Question

The monthly average salary of workers working in a production house is ₹ 10000 and its standard deviation is ₹ 2000. By assuming that the monthly salary of a worker follows normal distribution, estimate the maximum salary of 20% of the workers having lowest salary. Also estimate the minimum salary of 10 % of the workers having highest salary.

Answer 1

Here, X = Monthly salary; µ = 10000; σ = 2000

Maximum salary of 20 % workers having lowest salary:

Suppose, maximum salary = ₹ x₁

P[Z ≤ z₁] = 0.2
= P[- ∞ < Z ≤ 0] – P[z₁ ≤ Z ≤ 0]
= P[0 ≤ Z < ∞] – P[0 ≤ Z ≤ z₁]
∴ P[0 ≤ Z ≤ z₁] = 0.5000 – 0.2000
= 0.3000

From the table for area 0.2995, z₁ = – 0.84 and for area 0.3023,

From the table for area 0.2995, z₁ = – 0.84 and for area 0.3023,

z₁ = – 0.85 and for area \(\frac{0.2995+0.3023}{2} \)= 0.3009,
z₁ =\( \frac{(−0.84)+(−0.85)}{2}\) = – 0.845

(∵ z₁ is on left side.)

According to the above calculation, 0.3000 is very near to 0.2995, so we will take z₁ = – 0.84.

Now, z₁ = \(\frac{x_1−10000}{2000}\)

∴ – 0.84 = \(\frac{x_1−10000}{2000}\)

∴ – 0.84 × 2000 = x₁ – 10000
∴ – 1680 = x₁ – 10000
∴ 10000 – 1680 = x₁
∴ x₁ = 8320

Hence, the maximum salary of 20 % of workers having lowest salary is ₹ 8320.

The minimum salary of 10 % of workers having maximum salary:

Suppose, maximum salary = ₹ x₂

P[Z ≥ z₂] = 0.2
= P[0 ≤ Z < ∞] – P[0 ≤ Z ≤ z₂]
∴ P[0 ≤ Z ≤ z₂] = = P[0 ≤ Z < ∞] – P[Z ≥ z₂]
= 0.5000 – 0.1000
= 0.4000

From the table for area 0.2995, z₂ = 1.28 and for area 0.4015,

z₂ = 1.29 and for area\(\frac{ 0.3997+0.4015}{2 }\)= 0.4006,

z₂ = \(\frac{1.28+1.29}{2}\) = – 1.285

According to the above calculation, 0.4000 is near to 0.3997, so we will take z₂ = 1.28.

Now, z₂ =\(\frac{ x_2−10000}{2000}\)

∴ 1.28 = \(\frac{ x_2−10000}{2000}\)

∴ 1.28 × 2000 = x₂ – 10000
∴2560 = x₂ – 10000
∴ x₂ = 10000 + 2560
∴ x₂ = 12560

Hence, the maximum salary of 10 % of workers having maximum salary is ₹ 12560.