Q38E There are two traffic lights on ... [FREE SOLUTION]

Chapter 5: Q38E (page 229)

There are two traffic lights on a commuter's route to and from work. Let ${{\rm{X}}_{\rm{1}}}$ be the number of lights at which the commuter must stop on his way to work, and ${{\rm{X}}_{\rm{2}}}$ be the number of lights at which he must stop when returning from work. Suppose these two variables are independent, each with pmf given in the accompanying table (so ${{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}$ is a random sample of size ${\rm{n = 2}}$).
a. Determine the pmf of ${{\rm{T}}_{\rm{o}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}$.
b. Calculate ${{\rm{\mu }}_{{{\rm{T}}_{\rm{o}}}}}$. How does it relate to ${\rm{\mu }}$, the population mean?
c. Calculate ${\rm{\sigma }}_{{{\rm{T}}_{\rm{o}}}}^{\rm{2}}$. How does it relate to ${{\rm{\sigma }}^{\rm{2}}}$, the population variance?
d. Let ${{\rm{X}}_{\rm{3}}}$ and ${{\rm{X}}_{\rm{4}}}$ be the number of lights at which a stop is required when driving to and from work on a second day assumed independent of the first day. With ${{\rm{T}}_{\rm{o}}}{\rm{ = }}$ the sum of all four ${{\rm{X}}_{\rm{i}}}$ 's, what now are the values of ${\rm{E}}\left( {{{\rm{T}}_{\rm{a}}}} \right)$ and ${\rm{V}}\left( {{{\rm{T}}_{\rm{a}}}} \right)$?
e. Referring back to (d), what are the values of ${\rm{P}}\left( {{{\rm{T}}_{\rm{o}}}{\rm{ = 8}}} \right)$ and $\text{P}\left( {{\text{T}}_{\text{e}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)$ (Hint: Don't even think of listing all possible outcomes!)

Short Answer

Expert verified

(a) The pmf of ${{\rm{T}}_{\rm{o}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}$ is

(b) The value of ${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}$ is ${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}{\rm{ = 2}}{\rm{.2}}$ and the relation between ${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}$, and ${\rm{\mu }}$$ is ${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}{\rm{ = 2 \times \mu }}$.

(c) The value of ${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}$ is ${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}{\rm{ = 0}}{\rm{.98}}$ and the relation between ${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}$ and ${{\rm{\sigma }}^{\rm{2}}}$ is${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}{\rm{ = 2 \times }}{{\rm{\sigma }}^{\rm{2}}}$.

(d) The value of ${\rm{E}}\left( {{{\rm{T}}_{\rm{0}}}} \right)$ is ${\rm{E}}\left( {{{\rm{T}}_{\rm{0}}}} \right){\rm{ = 4}}{\rm{.4}}$and value of ${\rm{V}}\left( {{{\rm{T}}_{\rm{0}}}} \right)$ is${\rm{V}}\left( {{{\rm{T}}_{\rm{0}}}} \right){\rm{ = 1}}{\rm{.96}}$.

(e) The value of ${\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 8}}} \right)$ is ${\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 8}}} \right){\rm{ = 0}}{\rm{.0081}}$ and $\text{P}\left( {{\text{T}}_{\text{e}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)$ value of is $\text{P}\left( {{\text{T}}_{\text{0}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)\text{=0}\text{.0297}\text{. }$

Step by step solution

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Determine the pmf of ${{\rm{T}}_{\rm{o}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}$

(a) :

Random variable ${{\rm{T}}_{\rm{0}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}$ can take the following values

$\begin{aligned}\rm 0 + 0 &= 0;\\ \rm0 + 1 &= 1;\\ \rm 0 + 2 &= 2;\\ \rm 1 + 2 &= 3;\\ \rm 2 + 2 &= 4\end{aligned}$

The probabilities are calculated using independence. The following is true

$\begin{aligned}{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 0}}} \right)\rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 0,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 0}}} \right)\\ \rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 0}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 0}}} \right)\\ \rm &= 0{\rm{.2 \times 0}}{\rm{.2}}\\ \rm &= 0{\rm{.04}}\end{aligned}$

$\begin{aligned}{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 1}}} \right)\rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 0}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 0,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right)\\ \rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 0}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 0}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right)\\ \rm &= 0{\rm{.5 \times 0}}{\rm{.2 + 0}}{\rm{.2 \times 0}}{\rm{.5}}\\ \rm &= 0 {\rm{.2}}\end{aligned}$

