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a. Show that if X has a normal distribution with parameters \({\rm{\mu }}\) and\({\rm{\sigma }}\), then \({\rm{Y = aX + b}}\) (a linear function of X ) also has a normal distribution. What are the parameters of the distribution of Y (i.e., E(Y) and V(Y)) ? (Hint: Write the cdf of\({\rm{Y,P(Y}} \le {\rm{y)}}\), as an integral involving the pdf of X, and then differentiate with respect to y to get the pdf of Y.)

b. If, when measured in\(^{\rm{^\circ }}{\rm{C}}\), temperature is normally distributed with mean 115 and standard deviation 2 , what can be said about the distribution of temperature measured in\(^{\rm{^\circ }}{\rm{F}}\)?

Short Answer

Expert verified

(a) The parameters of Y are \(\underline {{\rm{E(Y) = a\mu + b}}} \) and \({\rm{V(Y) = (a\sigma }}{{\rm{)}}^{\rm{2}}}\).

(b) The distribution of the temperature measured in \(^{\rm{^\circ }}{\rm{F}}\) is normal with parameters \({\rm{E(F) = 239}}\) and \({\rm{V(F) = (3}}{\rm{.6}}{{\rm{)}}^{\rm{2}}}{\rm{ = 12}}{\rm{.96}}\).

Step by step solution

01

Definition of Concept

Probability: Probability means possibility. It is a branch of mathematics which deals with the occurrence of a random event.

02

Prove that Y has a normal distribution and Find the parameters of Y

(a)

Considering the given information:

The random variable X has a normal distribution with parameters\({\rm{\mu }}\)and\({\rm{\sigma }}\). A linear function of X is Y, such that\({\rm{Y = aX + b}}\).

The cumulative distribution function (cdf) of the random variable Y at y is:

\(\begin{array}{l}{F_Y}(y) = P(Y \le y)\\ = P(aX + b \le y)\\ = P(aX \le y - b)\\ = P\left( {X \le \frac{{y - b}}{a}} \right)\\ = {F_X}\left( {\frac{{y - b}}{a}} \right)\\ = \Phi \left( {\frac{{y - b}}{a}} \right),(X{\rm{ has a normal distribution }})\end{array}\)

Differentiating the cdf of Y with respect to y yields the probability density function (pdf) of \({\rm{Y,}}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\).

\(\begin{aligned}{{\rm{f}}_{\rm{Y}}}(y) &= \frac{{\rm{d}}}{{{\rm{dy}}}}{{\rm{F}}_{\rm{Y}}}{\rm{(y)}}\\&= \frac{{\rm{d}}}{{{\rm{dy}}}}{{\rm{F}}_{\rm{X}}}\left( {\frac{{{\rm{y - b}}}}{{\rm{a}}}} \right)\\ &= {{\rm{f}}_{\rm{X}}}\left( {\frac{{{\rm{y - b}}}}{{\rm{a}}}} \right){\rm{ \times }}\left( {\frac{{\rm{1}}}{{\rm{a}}}} \right)\\ &= \frac{{\rm{1}}}{{\rm{a}}}{{\rm{f}}_{\rm{X}}}\left( {\frac{{{\rm{y - b}}}}{{\rm{a}}}} \right)\end{aligned}\)

The following is the pdf of a normal random variable X with parameters \({\rm{\mu }}\)and \({\rm{\sigma }}\):

\({f_X}(x) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{(x - \mu )}^2}}}{{2{\sigma ^2}}}}},}&{ - \infty < x < \infty , - \infty < \mu < \infty ,\sigma > 0}\\{0,}&{{\rm{ otherwise}}{\rm{. }}}\end{array}} \right.\)

Replace \({\rm{x = }}\frac{{{\rm{y - b}}}}{{\rm{a}}}\) in the pdf of X :

\(\begin{aligned}{{\rm{f}}_{\rm{X}}}\left( {\frac{{{\rm{y - b}}}}{{\rm{a}}}} \right) &= \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \sigma }}} }}{{\rm{e}}^{{\rm{ - }}\frac{{{{\left( {\frac{{{\rm{y - b}}}}{{\rm{a}}}{\rm{ - \mu }}} \right)}^{\rm{2}}}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}}}\\ &= \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \sigma }}} }}{{\rm{e}}^{{\rm{ - }}\frac{{{{\left( {\frac{{{\rm{y - b - a\mu }}}}{{\rm{a}}}} \right)}^{\rm{2}}}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}}}\\ &= \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \sigma }}} }}{{\rm{e}}^{{\rm{ - }}\frac{{{{{\rm{(y - b - a\mu )}}}^{\rm{2}}}}}{{{\rm{2}}{{\rm{a}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}}}\end{aligned}\)

