Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: May 2015

Mean(m) = 3000

Standard deviation ($\sigma )=250$

Let X denote monthly salary of a worker.

Let $X_1$ be the minimum salary of the top 5\% workers.

Let $z_1$ be the corresponding pnv.

$$\mathrm{\therefore } P(X \gt X_1)=5%=0.05$$

$$\mathrm{\therefore } P(Z \gt Z_1)=0.05$$

$\mathrm{\therefore }$ Area between z=0 to z=5 = 0.5-0.05 = 0.45

From table $$z_1=1.645$$ $$\mathrm{\therefore } \frac{X_1-m}{\sigma } = 1.645$$ $$\mathrm{\therefore } \frac{X_1-3000}{250} = 1.645$$ $$X_1=3411.25$$

Hence the minimum salary of the top 5% workers = Rs 3411.25