Answer to Question #267792 in Statistics and Probability for Ann

Question

The distribution of monthly income of 3000 workers of a factory confirms normal law with mean Rs.900 and standard deviation Rs.100. Find i. the percentage of workers having more than Rs. 800 income ii. the number of workers having less than Rs. 600 income iii. the range of income of middle 80% of workers

Answer 1

Let X be a random variable represenrs the monthly income of randomly selected worker. The X ~ $N(900, 100^2)=900+100N(0,1)$

(i) $P(X>800)=P(900+100N(0,1)>800)=P(N(0,1)>-1)=0.84134$

So, there is 0.84134 * 100% $\approx$ 84% percent of workers with salary more than 800

(ii) $P(X<600)=P(900+100N(0,1)<600)=P(N(0,1)<-3)=0.00135$

So, there is 0.00135 * 100% = 0.135% percent of workers with salary less than 600

(iii) We have to find numbers a, b, such that

$P(900+100N(0,1)<b)=0.9\implies {\frac {b-900} {100}}=1.28\implies b=1028$ 

$P(900+100N(0,1)<a)=0.1\implies {\frac {a-900} {100}}=-1.28\implies a=772$

The range of income of middle 80% of workers is Rs. 772-1028