$\begin{aligned}{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 2}}} \right)\rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 2,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 0}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 0}}{\rm{.}}{{\rm{X}}_{\rm{2}}}{\rm{ = 2}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right)\\ \rm &= P \left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 2}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 0}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 0}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 2}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right)\\ \rm &= 0{\rm{.3 \times 0}}{\rm{.2 + 0}}{\rm{.2 \times 0}}{\rm{.3 + 0}}{\rm{.5 \times 0}}{\rm{.5}}\\ \rm &= 0{\rm{.37}}\end{aligned}$

$\begin{aligned}{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 3}}} \right)\rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 2}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 2}}{\rm{.}}{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right)\\ \rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 2}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 2}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right)\\ \rm &= 0{\rm{.5 \times 0}}{\rm{.3 + 0}}{\rm{.3 \times 0}}{\rm{.5}}\\ \rm &= 0 {\rm{.3}}\end{aligned}$

$\begin{aligned}{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 4}}} \right)\rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 2,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 2}}} \right)\\ \rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 2}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 2}}} \right)\\ \rm &= 0{\rm{.3 \times 0}}{\rm{.3}}\\ \rm &= 0{\rm{.09}}\end{aligned}$

The probability mass function of random variable ${{\rm{T}}_{\rm{0}}}$ is

Calculate${{\rm{\mu }}_{{{\rm{T}}_{\rm{o}}}}}$, relates to${\rm{\mu }}$, the population mean

(b):

The following is expectation of random variable ${{\rm{T}}_{\rm{0}}}$

$\begin{aligned}{\rm{E}}\left( {{{\rm{T}}_{\rm{0}}}} \right)\rm &= {{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}\\\rm &= 0 \times 0{\rm{.04 + 1 \times 0}}{\rm{.2 + 2 \times 0}}{\rm{.37 + 3 \times 0}}{\rm{.3 + 4 \times 0}}{\rm{.09}}\\\rm &= 2{\rm{.2}}{\rm{.}}\end{aligned}$

The expectations of random variable${{\rm{X}}_{\rm{1}}}$, and ${{\rm{X}}_{\rm{2}}}$ are the same. The following holds

$\begin{aligned}{\rm{E}}\left( {{{\rm{X}}_{\rm{1}}}} \right)\rm &= E\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ = \mu }}\\\rm &= 0 \times 0{\rm{.2 + 1 \times 0}}{\rm{.5 + 2 \times 0}}{\rm{.3}}\\\rm &= 1{\rm{.1}}{\rm{.}}\end{aligned}$

The relation between${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}$, and ${\rm{\mu }}$$ is

$\begin{aligned}{{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}\rm &= 2{\rm{.2}}\\\rm &= 2 \times 1{\rm{.1}}\\\rm &= 2 \times {{\rm{\mu }}_{\rm{.}}}\end{aligned}$

Therefore, the value of ${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}$ is ${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}{\rm{ = 2}}{\rm{.2}}$ and the relation between ${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}$, and ${\rm{\mu }}$ $ is ${{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}{\rm{ = 2 \times \mu }}$.

Calculate ${\rm{\sigma }}_{{{\rm{T}}_{\rm{o}}}}^{\rm{2}}$, relate to ${{\rm{\sigma }}^{\rm{2}}}$, the population variance

(c):

The Variance of ${\rm{X}}$, denoted by ${\rm{V(X)}}\left( {{\rm{\sigma }}_{\rm{X}}^{\rm{2}}} \right.$ or $\left. {{{\rm{\sigma }}^{\rm{2}}}} \right)$ is

$\begin{aligned}{\rm{V(X) = }}{{\rm{\sigma }}_{\rm{X}}}\rm &= E\left( {{{{\rm{(X - E(X))}}}^{\rm{2}}}} \right)\\\rm &= E\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}\end{aligned}$

The expectation of random variable ${\rm{T}}_{\rm{0}}^{\rm{2}}$ is

$\begin{aligned}{\rm{E}}\left( {{\rm{T}}_{\rm{0}}^{\rm{2}}} \right)\rm &= {{\rm{0}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.04 + }}{{\rm{1}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.2 + }}{{\rm{2}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.37 + }}{{\rm{3}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.3 + }}{{\rm{4}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.09}}\\\rm &= 5{\rm{.82}}\end{aligned}$

