M1 Study Notes: Applications of the Normal Distribution
Hello there! Welcome to the study notes for one of the most useful topics in statistics: Applications of the Normal Distribution.
Ever wondered how companies know how many T-shirts of each size to make, or how teachers can determine what a "good" score is on a new test? The answer often lies in the Normal Distribution, also famously known as the "bell curve".
In this chapter, we're going to move beyond the theory and learn how to use the normal distribution to solve real-world problems. We'll learn how to find probabilities (like the chance of a student scoring above 90) and how to find specific data points (like the score needed to be in the top 10%). Don't worry if this sounds tricky at first, we'll break it down step-by-step!
A Quick Review: What is Standardisation?
Before we jump into new problems, let's remember our most important tool: standardisation. Most real-life data is normally distributed but has its own unique mean ($$\mu$$) and standard deviation ($$\sigma$$). This is like every country having its own currency. To compare them easily, we need to convert them to a common standard.
In statistics, our "common standard" is the Standard Normal Distribution, which always has a mean of 0 and a standard deviation of 1. We denote this as $$Z \sim N(0, 1)$$.
The magic formula to convert any value (X) from a normal distribution $$N(\mu, \sigma^2)$$ into a standard value (Z) is the z-score formula:
$$Z = \frac{X - \mu}{\sigma}$$Where:
- X is your specific data point (e.g., a student's height).
- $$\mu$$ (mu) is the mean of the distribution.
- $$\sigma$$ (sigma) is the standard deviation of the distribution.
Common Mistake Alert!
Questions often give you the variance ($$\sigma^2$$), not the standard deviation ($$\sigma$$). Remember to always take the square root of the variance before using the formula! If $$X \sim N(100, 25)$$, then $$\mu = 100$$ and $$\sigma = \sqrt{25} = 5$$.
Part 1: Finding Probabilities from Values (The 'Forward' Problem)
This is the most common type of question. You are given a value (or values) of X, and you need to find a probability associated with it.
Example Question: The heights of students in a school are normally distributed with a mean of 165 cm and a standard deviation of 5 cm. What is the probability that a randomly chosen student is shorter than 172 cm?
Step-by-Step Guide to Finding Probabilities
Let's solve this together. The process is always the same: X → Z → Probability.
Step 1: Identify your variables.
We are given a normal distribution for height X, so $$X \sim N(165, 5^2)$$.
Mean, $$\mu = 165$$
Standard Deviation, $$\sigma = 5$$
Our value of interest is $$X = 172$$. We want to find $$P(X < 172)$$.
Step 2: Standardise your X-value into a Z-score.
Use the formula:
$$Z = \frac{X - \mu}{\sigma} = \frac{172 - 165}{5} = \frac{7}{5} = 1.4$$
So, finding $$P(X < 172)$$ is the exact same as finding $$P(Z < 1.4)$$.
Step 3: Sketch the curve (optional but highly recommended!).
Draw a quick bell curve. Mark the centre at 0. Mark your Z-score (1.4 is to the right of 0) and shade the area you need. In this case, you're shading everything to the left of 1.4. This helps you visualise the answer!
Step 4: Use the Standard Normal Distribution Table.
Look up the Z-score of 1.4 in your table. The table gives you the area to the left of the Z-score, which is exactly what $$P(Z < 1.4)$$ means.
From the table, $$P(Z < 1.4) = 0.9192$$.
So, the probability that a randomly chosen student is shorter than 172 cm is 0.9192 or 91.92%.
Handling Different Types of Probabilities
-
Greater than ( > )
e.g., Find $$P(X > 172)$$
We know the total area under the curve is 1. So, $$P(X > 172) = 1 - P(X < 172)$$.
From our calculation, this is $$1 - 0.9192 = 0.0808$$. -
Between two values ( < X < )
e.g., Find the probability a student's height is between 160 cm and 172 cm, i.e., $$P(160 < X < 172)$$.
