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3 years ago

CAN SOMEONE PLEASE PLEASE HELP ME!! ASAP

Mathematics

2 answers:

goldenfox [79]3 years ago

8 0

1st let's calculate r

16 - r > 8
-r > 8 - 16
-r > -8

r<8 (When changing the signs you flip the inequality)

r={5,6,7,8,9,10}, with the restriction of r<8 , the only solutions remaining to satisfy the condition of r<8 is the set {5,6,7}

Elanso [62]3 years ago

5 0

The answer correct answer is b

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The n candidates for a job have been ranked 1, 2, 3,..., n. Let X 5 the rank of a randomly selected candidate, so that X has pmf

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Question:

The n candidates for a job have been ranked 1, 2, 3,..., n. Let x = rank of a randomly selected candidate, so that x has pmf:

$p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.$

(this is called the discrete uniform distribution).

Compute E(X) and V(X) using the shortcut formula.

[Hint: The sum of the first n positive integers is $\frac{n(n +1)}{2}$ , whereas the sum of their squares is $\frac{n(n +1)(2n+1)}{6}$

Answer:

$E(x) = \frac{n+1}{2}$

$Var(x) = \frac{n^2 -1}{12}$ or $Var(x) = \frac{(n+1)(n-1)}{12}$

Step-by-step explanation:

Given

PMF

$p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.$

Required

Determine the E(x) and Var(x)

E(x) is calculated as:

$E(x) = \sum \limits^{n}_{i} \ x * p(x)$

This gives:

$E(x) = \sum \limits^{n}_{x=1} \ x * \frac{1}{n}$

$E(x) = \sum \limits^{n}_{x=1} \frac{x}{n}$

$E(x) = \frac{1}{n}\sum \limits^{n}_{x=1} x$

From the hint given:

$\sum \limits^{n}_{x=1} x =\frac{n(n +1)}{2}$

So:

$E(x) = \frac{1}{n} * \frac{n(n+1)}{2}$

$E(x) = \frac{n+1}{2}$

Var(x) is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Calculating: $E(x^2)$

$E(x^2) = \sum \limits^{n}_{x=1} \ x^2 * \frac{1}{n}$

$E(x^2) = \frac{1}{n}\sum \limits^{n}_{x=1} \ x^2$

Using the hint given:

$\sum \limits^{n}_{x=1} \ x^2 = \frac{n(n +1)(2n+1)}{6}$

So:

$E(x^2) = \frac{1}{n} * \frac{n(n +1)(2n+1)}{6}$

$E(x^2) = \frac{(n +1)(2n+1)}{6}$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = \frac{(n+1)(2n+1)}{6} - (\frac{n+1}{2})^2$

$Var(x) = \frac{(n+1)(2n+1)}{6} - \frac{n^2+2n+1}{4}$

$Var(x) = \frac{2n^2 +n+2n+1}{6} - \frac{n^2+2n+1}{4}$

$Var(x) = \frac{2n^2 +3n+1}{6} - \frac{n^2+2n+1}{4}$

Take LCM

$Var(x) = \frac{4n^2 +6n+2 - 3n^2 - 6n - 3}{12}$

$Var(x) = \frac{4n^2 - 3n^2+6n- 6n +2 - 3}{12}$

$Var(x) = \frac{n^2 -1}{12}$

Apply difference of two squares

$Var(x) = \frac{(n+1)(n-1)}{12}$

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