Normal Distribution Problems: Z‑Scores, Percentages & Parameters

Problem 4: Bank Savings Distribution

The average savings in the clients' accounts of a certain bank follow a normal distribution N(50,000, 15,000) (mean = 50,000 €, standard deviation = 15,000 €). The bank distributes its clients into three categories:

• Category A: Those clients that have less than 20,000 € in their savings account
• Category B: Those clients that have between 20,000 € and 80,000 € in their savings account
• Category C: Those clients that have more than 80,000 € in their savings account

Calculate:

4a) What percentage of clients belong to Category A?

Use the standard normal variable Z = (X - μ) / σ.

For X = 20,000 €:

Z = (20,000 - 50,000) / 15,000 = -30,000 / 15,000 = -2.000

The cumulative probability Φ(-2.00) ≈ 0.0228, so ~2.28% of clients belong to Category A.

4b) What percentage of clients belong to Category B?

For X = 80,000 €:

Z = (80,000 - 50,000) / 15,000 = 30,000 / 15,000 = 2.000

Φ(2.00) ≈ 0.97725. The proportion between 20,000 € and 80,000 € is Φ(2.00) - Φ(-2.00) ≈ 0.97725 - 0.0228 = 0.95445 ≈ 95.45%.

4c) What percentage of clients from Category C have more than 100,000 €?

First find the proportion of all clients in Category C (X > 80,000 €):

P(X > 80,000) = 1 - Φ(2.00) ≈ 1 - 0.97725 = 0.02275 (≈ 2.275%).

Now find P(X > 100,000):

Z = (100,000 - 50,000) / 15,000 = 50,000 / 15,000 ≈ 3.3333

Φ(3.33) ≈ 0.99957, so P(X > 100,000) ≈ 1 - 0.99957 = 0.00043.

The fraction of Category C clients with more than 100,000 € is

0.00043 / 0.02275 ≈ 0.0189 ≈ 1.89%.

Problem 5: Monthly Expenditure Classification

A company classifies its clients in three groups:

• Those that buy less than 25,000 €/month
• Those that buy between 25,000 € and 100,000 €/month
• Those that buy more than 100,000 €/month

The monthly expenditure of clients follows a normal distribution N(μ, σ). If 10% of the clients belong to the first group and 15% to the second group, what are the average (μ) and the standard deviation (σ)?

Given:

• P(X < 25,000) = 10% ⇒ Φ(z1) = 0.10 ⇒ z1 ≈ -1.28155
• P(X < 100,000) = 25% ⇒ Φ(z2) = 0.25 ⇒ z2 ≈ -0.67449

Set up the z equations:

-1.28155 = (25,000 - μ) / σ

-0.67449 = (100,000 - μ) / σ

From these two equations:

25,000 + 1.28155·σ = 100,000 + 0.67449·σ

Therefore (1.28155 - 0.67449)·σ = 75,000 ⇒ 0.60706·σ ≈ 75,000

So σ ≈ 75,000 / 0.60706 ≈ 123,584 € (rounded).

Now compute μ using either equation, e.g. μ = 25,000 + 1.28155·σ ≈ 25,000 + 1.28155·123,584 ≈ 183,372 € (rounded).

Problem 6: Interpretation of Distribution Functions and Formula

Think about the figure (not shown) and explain in detail. What can you say about each one of the distribution functions? Further, explain the following formula and write the final expression for the standard normal. Explain.

Interpretation (corrected and clarified):

• The blue distribution has the smallest dispersion (smallest variance or standard deviation) and therefore is the most concentrated around its mean.
• The red distribution has the largest dispersion (largest variance or standard deviation) and is the most spread out.
• The blue and red distributions share the same mean in the figure, while the green distribution has a lower mean (its center is shifted to the left).
• On the vertical axis we find the frequencies (probabilities / probability density), and on the horizontal axis we find the values of X.

The formula corresponds to the normal distribution probability density function. It has parameters μ (mean) and σ (standard deviation). The general form is:

f(x) = (1 / (σ √(2π))) · exp( - (x - μ)² / (2 σ²) )

For the standard normal distribution (μ = 0, σ = 1) this simplifies to the standard normal density:

φ(z) = (1 / √(2π)) · exp( - z² / 2 )

Explanation: the coefficient 1 / (σ √(2π)) normalizes the area under the curve to 1. The exponential term exp( - (x - μ)² / (2 σ²) ) determines the bell shape; values further from the mean are exponentially less likely. For the standard normal, the mean is zero and the standard deviation is one, so the expression is simplified by setting μ = 0 and σ = 1.