The variance of random variable ${{\rm{T}}_{\rm{0}}}$ is

$\begin{aligned}{\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}\rm &= E\left( {{\rm{T}}_{\rm{0}}^{\rm{2}}} \right){\rm{ - }}{\left( {{\rm{E}}\left( {{{\rm{T}}_{\rm{0}}}} \right)} \right)^{\rm{2}}}\\\rm &= 5{\rm{.82 - 2}}{\rm{.}}{{\rm{2}}^{\rm{2}}}\\\rm &= 0{\rm{.98}}\end{aligned}$

The expectation of random variable ${\rm{X}}_{\rm{1}}^{\rm{2}}$ is

$\begin{aligned}{\rm{E}}\left( {{\rm{X}}_{\rm{1}}^{\rm{2}}} \right)\rm &= {{\rm{0}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.2 + }}{{\rm{1}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.5 + }}{{\rm{2}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.3}}\\\rm &= 1{\rm{.7}}\end{aligned}$

The variance of random variable ${{\rm{X}}_{\rm{1}}}$ is

$\begin{aligned}{{\rm{\sigma }}^{\rm{2}}}\rm &= \sigma _{{{\rm{X}}_{\rm{1}}}}^{\rm{2}}\\\rm &= E\left( {{\rm{X}}_{\rm{1}}^{\rm{2}}} \right){\rm{ - }}{\left( {{\rm{E}}\left( {{{\rm{X}}_{\rm{1}}}} \right)} \right)^{\rm{2}}}\\\rm &= 1{\rm{.7 - 1}}{\rm{.}}{{\rm{1}}^{\rm{2}}}\\\rm &= 0{\rm{.49}}\end{aligned}$

The relation between ${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}$ and ${{\rm{\sigma }}^{\rm{2}}}$ is

${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}{\rm{ = 2 \times }}{{\rm{\sigma }}^{\rm{2}}}$

Therefore, the value of ${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}$ is ${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}{\rm{ = 0}}{\rm{.98}}$ and the relation between ${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}$ and ${{\rm{\sigma }}^{\rm{2}}}$ is ${\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}{\rm{ = 2 \times }}{{\rm{\sigma }}^{\rm{2}}}$.

Calculating values of ${\rm{E}}\left( {{{\rm{T}}_{\rm{a}}}} \right)$ and ${\rm{V}}\left( {{{\rm{T}}_{\rm{a}}}} \right)$

(d):

Random variable ${{\rm{T}}_{\rm{0}}}$ is now defined as

${{\rm{T}}_{\rm{0}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ + }}{{\rm{X}}_{\rm{3}}}{\rm{ + }}{{\rm{X}}_{\rm{4}}}{\rm{.}}$

The following holds

${\rm{E}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ = E}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ = E}}\left( {{{\rm{X}}_{\rm{3}}}} \right){\rm{ = E}}\left( {{{\rm{X}}_{\rm{4}}}} \right){\rm{ = 1}}{\rm{.1}}$

The expectation of random variable ${{\rm{T}}_{\rm{0}}}$ is

$\begin{aligned}{\rm{E}}\left( {{{\rm{T}}_{\rm{0}}}} \right)\rm & = E\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ + }}{{\rm{X}}_{\rm{3}}}{\rm{ + }}{{\rm{X}}_{\rm{4}}}} \right)\\\rm &= E\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + E}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ + E}}\left( {{{\rm{X}}_{\rm{3}}}} \right){\rm{ + E}}\left( {{{\rm{X}}_{\rm{4}}}} \right)\\\rm &= 4 \times 1{\rm{.1}}\\\rm &= 4{\rm{.4}}{\rm{.}}\end{aligned}$

Similarly, because of the independence, the following holds

$\begin{aligned}{\rm{V}}\left( {{{\rm{T}}_{\rm{0}}}} \right)\rm &= V\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ + }}{{\rm{X}}_{\rm{3}}}{\rm{ + }}{{\rm{X}}_{\rm{4}}}} \right)\\ \rm &= {\rm{V}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + V}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ + V}}\left( {{{\rm{X}}_{\rm{3}}}} \right){\rm{ + V}}\left( {{{\rm{X}}_{\rm{4}}}} \right)\\ \rm &= 4 \times 0{\rm{.49}}\\\rm &= 1{\rm{.96}}\end{aligned}$

this only holds when the random variables are independent!

