# uniform distribution of random independent variables

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• Nov 28th 2008, 02:27 PM
chrisc
uniform distribution of random independent variables
okay. my exam is approaching in a few days and I have one last question.

Let X1, X2, ... X7 be a set of seven independent random variables, each
having the uniform distribution of values from the [3, 6] interval (this is
called random independent sample of size 7 from this distribution).

(a) Find the expected value and standard deviation of their sum T ≡ X1 + X2 + ... + X7.
(b) What is the probability that more than three of these seven random
variables will have a value smaller than 1
3?
(c) Find the skewness and kurtosis of X1.

answers:

(a) E(x) = 7(6+3)/2 = 31.5
Var(x) = 7*[(6-3)^2]/2
standard deviation = 2.2913

(b) not sure

(c) would skewness be 0 since its uniform? what about kurtosis
• Nov 28th 2008, 08:28 PM
mr fantastic
Quote:

Originally Posted by chrisc
okay. my exam is approaching in a few days and I have one last question.

Let X1, X2, ... X7 be a set of seven independent random variables, each
having the uniform distribution of values from the [3, 6] interval (this is
called random independent sample of size 7 from this distribution).

(a) Find the expected value and standard deviation of their sum T ≡ X1 + X2 + ... + X7.
(b) What is the probability that more than three of these seven random
variables will have a value smaller than 1
3? Mr F says: The question is not clear to me. Is it meant to be 13, 1 or something else?
(c) Find the skewness and kurtosis of X1.

answers:

(a) E(x) = 7(6+3)/2 = 31.5
Var(x) = 7*[(6-3)^2]/2
standard deviation = 2.2913

(b) not sure

(c) would skewness be 0 since its uniform? what about kurtosis

(a) You're expected to know or derive that $E(X_i) = \frac{6+3}{2} = \frac{9}{2}$ and $Var(X_i) = \frac{(6-3)^2}{12} = \frac{3}{4}$.

Then:

$E\left(\sum_{i=1}^{7} X_i\right) = \sum_{i=1}^{7} E(X_i) = 7 \cdot \frac{9}{2} = \frac{63}{2}$.

$Var\left(\sum_{i=1}^{7} X_i\right) = \sum_{i=1}^{7} Var(X_i) = 7 \cdot \frac{3}{4} = \frac{21}{4}$ (summing the variances follows because the X's are independent).

(b) Is there a typo?

(c) What definition/type of kurtosis are you using?
• Nov 29th 2008, 10:46 AM
chrisc
a) okay, confirmed

b) was a typo, its 1/3, not 1

c) nothing else was specified outside of this question
• Nov 29th 2008, 01:29 PM
mr fantastic
Quote:

Originally Posted by chrisc
a) okay, confirmed

b) was a typo, its 1/3, not 1 Mr F says: Well if that's the case, what's the probability that any of the X's can be smaller than 1/3 .....?! Look at the interval over which the pdf is not equal to zero!

c) nothing else was specified outside of this question

c) That's why you go to your class notes or textbook, find the definition/type that's been used and then report the answer in this thread. I had thought that's what you'd do.
• Nov 30th 2008, 01:01 PM
chrisc
this is the one thing we were given about for kurtosis, part of our exponential stuff (not uniform like the question asks)

kurtosis:
[E(X4)−4μE(X3)+6μ2E(X2)−3μ4]/σ4 = ...
[24β4−4×6β4+6×2β4−3β4]/β4 = 9

does this help at all? if not, ill just hope this specific type of question doesnt make its way to the exam
• Dec 1st 2008, 04:45 AM
mr fantastic
Quote:

Originally Posted by chrisc
this is the one thing we were given about for kurtosis, part of our exponential stuff (not uniform like the question asks)

kurtosis:
[E(X4)−4μE(X3)+6μ2E(X2)−3μ4]/σ4 = ...
[24β4−4×6β4+6×2β4−3β4]/β4 = 9

does this help at all? if not, ill just hope this specific type of question doesnt make its way to the exam

Then calculate each of the bits in the formula:

$E(X^n) = \int_3^6 x^n \cdot 1 \, dx$.

$\mu = E(X_1) = \frac{9}{2}$.

$\sigma = s.d.(X_1) = \frac{\sqrt{3}}{2}$.

Now substitute into the formula.