Answer to Question #112647 in Statistics and Probability for Yaliwe Mwilwa

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

1) P(840<x<900)=P(x<900)-P(x<840)

Z1=(x-mean)/sd=(900-800)/85=1.176, using table P(x<900)=0.88

Z2=(840-800)/85=0.47, using table P(x<840)=0.681

P(840<x<900)=0.199=19.9%

2) P(x>905)=1-P(x<905)

z=(905-800)/85=1.235

using table: P(x>905)=1-0.8916=0.1084=10.84%

3) P(x<905)

z=1.235

P(x<905)=0.8916=89.16%

4) P(750<x<850)=P(x<850)-P(x<750)

Z1=(850-800)/85=0.588, using table P(x<850)=0.72181

Z2=(750-800)/85=-0.588, using table P(x<750)=0.27819

P(750<x<850)=0.4436=44.36%

5) P(700<x<790)=P(x<790)-P(x<700)

Z1=(790-800)/85=-0.1176, using table P(x<790)=0.45317

Z2=(700-800)/85=-1.176, using table P(x<700)=0.11970

P(700<x<790)=0.3335=33.35%

6)P(x>X)=0.1

z=(X-800)/85=1.28155 - using table

X=909

7)P(x<X)=0.1

z=(X-800)/85=-1.28155 - using table

X=691