Therefore, the value of ${\rm{E}}\left( {{{\rm{T}}_{\rm{0}}}} \right)$ is ${\rm{E}}\left( {{{\rm{T}}_{\rm{0}}}} \right){\rm{ = 4}}{\rm{.4}}$ and value of ${\rm{V}}\left( {{{\rm{T}}_{\rm{0}}}} \right)$ is${\rm{V}}\left( {{{\rm{T}}_{\rm{0}}}} \right){\rm{ = 1}}{\rm{.96}}$.

Calculating values of ${\rm{P}}\left( {{{\rm{T}}_{\rm{o}}}{\rm{ = 8}}} \right)$ and $\text{P}\left( {{\text{T}}_{\text{e}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)$

(e):

The only way that the total number of lights at which a stop is required is ${\rm{8}}$, is if at each way (each random variable) the required stops are ${\rm{2}}$. This stands because

${\rm{2 + 2 + 2 + 2 = 8}}$

and there is no other way to obtain 8 lights. Therefore, the following holds

$\begin{aligned}{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 8}}} \right)\rm &= P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 2,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 2,}}{{\rm{X}}_{\rm{3}}}{\rm{ = 2,}}{{\rm{X}}_{\rm{4}}}{\rm{ = 2}}} \right)\\\rm &= 0{\rm{.3 \times 0}}{\rm{.3 \times 0}}{\rm{.3 \times 0}}{\rm{.3}}\\\rm &= 0{\rm{.0081}}{\rm{.}}\end{aligned}$

In order to compute $\text{P}\left( {{\text{T}}_{\text{e}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)$ , the only probability that is left to compute is ${\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 7}}} \right)$. The only way to stop at ${\rm{7}}$ lights are combinations

${\rm{(2,2,2,1),}}\;\;\;{\rm{(2,2,1,2),}}\;\;\;{\rm{(2,1,2,2),}}\;\;\;{\rm{(1,2,2,2)}}$

Because the random variables are identically distributed, the following is true

$\begin{aligned}{c}{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 7}}} \right)\rm &= 4 \times P\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 2,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 2,}}{{\rm{X}}_{\rm{3}}}{\rm{ = 2,}}{{\rm{X}}_{\rm{4}}}{\rm{ = 1}}} \right)\\\rm &= 4 \times 0{\rm{.3 \times 0}}{\rm{.3 \times 0}}{\rm{.3 \times 0}}{\rm{.2}}\\\rm &= 0{\rm{.0216}}{\rm{.}}\end{aligned}$

The given random variable ${{\rm{T}}_{\rm{0}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ + }}{{\rm{X}}_{\rm{3}}}{\rm{ + }}{{\rm{X}}_{\rm{4}}}$ cannot take values bigger than ${\rm{8}}$ , therefore the following is true

$\begin{align}\text{P}\left( {{\text{T}}_{\text{0}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right) &\text{=P}\left( {{\text{T}}_{\text{0}}}\text{=7} \right)\text{+P}\left( {{\text{T}}_{\text{0}}}\text{=8} \right) \\ & \text{=0}\text{.0216+0}\text{.0081} \\ &\text{=0}\text{.0297}\text{.} \end{align}$

Therefore, the value of ${\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 8}}} \right)$ is ${\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ = 8}}} \right){\rm{ = 0}}{\rm{.0081}}$ and value of is

$\text{P}\left( {{\text{T}}_{\text{0}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)\text{=0}\text{.0297}\text{. }$

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Short Answer

Step by step solution

Definition

Determine the pmf of \({{\rm{T}}_{\rm{o}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\)

Calculate\({{\rm{\mu }}_{{{\rm{T}}_{\rm{o}}}}}\), relates to\({\rm{\mu }}\), the population mean

Calculate \({\rm{\sigma }}_{{{\rm{T}}_{\rm{o}}}}^{\rm{2}}\), relate to \({{\rm{\sigma }}^{\rm{2}}}\), the population variance

Calculating values of \({\rm{E}}\left( {{{\rm{T}}_{\rm{a}}}} \right)\) and \({\rm{V}}\left( {{{\rm{T}}_{\rm{a}}}} \right)\)

Calculating values of \({\rm{P}}\left( {{{\rm{T}}_{\rm{o}}}{\rm{ = 8}}} \right)\) and \(\text{P}\left( {{\text{T}}_{\text{e}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)\)

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