Question

There is no nice formula for the standard normal cdf\({\rm{\Phi (z)}}\), but several good approximations have been published in articles. The following is from "Approximations for Hand Calculators Using Small Integer Coefficients" (Mathematics of Computation, 1977: 214-222). For \({\rm{0 < z}} \le {\rm{5}}{\rm{.5}}\)

\(\begin{array}{l}{\rm{P(Z}} \ge {\rm{z) = 1 - \Phi (z)}}\\ \Leftrightarrow {\rm{.5exp}}\left\{ {{\rm{ - }}\left( {\frac{{{\rm{(83z + 351)z + 562}}}}{{{\rm{703/z + 165}}}}} \right)} \right\}\end{array}\)

The relative error of this approximation is less than \({\rm{.042\% }}\). Use this to calculate approximations to the following probabilities, and compare whenever possible to the probabilities obtained from Appendix Table A.3.

a. \({\rm{P(Z}} \ge {\rm{1)}}\)

b. \({\rm{P(Z < - 3)}}\)

c. \({\rm{P( - 4 < Z < 4)}}\)

d. \({\rm{P(Z > 5)}}\)

Short answer

(a) The approximation for the \({\rm{P(Z}} \ge {\rm{1)}}\) is\({\rm{0}}{\rm{.1587}}\).

(b) The approximation for the \({\rm{P(Z < - 3)}}\) is \({\rm{0}}{\rm{.00135}}\).

(c) The approximation for the \({\rm{P( - 4 < Z < 4)}}\) is \({\rm{0}}{\rm{.99997}}\).

(d) The approximation for the \({\rm{P(Z > 5)}}\) is \({\rm{0}}{\rm{.00000029}}\).

Step-by-step solution

Definition of Concept

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

Calculate approximations to the following probabilities

(a)

Considering the given information:

The standard normal cumulative density function can be approximated by,

\(P(Z \ge z) = \left\{ {\begin{array}{*{20}{l}}{0.5\exp \left\{ { - \left( {\frac{{(83z + 351)z + 562}}{{\frac{{703}}{z} + 165}}} \right)} \right\}0 \le z \le 5.5}\\{{\rm{ otherwise }}}\end{array}} \right.\)

The\(P(Z \ge 1)\)is obtained as shown below:

\(\begin{aligned}P(Z \ge 1) \approx 0.5 \times \exp \left\{ { - \left( {\frac{{(83(1) + 351)(1) + 562}}{{\frac{{703}}{1} + 165}}} \right)} \right\}\\ &= 0.5 \times \exp \left\{ { - \left( {\frac{{(83 + 351) + 562}}{{703 + 165}}} \right)} \right\}\\ &= 0.5 \times \exp \left\{ { - \left( {\frac{{996}}{{868}}} \right)} \right\}\\ &= 0.5 \times \exp \{ - (1.147)\} \\ &= 0.5 \times 0.3176\\ &= 0.15879\end{aligned}\)

Therefore, the \(P(Z \ge 1)\) is \({\rm{0}}{\rm{.15879}}\).

Calculate approximations to the following probabilities

(b)

Considering the given information:

It is a symmetric distribution because Z is an approximation of the standard normal variable.

Hence, \({\rm{P(Z < - 3) = P(Z > 3)}}\)

Thus,

\(\begin{aligned}P(Z < - 3) &= P(Z > 3) \\ &= 0 {\rm{.5 \times exp}}\left\{ {{\rm{ - }}\left( {\frac{{{\rm{(83(3) + 351)(3) + 562}}}}{{\frac{{{\rm{703}}}}{{\rm{3}}}{\rm{ + 165}}}}} \right)} \right\}\\ &= 0 {\rm{.5 \times exp}}\left\{ {{\rm{ - }}\left( {\frac{{{\rm{(249 + 351)(3) + 562}}}}{{\frac{{{\rm{703}}}}{{\rm{3}}}{\rm{ + 165}}}}} \right)} \right\}\\ &= 0 {\rm{.5 \times exp}}\left\{ {{\rm{ - }}\left( {\frac{{{\rm{2362}}}}{{{\rm{399}}{\rm{.33}}}}} \right)} \right\}\\ &= 0 {\rm{.5 \times exp\{ - (5}}{\rm{.91)\} }}\\ &= 0{\rm{.5 \times 0}}{\rm{.002699}}\\ &= 0 {\rm{.00135}}\end{aligned}\)

Therefore, the \({\rm{P(Z < - 3)}}\) is\({\rm{0}}{\rm{.00135}}\).

Calculate approximations to the following probabilities

(c)

Considering the given information:

The range of Z is\(0 \le z \le 5.5\), thus\(P( - 4 < Z \le 0) = 0\)

The\({\rm{P( - 4 < Z < 4)}}\)is obtained as shown below:

\(\begin{aligned}P( - 4 < Z < 4) &= P( - 4 < Z \le 0) + P(0 < Z < 4)\\ &= 0 + P(0 < Z < 4\\ &= 1 - P(Z \ge 4)\\ &= 1 - 0.5 \times \exp \left\{ { - \left( {\frac{{(83(4) + 351)(4) + 562}}{{\frac{{703}}{4} + 165}}} \right)} \right\}\\ &= 1 - 0.5 \times \exp \left\{ { - \left( {\frac{{(332 + 351)(4) + 562}}{{\frac{{703}}{4} + 165}}} \right)} \right\}\\ &= 1 - 0.5 \times \exp \left\{ { - \left( {\frac{{3.294}}{{340.75}}} \right)} \right\}\\ &= 1 - (0.5 \times \exp \{ - (9.6669)\} )\\ &= 1 - (0.5 \times 0.0000633)\\ &= 1 - 0.0000317\\ &= 0.99997\end{aligned}\)

Therefore, the \({\rm{P( - 4 < Z < 4)}}\) is \({\rm{0}}{\rm{.99997}}\).

Calculate approximations to the following probabilities

(d)

Considering the given information:

The\({\rm{P(Z > 5)}}\)is obtained as shown below:

\(\begin{aligned}P(Z > 5) \approx 0.5 \times \exp \left\{ { - \left( {\frac{{(83(5) + 351)(5) + 562}}{{\frac{{703}}{5} + 165}}} \right)} \right\}\\ &= 0.5 \times \exp \left\{ { - \left( {\frac{{(415 + 351)(4) + 562}}{{\frac{{703}}{5} + 165}}} \right)} \right\}\\ &= 0.5 \times \exp \left\{ { - \left( {\frac{{4.392}}{{305.6}}} \right)} \right\}\\ &= 0.5 \times \exp \{ - (14.37)\} \\ &= 0.5 \times 0.000000573\\ &= 0.00000029\end{aligned}\)

Therefore, the \({\rm{P(Z > 5)}}\)is\({\rm{0}}{\rm{.00000029}}\).