91Ó°ÊÓ

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

(a) The probability that maximum speed is at most \({\rm{50\;km/h}}\) is denoted as\(P(X \le 50)\). Standardizing gives:

\(X \le 50\)if and only if

\(\frac{{X - 46.8}}{{1.75}} \le \frac{{50 - 46.8}}{{1.75}}\frac{{X - 46.8}}{{1.75}} \le \frac{{3.2}}{{1.75}}Z \le 1.83\)

Thus

\(P(X \le 50) = P(Z \le 1.83)\)

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

\(P(X \le 50) = P(Z \le 1.83) = \phi (1.83)\)

To get\(\phi (1.83)\), we check Appendix Table \({\rm{A}}{\rm{. 3}}\)at the intersection of the row marked \({\rm{1}}{\rm{.8}}\) and the column marked .03 The number there is\({\rm{0}}{\rm{.9664}}\), so

\(\phi (1.83) = 0.9664\)

Hence

\(P(X \le 50) = 0.9664\)

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

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

(b) The probability that maximum speed is at least \({\rm{48\;km/h}}\)is denoted as\(P(X \ge 48)\). Standardizing gives:

\(X \ge 48\)

if and only if

\(\frac{{X - 46.8}}{{1.75}} \ge \frac{{48 - 46.8}}{{1.75}}\frac{{X - 46.8}}{{1.25}} \ge \frac{{1.2}}{{1.75}}Z \ge 0.69\)

Thus

\(P(X \ge 48) = P(Z \ge 0.69)\)Here \({\rm{Z}}\)is a standard normal distribution \(rv\) with \(cdf\) \(\phi (z)\). Hence

\(P(X \ge 48) = P(Z \ge 0.69) = 1 - \phi (0.69)\)

To get\(\phi (0.69)\), we check Appendix Table \({\rm{A}}{\rm{.3}}\), from there

\(\begin{array}{l}\phi (0.69) = 0.7549\\1 - \phi (0.69) = 1 - 0.7549 = 0.2451\end{array}\)

Hence

\(P(X \ge 48) = 0.2451\)

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

\(P(X \ge a) = 1 - F(a)\)

(c) The probability that maximum speed differs from the mean value by at most \({\rm{1}}{\rm{.5}}\)standard deviations \({\rm{( = 1}}{\rm{.5 \times 1}}{\rm{.75 = 2}}{\rm{.625)}}\)is denoted as\(P(|X - 46.8| \le 2.625)\), Standardizing gives:

\(|X - 46.8| \le 2.625{\rm{ }} - 2.625 \le X - 46.8 \le 2.625\)

if and only if

\(\frac{{ - 2.625}}{{1.75}} \le \frac{{X - 46.8}}{{1.75}} \le \frac{{2.625}}{{1.75}} - 1.5 \le \frac{{X - 46.8}}{{1.75}} \le 1.5 - 1.5 \le Z \le 1.5\)

Thus

\(P(|X - | \le 2.625) = P( - 1.5 \le Z \le 1.5)\)

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

\(\begin{array}{l}P(|X - 46.8| \le 2.625)\\ = P( - 1.5 \le Z \le 1.5)\\ = \phi (1.5) - \phi ( - 1.5)\end{array}\)

To get\(\phi (1.5)\), we check Appendix Table A. \({\rm{3}}\) at the intersection of the row marked \({\rm{1}}{\rm{.5}}\)and the column marked .00 The number there is\({\rm{0}}{\rm{.9918}}\), so

\(\phi (1.5) = 0.9332\)

Using symmetry of standard normal distribution curve, we can write:

\(\phi ( - 1.5) = 1 - \phi (1.5) = 0.0668\)

Hence

\(\begin{array}{l}P(|X - 46.8| \le 2.625) = 0.9332 - 0.0668\\P(|X - 46.8| \le 2.625) = 0.8664\end{array}\)

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

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