The monthly income of 5000 workers at a plant are distribute

The monthly income of 5000 workers at a plant are distributed normally. Suppose the mean monthly income is $1250 and the standard deviation is $200.

a) What is the minimum salary of a worker who is in top 10%?

b) What is the minimum salary of a worker who is in top 30%?

c) What percentage of the workers earn between $750 and $1500 per month?

d) What percentage of the workers earn less than $1750 per month?

Solution

Mu = 1250

Standard Devition = 200

a)

P-value = 0.1

Therefore z value = -1.28

z = (X-bar - Mu)/(SD)

=> -1.28 = (X-bar - 1250)/ (200)

=>X-bar = -1.28 * (200) + 1250 = $994 Answer

b)

P-value = 0.3

Therefore z value = -0.52

z = (X-bar - Mu)/(SD)

=> -0.52 = (X-bar - 1250)/ (200)

=>X-bar = -0.52 * (200) + 1250 = $1146 Answer

c)

z1 = (1500-1250)/(200) = 1.25 ; P-value = 0.8944

z2 = (750-1250)/(200) = -2.5 ; P- value = 0.0062

P(750<X<1500) = 0.8944 - 0.0062 = 0.8882

Therefore,

Percentage of workers = 88.82% Answer

d)

z (X<1750) = (1750-1250)/(200) = 2.5  

P-value = 0.9938

Therefore,

Percentage of workers = 99.38% Answer