Question

A normal variable X has mean 400 and variance 900. FInd the fourth decile and 90th percentile for this distribution and also interpret the values.

Answer 1

Here, µ = 400; σ² = 900

∴ σ = 30
Fourth Decile D₄:

40% of the observations are less than D₄.

P[X ≤ D₄] = P[Z ≤ Z₁] = 0.40
∴ P[z₁ ≤ Z ≤ 0] = P[- ∞ < Z ≤ 0] – P[Z ≤ z₁]
= 0.50 – 0.40
= 0.10

Hence, the fourth decile D4 obtained is 392.35.

Interpretation: 40 % of the observations of the given distribution are less than 392.35.

90th Percentile P_9o:
90% of the observations are less than P_9o.

P[X ≤ P_9o] = P[Z ≤ Z₂] = 0.90
∴ P[0 ≤ Z ≤ z₂] = P[Z ≤ z₂] – P[- ∞ < Z ≤ 0]
= 0.90 – 0.50
= 0.40
For area 0.3997 ≈ 0.40, z₂ = 1.28

Now, z₂ = \(\frac{P90−400}{30}\)

∴ 1.28 = \(\frac{P90−400}{30}\)

∴ 1.28 × 30 = P_9o – 400∴ 38.4 = P_9o – 400

∴ P_9o = 38.4 + 400

∴ P_9o = 438.4

Hence, the 90th percentile P_9o obtained is 438.4.
Interpretation: 90 % of the observations of the given distribution are less than 438.4.