91影视

An unknown number, unknown value, or unknown quantity is represented by a variable, which is an alphabet or word. In the context of algebraic expressions or algebra, the variables are particularly useful.

(a)Standardization results in:

\({\rm{X}} \le {\rm{15}}\)

only if and only if

\(\begin{array}{c}\frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le \frac{{{\rm{15 - 15}}}}{{{\rm{1}}{\rm{.25}}}}\\\frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le {\rm{0}}\\{\rm{Z}} \le {\rm{0}}\end{array}\)

Thus

\({\rm{P(X}} \le {\rm{15) = P(Z}} \le {\rm{0)}}\)

\({\rm{Z}}\)is a standard normal distribution\({\rm{rv}}\)with cdf\(\phi {\rm{(z)}}\). Hence

\(\begin{array}{c}{\rm{P(X}} \le {\rm{15) = P(Z}} \le {\rm{0)}}\\{\rm{ = }}\phi {\rm{(0)}}\end{array}\)

To find\(\phi {\rm{(0)}}\), look in Appendix Table\({\rm{A}}{\rm{.3}}\)at the intersection of the\({\rm{0}}{\rm{.0}}\)row and the\({\rm{.00}}\)column. There's a number\({\rm{0}}{\rm{.5}}\), so

\(\phi {\rm{(0) = 0}}{\rm{.5}}\)

Hence,

\({\rm{P(X}} \le {\rm{15) = 0}}{\rm{.5}}\)

If\({\rm{Z}}\)is a continuous\({\rm{rv}}\)with cdf, then the following proposition is\(\phi {\rm{(z)}}\). Then, using\({\rm{a < b}}\), for any two numbers\({\rm{a}}\)and\({\rm{b}}\),

\({\rm{P(a}} \le {\rm{Z}} \le {\rm{b) = }}\phi {\rm{(b) - }}\phi {\rm{(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.5}}\).

(b)Standardization results in:

\({\rm{X}} \le {\rm{17}}{\rm{.5}}\)

only if and only if

\(\begin{array}{c}\frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le \frac{{{\rm{17}}{\rm{.5 - 15}}}}{{{\rm{1}}{\rm{.25}}}}\\\frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le \frac{{{\rm{2}}{\rm{.5}}}}{{{\rm{1}}{\rm{.25}}}}\\{\rm{Z}} \le {\rm{2}}\end{array}\)

Thus

\({\rm{P(X}} \le {\rm{17}}{\rm{.5) = P(Z}} \le {\rm{2)}}\)

\({\rm{Z}}\)is a standard normal distribution\({\rm{rv}}\)with cdf\(\phi {\rm{(z)}}\). Hence

\(\begin{array}{c}{\rm{P(X}} \le {\rm{17}}{\rm{.5) = P(Z}} \le {\rm{2)}}\\{\rm{ = }}\phi {\rm{(2)}}\end{array}\)

To find\(\phi {\rm{(2)}}\), look in Appendix Table\({\rm{A}}{\rm{.3}}\)at the intersection of the\({\rm{2}}{\rm{.0}}\)row and the\({\rm{.00}}\)column. There's a number\({\rm{0}}{\rm{.9772}}\), so

\(\phi {\rm{(2) = 0}}{\rm{.9772}}\)

Hence,

\({\rm{P(X}} \le {\rm{17}}{\rm{.5) = 0}}{\rm{.9772}}\)

Therefore, the value is \({\rm{0}}{\rm{.9772}}\).

(c)Standardization results in:

\({\rm{X}} \ge {\rm{10}}\)

only if and only if

\(\begin{array}{c}\frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \ge \frac{{{\rm{10 - 15}}}}{{{\rm{1}}{\rm{.25}}}}\\\frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \ge \frac{{{\rm{ - 5}}}}{{{\rm{1}}{\rm{.25}}}}\\{\rm{Z}} \ge {\rm{ - 4}}\end{array}\)

Thus

\({\rm{P(X}} \ge {\rm{10) = P(Z}} \ge {\rm{ - 4)}}\)

\({\rm{Z}}\)is a standard normal distribution\({\rm{rv}}\)with cdf\(\phi {\rm{(z)}}\). Hence

\(\begin{array}{c}{\rm{P(X}} \ge {\rm{10) = P(Z}} \ge {\rm{ - 4)}}\\{\rm{ = 1 - }}\phi {\rm{( - 4)}}\end{array}\)

To find\(\phi {\rm{( - 4)}}\), look in Appendix Table\({\rm{A}}{\rm{.3}}\)but last entry, however, is for\({\rm{z = - 3}}{\rm{.4}}\), which equals\({\rm{0}}{\rm{.0002}}\). As a result,\(\phi {\rm{( - 4)}} \approx {\rm{0}}{\rm{.0001}}\).

\(\begin{array}{c}{\rm{1 - }}\phi {\rm{( - 4) = 1 - 0}}{\rm{.0001}}\\{\rm{ = 0}}{\rm{.9999}}\end{array}\)

Hence,

\({\rm{P(X}} \ge {\rm{10) = 0}}{\rm{.9999}}\)

Let\({\rm{X}}\)be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F}}\)as parameters\({\rm{(x)}}\). After that, for any number a,

\({\rm{P(X}} \ge {\rm{a) = 1 - F(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.9999}}\).

