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Suppose X has an exponential distribution with mean equal to 11. Determine the following: (a) (Round your answer to 3 decimal places.) (b) (Round your answer to 3 decimal places.) (c) (Round your answer to 3 decimal places.) (d) Find the value of x such that . (Round your answer to 2 decimal places.)

Sagot :

MrRoyal

Answer:

[tex]P(X > 11) = 0.368[/tex]

[tex]P(X > 22) = 0.135[/tex]

[tex]P(X > 33) = 0.050[/tex]

[tex]x = 33[/tex]

Step-by-step explanation:

Given

[tex]E(x) = 11[/tex] --- Mean

Required (Missing from  the question)

[tex](a)\ P(X>11)[/tex]

[tex](b)\ P(X>22)[/tex]

[tex](c)\ P(X>33)[/tex]

(d) x such that [tex]P(X <x)=0.95[/tex]

In an exponential distribution:

[tex]f(x) = \lambda e^{-\lambda x}, x \ge 0[/tex] --- the pdf

[tex]F(x) = 1 - e^{-\lambda x}, x \ge 0[/tex] --- the cdf

[tex]P(X > x) = 1 - F(x)[/tex]

In the above equations:

[tex]\lambda = \frac{1}{E(x)}[/tex]

Substitute 11 for E(x)

[tex]\lambda = \frac{1}{11}[/tex]

Now, we solve (a) to (d) as follows:

Solving (a): P(X>11)

[tex]P(X > 11) = 1 - F(11)[/tex]

Substitute 11 for x in [tex]F(x) = 1 - e^{-\lambda x}[/tex]

[tex]P(X > 11) = 1 - (1 - e^{-\frac{1}{11}* 11})[/tex]

[tex]P(X > 11) = 1 - (1 - e^{-\frac{11}{11}})[/tex]

[tex]P(X > 11) = 1 - (1 - e^{-1})[/tex]

Remove bracket

[tex]P(X > 11) = 1 - 1 + e^{-1}[/tex]

[tex]P(X > 11) = e^{-1}[/tex]

[tex]P(X > 11) = 0.368[/tex]

Solving (b): P(X>22)

[tex]P(X > 22) = 1 - F(22)[/tex]

Substitute 22 for x in [tex]F(x) = 1 - e^{-\lambda x}[/tex]

[tex]P(X > 22) = 1 - (1 - e^{-\frac{1}{11}* 22})[/tex]

[tex]P(X > 22) = 1 - (1 - e^{-\frac{22}{11}})[/tex]

[tex]P(X > 22) = 1 - (1 - e^{-2})[/tex]

Remove bracket

[tex]P(X > 22) = 1 - 1 + e^{-2}[/tex]

[tex]P(X > 22) = e^{-2}[/tex]

[tex]P(X > 22) = 0.135[/tex]

Solving (c): P(X>33)

[tex]P(X > 33) = 1 - F(33)[/tex]

Substitute 33 for x in [tex]F(x) = 1 - e^{-\lambda x}[/tex]

[tex]P(X > 33) = 1 - (1 - e^{-\frac{1}{11}* 33})[/tex]

[tex]P(X > 33) = 1 - (1 - e^{-\frac{33}{11}})[/tex]

[tex]P(X > 33) = 1 - (1 - e^{-3})[/tex]

Remove bracket

[tex]P(X > 33) = 1 - 1 + e^{-3}[/tex]

[tex]P(X > 33) = e^{-3}[/tex]

[tex]P(X > 33) = 0.050[/tex]

Solving (d): x when [tex]P(X <x)=0.95[/tex]

Here, we make use of:

[tex]P(X<x) = F(x)[/tex]

Substitute [tex]F(x) = 1 - e^{-\lambda x}[/tex]

[tex]P(X<x) = 1 - e^{-\lambda x}[/tex]

So, we have:

[tex]0.95 = 1 - e^{-\lambda x}[/tex]

Subtract 1 from both sides

[tex]0.95 -1= 1-1 - e^{-\lambda x}[/tex]

[tex]-0.05=- e^{-\lambda x}[/tex]

Reorder the equation

[tex]e^{-\lambda x} = 0.05[/tex]

Substitute 1/11 for [tex]\lambda[/tex]

[tex]e^{-\frac{1}{11} x} = 0.05[/tex]

Solve for x:

[tex]x = -\frac{1}{1/11}\ ln(0.05)[/tex]

[tex]x = -11\ ln(0.05)[/tex]

[tex]x = 32.9530550091[/tex]

[tex]x = 33[/tex] --- approximated

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