Question

The monthly income of a group of persons follows normal distribution with mean ₹ 20000 and standard deviation ₹ 5000. If the minimum monthly income of 50 richest person is ₹ 31625, then how many persons are in the group ? Also, what is the maximum income of 50 persons having lowest monthly income ?

Answer 1

Hence, µ = 20000; σ = 5000;

X = Monthly income

The minimum monthly income of 50 persons having maximum income is ₹ 31625.

Now,
P[Z₂ ≥ 2.33] = P[0 ≤ Z < ∞] – P[0 ≤ Z ≤ 2.33]
= 0.5000 – 0.4901
= 0.0099

∴ For area = 0.0099, the number of persons = 50

For area = 1.00, the number of persons

= \(\frac{50}{0.0099}\)

≈ 5051

Hence, the total number of persons in a group will be 5051.
Suppose, the maximum income of 50 persons = ₹ x₁

Now, P[Z ≤ z₁] = 0.01
∴ P[z₁ ≤ Z ≤ 0] = P[- ∞ < Z ≤ 0] – P[Z ≤ z₁]
= 0.5000 – 0.01
= 0.4900
For area 0.4901 ≈ 0.49, z₁ = – 2.33

Now, z₁ = \(\frac{x1−20000}{5000}\)

∴ – 2.33 = \(\frac{x1−20000}{5000}\)

∴ – 2.33 × 5000 = x₁ – 20000
∴ – 11650 = x₁ – 20000
∴ x₁ = 20000 – 11650
∴ x₁ = 8350

Hence, the maximum income of 50 persons having lowest income obtained is ₹ 8350.