Question

The standard normal probability function can be approximated as
$F\left(X_N\right)=\frac{1}{1+exp(-1.7255X_N|X_N{|}^{0.12})},\text{where}{X}_N=$ standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 cm and 102 cm is

• A. 66.7%
• B. 50.0%
• C. 33.3%
• D. 16.7%

Answer 1

The correct option is D 16.7%

Given standard normal probability function
$F\left(X_N\right)=\frac{1}{1+exp(-1.7255X_N|X_N{|}^{0.12})}$

$X_N=\text{Standard normal deviate mean}$

$\left(\mu \right)=102cm$

$S.D\left(\sigma \right)=27cm$

We are required to find

$P(90\le x\le 102)=P(\frac{90-\mu }{\sigma }\le X_N\le \frac{102-\mu }{\sigma })$

$=P(\frac{90-102}{27}\le X_N\le 0)$

$=P(-0.44\le X_N\le 0)$

$=F\left(0\right)-F(-0.44)$

$=\frac{1}{1+exp\left(0\right)}-\frac{1}{1+exp[-1.7255(-0.44)|-0.44{|}^{0.12}]}$

$=\frac{1}{2}-\frac{1}{1+exp[1.7255\times 0.44\times (0.44{)}^{0.12}]}$

$=\frac{1}{2}-\frac{1}{1+exp\left(0.688\right)}$

$=\frac{1}{2}-\frac{1}{2.9897}$

= 16.55% (approx. value)