Answer to Question #120060 in Statistics and Probability for FRED

Question

The weekly incomes of a large group of Fire officers are normally distributed with a
mean of K800 and a standard deviation of K85.
(i) What is the probability of finding a Fire officers with a weekly income of
between K840 and K900?
(ii) What is the percent of Fire officers that earn more than K905?
(iii) What is the percent of Fire officers that earn less than K905?
(iv) What is the probability of finding a Fire officer with weekly income of
between K750 and K850?
(v) What is the probability of finding a Fire officer with a weekly income of
between K700 and K790?
(vi) Above what income would the top 10% of the officers earn?
(vii) Below what income would the lowest 10% of the officers earn?

Answer 1

Let X denote the random variable for weekly incomes of fire officers .

X~N(800, 85²)

$Z=\frac {x-\mu} {\sigma}$

i) P(840<X<900)

=P(X<900)-P(X<840)

$z_1=\frac{840 - 800}{85}$ =0.4706

$\Phi(0.4706)=0.6810$

$z_2=\frac{900 - 800}{85}$ =1.176

$\Phi(1.176)=0.8803$

P(840<X<900)=0.8803-0.6810

=0.1993

ii) P(X>905)

=1-P(x<905)

$z=\frac{905 - 800}{85}$ =1.2352

$\Phi (1.2352)=0.8916$

=1-0.8916

=0.1084

iii) P(X<905)

$z=\frac{905 - 800}{85}$ =1.2352

$\Phi (1.2352)=0.8916$

=0.8916

iv) P(750<X<850)

=P(X<850)-P(X<750)

= $z_1=\frac{750 - 800}{85}=-0.5882$

$\Phi (-0.5882)= 0.2782$

$z_2=\frac{850 - 800}{85}=0.5882$

$\Phi (-0.5882)= 0. 7218$

=0.7218-0.2782

=0.4436

v) P(700<X<790)

P(X<790)-P(X<700)

$z_1=\frac{700 - 800}{85}=-1.1764$

$\Phi(-1.1764) =0.1197$

$z_2=\frac{790 - 800}{85}=-0.1176$

$\Phi(-0.1176)=0.4532$

=0.4532-0.1197

=0.3334

vi) P(X>x) =0.1

$\Phi^{-1}(1-0.1)= 1.2816$

$1.2816 =\frac {x-800}{85}$

x=K908. 94

vii) P(X<0.10)

$\Phi^{-1}(0.1)=-1.2816$

$-1.2816 =\frac {x-800}{85}$

x=K691. 07