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,

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\\
\hline
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! :)

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

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

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

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