\(\begin{aligned}{\rm{ Replace }}{{\rm{f}}_{\rm{X}}}\left( {\frac{{{\rm{y - b}}}}{{\rm{a}}}} \right) &= \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \sigma }}} }}{{\rm{e}}^{{\rm{ - }}\frac{{{{{\rm{(y - b - a)}}}^{\rm{2}}}}}{{{\rm{2}}{{\rm{a}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}}}{\rm{ in the expression for }}{{\rm{f}}_{\rm{Y}}}{\rm{(y) : }}\\{{\rm{f}}_{\rm{Y}}} (y) &= \frac{{\rm{1}}}{{\rm{a}}}{{\rm{f}}_{\rm{X}}}\left( {\frac{{{\rm{y - b}}}}{{\rm{a}}}} \right)\\ &= \frac{{\rm{1}}}{{\rm{a}}}{\rm{ \times }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \sigma }}} }}{{\rm{e}}^{{\rm{ - }}\frac{{{{{\rm{(y - b - a\mu )}}}^{\rm{2}}}}}{{{\rm{2}}{{\rm{a}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}}}\\ &= \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi }}} {\rm{a\sigma }}}}{{\rm{e}}^{{\rm{ - }}\frac{{{{{\rm{(y - b - a\mu )}}}^{\rm{2}}}}}{{{\rm{2}}{{\rm{a}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}}}\\& = \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi (a\sigma )}}} }}{{\rm{e}}^{{\rm{ - }}\frac{{{{{\rm{(y - (a\mu + b))}}}^{\rm{2}}}}}{{{\rm{2(a\sigma }}{{\rm{)}}^{\rm{2}}}}}}}\end{aligned}\)

Here, a and b are constants and do not affect the range of Y. As a result, the range of Y is dependent on that of X only and is the same as that of X.

Thus, the pdf of Y is:

\({f_Y}(y) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {2\pi (a\sigma )} }}{e^{ - \frac{{{{(y - (a\mu + b))}^2}}}{{2{{(a\sigma )}^2}}}}}}&{, - \infty < y < \infty , - \infty < \mu < \infty ,\sigma > 0,a \ne 0}\\{0,}&{{\rm{ otherwise}}{\rm{. }}}\end{array}} \right.\)

A careful inspection of \({{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\) reveals that this is the pdf of a normal random variables, with parameters \({\rm{(a\mu + b)}}\) and \({\rm{(a\sigma )}}{\rm{.}}\)

Hence, Y has a normal distribution.

Now, it the known that the parameters of the normal random variable X are \({\rm{\mu }}\) and\({\rm{\sigma }}\) , where \({\rm{\mu = E(X)}}\) and\({\rm{\sigma = }}\sqrt {{\rm{V(X)}}} \).

Thus, here,

\({\rm{E(Y) = a\mu + b}}\underline {{\rm{V(Y) = (a\sigma }}{{\rm{)}}^{\rm{2}}}} \)

Therefore, the parameters of Y are \(\underline {{\rm{E(Y) = a\mu + b}}} \) and\(\underline {{\rm{V(Y) = (a\sigma }}{{\rm{)}}^{\rm{2}}}} \).

03

About the distribution of temperature measured in \(^{\rm{^\circ }}{\rm{F}}\)

(b)

Considering the given information:

Use the conversion formula between temperatures measured in \(^{\rm{^\circ }}{\rm{C}}\) and \(^{\rm{^\circ }}{\rm{F}}\) :

\(\frac{{\rm{C}}}{{\rm{5}}}{\rm{ = }}\frac{{{\rm{F - 32}}}}{{\rm{9}}}\), where C and F are the random variables denoting the temperature values measured in \(^{\rm{^\circ }}{\rm{C}}\) and \(^{\rm{^\circ }}{\rm{F}}\) respectively.

Here, it is known that C has a normal distribution, with parameters \({\rm{\mu = 115}}\) and \({\rm{\sigma = 2}}\).

Consider the following conversion formula:

\(\begin{array}{c}\frac{{\rm{C}}}{{\rm{5}}}{\rm{ = }}\frac{{{\rm{F - 32}}}}{{\rm{9}}}\\{\rm{F - 32 = }}\frac{{\rm{9}}}{{\rm{5}}}{\rm{C}}\quad \\{\rm{ = 1}}{\rm{.8C}}\\{\rm{F = 1}}{\rm{.8C + 18}}\end{array}\)

From part\({\rm{a}}\), if \({\rm{Y = aX + b}}\) and X has a normal distribution with parameters \({\rm{\mu }}\) and\({\rm{\sigma }}\) , then Y also has a normal distribution with parameters \({\rm{E(Y) = a\mu + b}}\) and\({\rm{V(Y) = (a\sigma }}{{\rm{)}}^{\rm{2}}}\).

Here, \({\rm{Y = F,a = 1}}{\rm{.8,b = 32}}\).

Thus, the distribution of the temperature measured in \(^{\rm{^\circ }}{\rm{F}}\) is normal.

Putting \({\rm{\mu = 115}}\) and \({\rm{\sigma = 2}}\) in the relationship between F and C :

\(\begin{aligned} E(F) &= a\mu + b\\ &= 1 {\rm{.8\mu + 32}}\\ &= (1 {\rm{.8 \times 115) + 32}}\\ & = 207 + 32 \\{\rm{ = 239}}\\ V(Y) &= (a\sigma {{\rm{)}}^{\rm{2}}}\\ &= (1{\rm{.8\sigma }}{{\rm{)}}^{\rm{2}}}\\ &= (1 {\rm{.8 \times 2}}{{\rm{)}}^{\rm{2}}}\\&= (3 {\rm{.6}}{{\rm{)}}^{\rm{2}}}\\ &= 12{\rm{.96}}{\rm{.}}\end{aligned}\)

Therefore, the distribution of the temperature measured in \(^{\rm{^\circ }}{\rm{F}}\) is normal with parameters \({\rm{E(F) = 239}}\) and \({\rm{V(F) = (3}}{\rm{.6}}{{\rm{)}}^{\rm{2}}}{\rm{ = 12}}{\rm{.96}}\).

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