91影视

The term "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.

(a) The probability that the concentration exceeds \({\rm{0}}{\rm{.50}}\)is denoted as\({\rm{P(X > 0}}{\rm{.5)}}\). Standardizing gives:

\({\rm{X > 0}}{\rm{.5}}\)if and only if

\(\frac{{{\rm{X - 0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.06}}}}{\rm{ > }}\frac{{{\rm{0}}{\rm{.5 - 0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.06}}}}\frac{{{\rm{X - 0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.06}}}}{\rm{ > }}\frac{{{\rm{0}}{\rm{.2}}}}{{{\rm{0}}{\rm{.06}}}}{\rm{Z > 3}}{\rm{.33}}\)

Thus

\({\rm{P(X > 0}}{\rm{.5) = P(Z > 3}}{\rm{.33)}}\)

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

\(P(X > 0.5) = P(Z > 3.33) = 1 - \phi (3.33)\)

To get\(\phi (3.33)\), we check Appendix Table \({\rm{A}}{\rm{.3}}\), from there

\(\begin{array}{l}\phi (3.33) = 0.9996\\1 - \phi (3.33) = 1 - 0.9996 = 0.0004\end{array}\)

Hence

\({\rm{P(X > 0}}{\rm{.5) = 0}}{\rm{.0004}}\)

Proposition: Let \({\rm{X}}\)be a continuous \({\rm{rv}}\) with pdf \({\rm{f(x)}}\) and \({\rm{cof F}}\left( {\rm{x}} \right)\). Then for any number a,

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

b)

The probability that the concentration is at most \({\rm{0}}{\rm{.20}}\)is denoted as\(P(X \le 0.2)\). Standardizing gives:

\(X \le 0.2\)

if and only if

\(\frac{{X - 0.3}}{{0.06}} \le \frac{{0.2 - 0.3}}{{0.06}}\frac{{X - 0.3}}{{0.06}} \le \frac{{ - 0.1}}{{0.06}}Z \le - 1.67\)

Thus

\(P(X \le 0.2) = P(Z \le - 1.67)\)

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

\(P(X \le 0.2) = P(Z \le - 1.67) = \phi ( - 1.67)\)

To get\(\phi ( - 1.67)\), we check Appendix \({\rm{Table A}}{\rm{. 3}}\)at the intersection of the row marked \({\rm{ - 1}}{\rm{.6}}\)and the column marked\({\rm{.07}}\), from there

\(\phi ( - 1.67) = 0.475\)

Hence

\(P(X \le 0.2) = 0.475\)

Proposition: If \({\rm{Z}}\)be a continuous \({\rm{\;rv}}\) with \(cdf\phi (z)\). Then for any two numbers a and \({\rm{b}}\) with\({\rm{a < b}}\),

\(P(a \le Z \le b) = \phi (b) - \phi (a)\)

c)

The largest \({\rm{5\% }}\)of all concentration value denote all the values greater than \({{\rm{z}}_{{\rm{0}}{\rm{.05}}}}\)

If we recall, according to the definition, \({{\rm{z}}_{\rm{\alpha }}}\)is the \({\rm{100(1 - \alpha }}{{\rm{)}}^{{\rm{th }}}}\)percentile of the standard normal distribution.

\({{\rm{z}}_{{\rm{0}}{\rm{.05}}}}\)means that area to the right of \({{\rm{z}}_{{\rm{0}}{\rm{.05}}}}\)under standard normal distribution curve is \({\rm{0}}{\rm{.05}}\)

we can also say that:

\(\begin{array}{l}\phi \left( {{z_{0.05}}} \right) = 1 - 0.05\\\phi \left( {{z_{0.05}}} \right) = 0.95\end{array}\)

Where \(\phi (z)\)is the \({\rm{cdf}}\) of standard normal distributed \({\rm{rv}}\) \({\rm{z}}\).

We check Appendix Table \({\rm{A}}{\rm{. 3}}\)to see if \(\phi (z)\)is equal to \({\rm{0}}{\rm{.95}}\)for any \({\rm{z}}\).

The two closest values to \({\rm{0}}{\rm{.95}}\)are \({\rm{0}}{\rm{.9495}}\)and \({\rm{0}}{\rm{.9505}}\)which belong to \({\rm{z}}\)-values of \({\rm{1}}{\rm{.64}}\)and \({\rm{1}}{\rm{.65}}\)respectively. Since both values are equidistant from \({\rm{0}}{\rm{.95}}\), Hence

\({{\rm{z}}_{{\rm{0}}{\rm{.05}}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.64 + 1}}{\rm{.65}}}}{{\rm{2}}}{\rm{ = 1}}{\rm{.645}}\)

Let \({{\rm{X}}_{{\rm{0}}{\rm{.05}}}}\)denote the value of rv corresponding to \({\rm{z - }}\)value of\({{\rm{Z}}_{{\rm{0}}{\rm{.05}}}}\). Then

\(\begin{array}{c}\frac{{{{\rm{X}}_{{\rm{0}}{\rm{.05}}}}{\rm{ - 0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.06}}}}{\rm{ = }}{{\rm{Z}}_{{\rm{0}}{\rm{.05}}}}\\\frac{{{{\rm{X}}_{{\rm{0}}{\rm{.05}}}}{\rm{ - 0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.06}}}}{\rm{ = 1}}{\rm{.645}}\\{{\rm{X}}_{{\rm{0}}{\rm{.05}}}}{\rm{ = 0}}{\rm{.3 + (0}}{\rm{.06)(1}}{\rm{.645)}}\\{{\rm{X}}_{{\rm{0}}{\rm{.05}}}}{\rm{ = 0}}{\rm{.3987}}\end{array}\)

Hence the largest \({\rm{5\% }}\)of all concentration values are greater than \({\rm{0}}{\rm{.3987mg/c}}{{\rm{m}}^{\rm{3}}}{\rm{.}}\)