(d)Standardization results in:

\({\rm{14}} \le {\rm{X}} \le {\rm{18}}\)

only if and only if

\(\begin{array}{c}\frac{{{\rm{14 - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le \frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le \frac{{{\rm{18 - 15}}}}{{{\rm{1}}{\rm{.25}}}}\\{\rm{ - 0}}{\rm{.8}} \le \frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le {\rm{2}}{\rm{.4}}\\{\rm{ - 0}}{\rm{.8}} \le {\rm{Z}} \le {\rm{2}}{\rm{.4}}\end{array}\)

Thus

\({\rm{P(14}} \le {\rm{X}} \le {\rm{18) = P( - 0}}{\rm{.8}} \le {\rm{Z}} \le {\rm{2}}{\rm{.4)}}\)

\({\rm{Z}}\)is a standard normal distribution\({\rm{rv}}\)with cdf\(\phi {\rm{(z)}}\). Hence

\(\begin{array}{c}{\rm{P(14}} \le {\rm{X}} \le {\rm{18) = P( - 0}}{\rm{.8}} \le {\rm{Z}} \le {\rm{2}}{\rm{.4)}}\\{\rm{ = }}\phi {\rm{(2}}{\rm{.4) - }}\phi {\rm{( - 0}}{\rm{.8)}}\end{array}\)

To find\(\phi {\rm{(2}}{\rm{.4)}}\), look in Appendix Table\({\rm{A}}{\rm{.3}}\)at the intersection of the\({\rm{2}}{\rm{.4}}\)row and the\({\rm{.00}}\)column. There's a number\({\rm{0}}{\rm{.9918}}\), so

\(\phi {\rm{(2}}{\rm{.4) = 0}}{\rm{.9918}}\)

And

\(\phi {\rm{( - 0}}{\rm{.8) = 0}}{\rm{.2119}}\)

Hence,

\(\begin{array}{l}{\rm{P(14}} \le {\rm{X}} \le {\rm{18) = 0}}{\rm{.9918 - 0}}{\rm{.2119}}\\{\rm{P(14}} \le {\rm{X}} \le {\rm{18) = 0}}{\rm{.7799}}\end{array}\)

If\({\rm{Z}}\)is a continuous\({\rm{rv}}\)with cdf, then the following proposition is\(\phi {\rm{(z)}}\). Then, using\({\rm{a < b}}\), for any two numbers\({\rm{a}}\)and\({\rm{b}}\),

\({\rm{P(a}} \le {\rm{Z}} \le {\rm{b) = }}\phi {\rm{(b) - }}\phi {\rm{(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.7799}}\).

(e)Standardization results in:

\(\begin{array}{l}\left| {{\rm{X - 15}}} \right| \le {\rm{3}}\\{\rm{ - 3}} \le {\rm{X - 15}} \le {\rm{3}}\end{array}\)

only if and only if

\(\begin{array}{c}\frac{{{\rm{ - 3}}}}{{{\rm{1}}{\rm{.25}}}} \le \frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le \frac{{\rm{3}}}{{{\rm{1}}{\rm{.25}}}}\\{\rm{ - 2}}{\rm{.4}} \le \frac{{{\rm{X - 15}}}}{{{\rm{1}}{\rm{.25}}}} \le {\rm{2}}{\rm{.4}}\\{\rm{ - 2}}{\rm{.4}} \le {\rm{Z}} \le {\rm{2}}{\rm{.4}}\end{array}\)

Thus

\({\rm{P(}}\left| {{\rm{X - 15}}} \right| \le {\rm{3) = P( - 2}}{\rm{.4}} \le {\rm{Z}} \le {\rm{2}}{\rm{.4)}}\)

\({\rm{Z}}\)is a standard normal distribution\({\rm{rv}}\)with cdf\(\phi {\rm{(z)}}\). Hence

\(\begin{array}{c}{\rm{P(}}\left| {{\rm{X - 15}}} \right| \le {\rm{3) = P( - 2}}{\rm{.4}} \le {\rm{Z}} \le {\rm{2}}{\rm{.4)}}\\{\rm{ = }}\phi {\rm{(2}}{\rm{.4) - }}\phi {\rm{( - 2}}{\rm{.4)}}\end{array}\)

To find\(\phi {\rm{(2}}{\rm{.4)}}\), look in Appendix Table\({\rm{A}}{\rm{.3}}\)at the intersection of the\({\rm{2}}{\rm{.4}}\)row and the\({\rm{.00}}\)column. There's a number\({\rm{0}}{\rm{.9918}}\), so

\(\phi {\rm{(2}}{\rm{.4) = 0}}{\rm{.9918}}\)

We can write: using the symmetry of the standard normal distribution curve:

\(\begin{array}{c}\phi {\rm{( - 2}}{\rm{.4) = 1 - }}\phi {\rm{(2}}{\rm{.4)}}\\{\rm{ = 0}}{\rm{.0082}}\end{array}\)

Hence,

\(\begin{array}{l}{\rm{P(|X - 15|}} \le {\rm{3) = 0}}{\rm{.9918 - 0}}{\rm{.0082}}\\{\rm{P(|X - 15|}} \le {\rm{3) = 0}}{\rm{.9836}}\end{array}\)

Therefore, the value is \({\rm{0}}{\rm{.9836}}\).