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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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nobillionaireNobley
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.

[tex]\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}[/tex]



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

[tex]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[/tex]

Therefore, the expected value of x is 3.01



Part C:

The expected value of [tex]x^2[/tex] is given by

[tex]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[/tex]

Therefore, the expected value of [tex]\bold{x^2} [/tex] is 12.45



Part D:

The variance of x is given by

[tex]Var(x)=E(x^2)-(E(x))^2 \\ \\ =12.45 - (3.01)^2=12.45-9.06 \\ \\ =3.39[/tex]

Therefore, the variance of x is 3.39.



Part E

The standard deviation of x is given by

[tex] \sqrt{Var(x)} = \sqrt{3.39} =1.84[/tex]

Therefore, the standard deviation of x is 1.84.



Part F:

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

[tex]Var(aX)=a^2Var(X)[/tex]

Thus, the variance of 3x is given by

[tex]Var(3X)=3^2Var(X)=9(3.39)=30.51[/tex]

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

[tex]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[/tex]

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

[tex](0.77)^5=0.27[/tex]

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

[tex]Var(x^2)=E((x^2)^2)-(E(x^2))^2=E(x^4)-(E(x^2))^2[/tex]

[tex]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[/tex]

Thus,

[tex]Var(x^2)=296.37-(12.45)^2=296.37-155.00=141.37[/tex]

Therefore, the variance of the random variable [tex]\bold{x^2}[/tex] is 141.37

Answer:.1

Step-by-step explanation:

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