Mathematics, 09.07.2019 11:20 alyssahockett4

8. based upon a long period of record keeping the following represents the probability distribution of the number of times the john jay wifi network is slow during a week. we call the random variable x. x 0 1 2 3 4 5 6 p(x) .08 .17 .21 k .21 k .13 a. what is the value of k? b. calculate the expected value of x. c. calculate the expected value of x^2 . d. calculate the variance of x. e. calculate the standard deviation of x f. calculate the variance of 3x. g. suppose that network slowness is independent from week to week. what is the probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks? h. calculate the variance of the random variable x^2

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Given a table representing the probability distribution of the number of times the John Jay wifi network is slow during a week. We call the random variable x.

$\begin{tabular}
{|c|c|c|c|c|c|c|c|}
x&0&1&2&3&4&5&6\\[1ex]
p(x)&.08&.17& .21& k& .21& k& .13
\end{tabular}$

Part A:

The total value of p(x) = 1.

Thus,

.08 + .17 + .21 + k + .21 + k + .13 = 1

0.8 + 2k = 1

2k = 1 - 0.8 = 0.2

k = 0.2 / 2 = 0.1

Therefore, the value of k is 0.1

Part B:

The expected value of x is given by

$E(x)=\Sigma xp(x) \\ \\ =0(0.08)+1(0.17)+2(0.21)+3(0.1)+4(0.21)+5(0.1)+6(0.13) \\ \\ =0+0.17+0.42+0.3+0.84+0.5+0.78=3.01$

Therefore, the expected value of x is 3.01

Part C:

The expected value of x^2

is given by

$E(x^2)=\Sigma x^2p(x) 
\\ \\ =0^2(0.08)+1^2(0.17)+2^2(0.21)+3^2(0.1)+4^2(0.21)+5^2(0.1)+6^2(0.13) \\ \\ 
=0(0.08)+1(0.17)+4(0.21)+9(0.1)+16(0.21)+25(0.1)+36(0.13) \\ \\ =0+0.17+0.84+0.9+3.36+2.5+4.68=12.45$

Therefore, the expected value of $\bold{x^2}$ is 12.45

Part D:

The variance of x is given by

$Var(x)=E(x^2)-(E(x))^2 \\ \\ =12.45 - (3.01)^2=12.45-9.06 \\ \\ =3.39$

Therefore, the variance of x is 3.39.

Part E

The standard deviation of x is given by

$\sqrt{Var(x)} = \sqrt{3.39} =1.84$

Therefore, the standard deviation of x is 1.84.

Part F:

The variance of ax, where a is a constant is given by

Var(aX)=a^2Var(X)

Thus, the variance of 3x is given by

Var(3X)=3^2Var(X)=9(3.39)=30.51

Therefore, the variance of 3x is 30.51.

Part G:

The probability that the network has no more that 4 slow times in one week is given by

$P(x\leq4)=P(0)+P(1)+P(2)+P(3)+P(4) \\ \\ =0.08+0.17+0.21+0.1+0.21=0.77$

Since, the network slowness is independent from week to week, the probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks is given by

(0.77)^5=0.27

Therefore, the probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks is 0.27

Part H:

The variance of x^2 is given by

Var(x^2)=E((x^2)^2)-(E(x^2))^2=E(x^4)-(E(x^2))^2

Var(x^2)=E((x^2)^2)-(E(x^2))^2=E(x^4)-(E(x^2))^2

$E(x^4)=\Sigma x^4p(x) \\ \\ =0^4(0.08)+1^4(0.17)+2^4(0.21)+3^4(0.1)+4^4(0.21)+5^4(0.1) \\ +6^4(0.13) \\ \\ =0(0.08)+1(0.17)+16(0.21)+81(0.1)+256(0.21)+625(0.1)\\+1,296(0.13) \\ \\ =0+0.17+3.36+8.1+53.76+62.5+168.48=296.37$

Thus,

Var(x^2)=296.37-(12.45)^2=296.37-155.00=141.37

Var(x^2)=296.37-(12.45)^2=296.37-155.00=141.37

Therefore, the variance of the random variable $\bold{x^2}$ is 141.37

Answer from: Quest

5 pm

step-by-step explanation:

Answer from: Quest

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b): y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4 perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

Which equation represents a line perpendicular to the line shown on the graph?

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