, above poster is wrong.

For d), Variance is given by:

$\sigma^2 (x) = [(1\times\frac{1}{15}) + (4\times\frac{2}{15}) + (9\times\frac{3}{15}) + (16\times\frac{4}{15}) + (25\times\frac{5}{15})] - [(1\times\frac{1}{15}) + (2\times\frac{2}{15})+(3\times\frac{3}{15})+(4 \times \frac{4}{15}) + (5\times\frac{5}{15})]^2 = 1.55$

Meaning that the above poster is wrong again.

And:

$\sigma(x) = \sqrt{1.55} = 1.25$

duffer987 · Sep 24, 2010, 12:51 PM

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, and 5. For example, then, P(x = 2) = p(2) = 2=15.
a. Write out the probability distribution of x.
p(x) = x 15
p(1) = 1/15
p(2) = 2/15
p(3) = 3/15
p(4) = 4/15
p(5) = 5/15
b. Show that the probability distribution of x satisfies the properties of a discrete probability distribution.
p(1) + p(2) + p(3) + p(4) + p(5) = x

(1 + 2 + 3 + 4 + 5)
15

x = 1
c. Calculate the mean of x.
µx = ∑ xp(x)
µx = 1p(1) + 2p(2) + 3p(3) + 4p(4) + 5p(5)
µx = 1(.067) + 2(.133) + 3(.20) + 4(.267) + 5(.333)
µx = .067 + .266 + .60 + 1.068 + 1.665
µx = 3.666

Unknown008 · Sep 24, 2010, 01:01 PM

Well, you are not being really helpful reposting what you did earlier and correcting what I already said what was to be done... :rolleyes: