Answer to Question #110545 in Statistics and Probability for kimote

Question

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

Answer 1

Let $X=$ the weekly income: $X\sim N(\mu,\sigma^2)$

Given that $\mu=800,\sigma=85$

$$Z={X-\mu \over \sigma}\sim N(0,1)$$

(i) The probability of finding a middle manager with a weekly income of between K840 and K900 is

$$P(840<X<900)=P(X<900)-P(X\leq840)=$$ $$=P(Z<{900-800 \over85})-P(X\leq{840-800 \over85})\approx$$ $$\approx P(Z<1.176471)-P(X\leq0.470588)\approx$$ $$\approx0.88030-0.68103\approx0.1993$$

(ii) The percent of middle managers that earn more than K905 is 

$$P(X>905)=1-P(X\leq905)=$$ $$=1-P(Z\leq{905-800 \over85})\approx1-P(Z\leq1.23529)\approx$$ $$\approx1-0.89164\approx0.1084\approx10.84\%$$

(iii) The percent of middle managers that earn less than K905 is  

$$P(X<905)=P(Z<1.23529)\approx$$ $$\approx0.89164\approx89.16\%$$

(iv) The probability of finding a middle manager with weekly income of between K750 and K850 is

$$P(750<X<850)=P(X<850)-P(X\leq750)=$$ $$=P(Z<{850-800 \over85})-P(X\leq{750-800 \over85})\approx$$ $$\approx P(Z<0.588235)-P(X\leq-0.588235)\approx$$ $$\approx0.72181-0.27819\approx0.4436\approx44.36\%$$

(v) The probability of finding a middle manager with a weekly income of between K700 and K790 is

$$P(700<X<790)=P(X<790)-P(X\leq700)=$$ $$=P(Z<{790-800 \over85})-P(X\leq{700-800 \over85})\approx$$ $$\approx P(Z<-0.117647)-P(X\leq-1.176471)\approx$$ $$\approx0.45317-0.11970\approx0.3335$$

(vi) Above what income would the top 10% of the managers earn?  

$$P(Z>z^*)=0.1$$

$$z^*={x^*-800 \over 85}\approx1.28155$$

$$x^*\approx909$$

(vii) Below what income would the lowest 10% of the managers earn?

$$P(Z<z^*)=0.1$$

$$z^*={x^*-800 \over 85}\approx-1.28155$$

$$x^*\approx691$$