91影视

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.

Given: \({{\rm{f}}_{\rm{1}}}\)and \({{\rm{f}}_{\rm{2}}}\)are normal pdfs:

\(\begin{array}{l}{\rm{f(x) = p}}{{\rm{f}}_{\rm{1}}}\left( {{\rm{x;}}{{\rm{\mu }}_{\rm{1}}}{\rm{,\sigma }}} \right){\rm{ + (1 - p)}}{{\rm{f}}_{\rm{2}}}\left( {{\rm{x;}}{{\rm{\mu }}_{\rm{2}}}{\rm{,\sigma }}} \right)\\{\rm{p = 0}}{\rm{.35}}\\{{\rm{\mu }}_{\rm{1}}}{\rm{ = 4}}{\rm{.4}}\\{{\rm{\mu }}_{\rm{2}}}{\rm{ = 5}}{\rm{.0}}\\{\rm{\sigma = 0}}{\rm{.27}}\end{array}\)

For the linear combination\({\rm{W = a}}{{\rm{X}}_{\rm{1}}}{\rm{ + b}}{{\rm{X}}_{\rm{2}}}\), the mean then was the following property:

\({{\rm{\mu }}_{\rm{W}}}{\rm{ = a}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + b}}{{\rm{\mu }}_{\rm{2}}}\)

Use this property with \({\rm{X = p}}{{\rm{X}}_{\rm{1}}}{\rm{ + (1 - p)}}{{\rm{X}}_{\rm{2}}}{\rm{,a = p}}\)and\({\rm{b = 1 - p}}\):

\({{\rm{\mu }}_{\rm{X}}}{\rm{ = p}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + (1 - p)}}{{\rm{\mu }}_{\rm{2}}}\)

Fill in the known values and evaluate:

\({{\rm{\mu }}_{\rm{X}}}{\rm{ = 0}}{\rm{.35(4}}{\rm{.4) + (1 - 0}}{\rm{.35)(5}}{\rm{.0) = 4}}{\rm{.79}}\)

b) We want to determine the probability \(P(4拢X拢7)\). The probability is equal to the integral of the pdf over the corresponding interval:

\(\begin{aligned}P(4拢X拢7)&= \int_4^7 f (x)dx \\&= \int_4^7 {\left( {p{f_1}\left( {x;{\mu _1},\sigma } \right) + (1 - p){f_2}\left( {x;{\mu _2},\sigma } \right)} \right)} dx \\&= \left. {p\int_4^7 {{f_1}} \left( {x;{\mu _1},\sigma } \right)dx + (1 - p)\int_4^7 {{f_2}} \left( {x;{\mu _2},\sigma } \right)} \right)dx \\&= pP\left( {4拢{X_1}拢7} \right) + (1 - p)P\left( {4拢{X_2}拢7} \right) \\\end{aligned} \)

The standardized score is the value\({\rm{x}}\)decreased by the mean and then divided by the standard deviation.

\(\begin{array}{l}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{4 - 4}}{\rm{.4}}}}{{{\rm{0}}{\rm{.27}}}}{\rm{\gg - 1}}{\rm{.48}}\\{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{7 - 4}}{\rm{.4}}}}{{{\rm{0}}{\rm{.27}}}}{\rm{\gg 9}}{\rm{.63}}\\{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{4 - 5}}{\rm{.0}}}}{{{\rm{0}}{\rm{.27}}}}{\rm{\gg - 3}}{\rm{.70}}\\{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{7 - 5}}{\rm{.0}}}}{{{\rm{0}}{\rm{.27}}}}{\rm{\gg 7}}{\rm{.41}}\end{array}\)

Determine the corresponding probability using the normal probability table in the appendix:

\(\begin{aligned}P\left( {4拢{X_1}拢7} \right) &= P( - 1.48 < Z < 9.63) \\&= P(Z < 9.63) - P(Z < - 1.48) \\\gg& 1 - 0.0694 \\&= 0.9306 \\P\left( {4拢{X_2}拢7} \right) &= P( - 3.70 < Z < 7.41) \\&= P(Z < 7.41) - P(Z < - 3.70) \\\gg& 1 - 0 \\ &= 1 \\\end{aligned} \)

Determine the corresponding probability for\({\rm{X}}\):

\(\begin{aligned}P(4拢X\\拢7) &= pP\left( {4拢{X_1}拢7} \right) + (1 - p)P\left( {4拢{X_2}拢7} \right) \\&= 0.35(0.9306) + (1 - 0.35)(1) \\&= 0.97571 \\&= 97.571\% \\\end{aligned} \)

c) For the linear combination\({\rm{W = a}}{{\rm{X}}_{\rm{1}}}{\rm{ + b}}{{\rm{X}}_{\rm{2}}}\), the mean then was the following property:

\({{\rm{\mu }}_{\rm{W}}}{\rm{ = a}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + b}}{{\rm{\mu }}_{\rm{2}}}\)

Use this property with \({\rm{X = p}}{{\rm{X}}_{\rm{1}}}{\rm{ + (1 - p)}}{{\rm{X}}_{\rm{2}}}{\rm{,a = p}}\) and \({\rm{b = 1 - p}}\) :

\({{\rm{\mu }}_{\rm{X}}}{\rm{ = p}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + (1 - p)}}{{\rm{\mu }}_{\rm{2}}}\)

Fill in the known values and evaluate:

\({{\rm{\mu }}_{\rm{X}}}{\rm{ = 0}}{\rm{.35(4}}{\rm{.4) + (1 - 0}}{\rm{.35)(5}}{\rm{.0) = 4}}{\rm{.79}}\)

We want to determine the probability\({\rm{P}}\left( {{\rm{X < }}{{\rm{\mu }}_{\rm{X}}}} \right)\). The probability is equal to the integral of the pdf over the corresponding interval:

\(\begin{aligned}P\left( {X < {\mu _X}} \right) &= \int_{ - 楼}^{\mu X} f (x)dx \\&= \int_{ - 楼}^{{\mu _X}} {\left( {p{f_1}\left( {x;{\mu _1},\sigma } \right) + (1 - p){f_2}\left( {x;{\mu _2},\sigma } \right)} \right)} dx \\\left. {= p\int_{ - 楼}^{{\mu _X}} {{f_1}} \left( {x;{\mu _1},\sigma } \right)dx + (1 - p)\int_{ - 楼}^{{\mu _X}} {{f_2}} \left( {x;{\mu _2},\sigma } \right)} \right)dx \\&= pP\left( {{X_1} < {\mu _X}} \right) + (1 - p)P\left( {{X_2} < {\mu _X}} \right) \\\end{aligned} \)

The standardized score is the value \({\rm{x}}\) decreased by the mean and then divided by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{{\rm{\mu }}_{\rm{X}}}{\rm{ - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{4}}{\rm{.79 - 4}}{\rm{.4}}}}{{{\rm{0}}{\rm{.27}}}}\\{\rm{\gg 1}}{\rm{.44}}\\{\rm{z = }}\frac{{{{\rm{\mu }}_{\rm{X}}}{\rm{ - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{4}}{\rm{.79 - 5}}{\rm{.0}}}}{{{\rm{0}}{\rm{.27}}}}\\{\rm{\gg - 0}}{\rm{.77}}\end{array}\)

Determine the corresponding probability using the normal probability table in the appendix:

\(\begin{aligned}P\left( {{X_1} < {\mu _X}} \right) &= P(Z < 1.44) \\&= 0.9251 \\P\left( {{X_2} < {\mu _X}} \right) &= P(Z < - 0.77) \\&= 0.2206 \\\end{aligned} \)

Determine the corresponding probability for\({\rm{X}}\):

\(\begin{aligned}P\left( {X < {\mu _X}} \right) &= pP\left( {{X_1} < {\mu _X}} \right) + (1-p)P\left( {{X_2} < {\mu _X}} \right) \\&= 0.35(0.9251) + (1 - 0.35)(0.2206) \\&= 0.467175 \\&= 46.7175\% \\ \end{aligned} \)