Uniform and Normal Distributions

Introduction: Uniform and normal distributions are essential concepts in probability and statistics, representing different patterns of random variables’ probabilities. In this series of questions, we’ll explore these distributions through numerical examples and provide step-by-step solutions.

Continuous Probability Distribution with Example

A continuous probability distribution deals with variables that can take on any real value within a range. Let’s dive into this concept with an example using the normal distribution, often called the bell curve.

Example: Exam Scores

Imagine you have a class of 100 students, and you want to understand how their scores are distributed in a challenging math exam. You record their scores and notice that they form a bell-shaped curve.

- Average Score (Mean): The peak of the curve represents the average score, let’s say it’s 75 out of 100.

• Spread (Standard Deviation): The curve’s width represents the variation in scores. A smaller spread means scores are clustered around the mean, while a larger spread means scores are more scattered.

Now, you can use this distribution to answer questions like:

1. Probability of Scoring Below 70: What’s the chance a random student scored less than 70? You calculate it using the normal distribution, finding it’s 15%.

2. Probability of Scoring Above 85: What’s the likelihood of scoring above 85? Again, using the normal distribution, you find it’s 10%.

3. Range of Scores: You can estimate that about 68% of students scored between 65 and 85, approximately.

In essence, continuous probability distributions help you make numerical predictions and understand how data is spread out. They’re practical tools in fields like statistics, science, and finance, aiding in decision-making and problem-solving based on real-world data.

Let’s explore other practical examples:

• *Question 1: Uniform Distribution**

Question: What is the probability of a random variable X lying between -1.5 and +2.5, i.e., P(-1.5 < X < 2.5)?

Options: A) 0.1 B) 0.4

Answer: B) 0.4

• Explanation:
• Calculate the area under the PDF curve between -1.5 and 2.5.
• In a uniform distribution, the PDF curve’s height is constant.
• Since the width of the interval is 4 units and the constant PDF value is 0.1, the probability is 4 * 0.1 = 0.4.

The normal distribution, also known as the Gaussian distribution, owes its discovery to Carl Friedrich Gauss in 1809. Gauss’s quest was to formulate a probability distribution that could model the errors inherent in astronomical observations. These errors, referred to as astronomical errors, occur when astronomers measure various phenomena like the distances in space.

For instance, Gauss’s work revealed that when an astronomer attempts to gauge the distance between Earth and Uranus, there’s always some degree of error involved. These errors conform to a normal distribution, characterized by a mean (µ) of 0 kilometers and a standard deviation (σ) of 1,000 kilometers.

*Question:What is the probability of the astronomer overestimating the distance by 2,330 km or more?**

Options: A) 1% B) 2% C) 1.5%

Answer: A) 1%

Explanation:

• Astronomical errors follow a normal distribution with µ = 0 km and σ = 1,000 km.
• We need to find P(X > 2330).
• Convert it to a standard normal random variable Z: P(Z > 2.33).
• Calculate using a Z-table: P(Z > 2.33) = 1 — P(Z < 2.33) = 1–0.9901 ≈ 0.0099 or 0.99%.

Question: What is the probability that the astronomer under- or over-estimates the distance by less than 500 km?

Options: A) 30.85% B) 69.15% C) 38.30%

Answer: C) 38.30%

• **Explanation:***
• The same normal distribution with µ = 0 km and σ = 1,000 km is used.
• We want P(-500 < X < 500).
• Convert to standard normal random variable Z: P(-0.5 < Z < 0.5).
• Calculate using a Z-table: P(-0.5 < Z < 0.5) = P(Z < 0.5) — P(Z < -0.5) = 0.6915–0.3085 = 0.3830, or 38.30%.

In light of severe rainfall and subsequent floods in Kerala, the state government seeks to assess the probability of such extreme rainfall events occurring during August. This analysis will aid in the redesign of current infrastructure and preparation for future events. The rainfall data is assumed to follow a normal distribution, with an average (μ) of 1600 mm and a standard deviation (σ) of 400 mm.

Question:

Given that floods were caused by a rainfall of 2200 mm, what is the probability that Kerala will experience more than 2200 mm of rainfall in this period?

Options:

A) 3.5%

B) 4.8%

C) 1.2%

D) 6.7%

Answer: 6.7%

Explanation:

To solve this, we’ll use the Z-score formula: Z = (X — μ) / σ, where X is the value (2200 mm), μ is the mean (1600 mm), and σ is the standard deviation (400 mm). Calculating the Z-score: (2200–1600) / 400 = 1.5.

Now, we want to find P(X > 2200), which is equivalent to P(Z > 1.5). Using the Z-table, we find P(Z > 1.5) is approximately 0.0668 or 6.7%.

Question:

To ensure there’s only a 3% chance of experiencing similar or heavier rains in the future, what should be the cutoff rainfall value for infrastructure redesign?

Options:

A) 2433 mm

B) 2352 mm

C) 2295 mm

D) 2330 mm

Answer: 2352 mm

Explanation:

For this scenario, we set up the equation P(Z ≥ z) = 0.03, where z is the Z-score. Solving for z gives us z ≈ 1.88. Now, we use the Z-score formula (x — μ) / σ and plug in the values:

(x — 1600) / 400 = 1.88.

Solving for x:

x = 1600 + (1.88 * 400) = 2352 mm.

So, the infrastructure should be redesigned for rainfall above 2352 mm to ensure a 3% or lower chance of such events in the future.