First, convert BOTH X-values to Z-scores.
$$Z_1 = \frac{160 - 165}{5} = -1.0$$ $$Z_2 = \frac{172 - 165}{5} = 1.4$$ So we need $$P(-1.0 < Z < 1.4)$$.
The logic is: (Area to the left of the big Z-score) - (Area to the left of the small Z-score).
$$P(-1.0 < Z < 1.4) = P(Z < 1.4) - P(Z < -1.0)$$
Using the table and symmetry ($$P(Z < -1.0) = P(Z > 1.0) = 1 - P(Z < 1.0)$$), we get:
$$0.9192 - (1 - 0.8413) = 0.9192 - 0.1587 = 0.7605$$
Key Takeaway for Part 1
To find a probability, always convert your X-value(s) to Z-score(s) first. Then use the standard normal table. Remember that the table gives you the area to the left. Use sketching and the 'total area = 1' rule to find any other area you need.
Part 2: Finding Values from Probabilities (The 'Inverse' Problem)
Now we flip the problem around. You are given a probability (an area under the curve), and you need to find the specific X-value that corresponds to it.
Example Question: The scores on a test are normally distributed with a mean of 70 and a variance of 64. To get an 'A' grade, a student must score in the top 10%. What is the minimum score needed to get an 'A'?
Step-by-Step Guide to Finding Values
This time, the process is: Probability → Z → X.
Step 1: Identify your variables and the required area.
Scores X are distributed as $$X \sim N(70, 64)$$.
Mean, $$\mu = 70$$
Variance, $$\sigma^2 = 64 \implies$$ Standard Deviation, $$\sigma = \sqrt{64} = 8$$. (Don't forget this step!)
We are looking for the score 'x' such that the probability of being *above* it is 10% (or 0.10).
So, we need to find 'x' where $$P(X > x) = 0.10$$.
Step 2: Sketch the curve and find the area the table can read.
Draw the bell curve. The "top 10%" is a small area on the far right. The table reads the area to the *left*. If the area to the right of our z-score is 0.10, then the area to the left is $$1 - 0.10 = 0.90$$.
So we need to find the z-score where $$P(Z < z) = 0.90$$.
Step 3: Use the Standard Normal Table in reverse.
Look inside the table for the probability value closest to 0.9000. You'll find values like 0.8997 (for Z=1.28) and 0.9015 (for Z=1.29). 0.8997 is closer. So, we'll use the corresponding Z-score.
Z = 1.28 (approximately).
Step 4: Un-standardise your Z-score to find the X-value.
Rearrange the Z-score formula to solve for X: $$X = \mu + Z\sigma$$
Now, plug in your values:
$$X = 70 + (1.28)(8) = 70 + 10.24 = 80.24$$
So, the minimum score needed to get an 'A' grade is 80.24.
Key Takeaway for Part 2
To find a data value (X) from a probability, first figure out the area to the *left* of your point. Use the table in reverse to find the Z-score for that area, and then use the "un-standardise" formula $$X = \mu + Z\sigma$$ to convert back to your final answer.
Part 3: Putting It All Together - Word Problems
This is where you show off your skills! Real-world problems won't say "find P(X > 50)". You need to read the story, pull out the key information, and decide whether it's a 'Forward' or 'Inverse' problem.
Problem-Solving Strategy
- Read carefully: What is the context? (e.g., weights, scores, time).
- Extract the numbers: What is the mean ($$\mu$$)? What is the variance ($$\sigma^2$$) or standard deviation ($$\sigma$$)?
- Identify the Goal:
- Are you given an X-value and asked for a probability, percentage, or proportion? → It's a Forward Problem (X → Z → P).
- Are you given a probability, percentage (like 'top 5%'), or proportion and asked for a specific value, score, or measurement? → It's an Inverse Problem (P → Z → X).
- Execute the Plan: Follow the steps from Part 1 or Part 2.
- Answer the question: Write a concluding sentence that makes sense in the context of the problem. e.g., "Therefore, the minimum weight is 10.5 kg."
Did you know?
The normal distribution is sometimes called the Gaussian distribution, named after the famous mathematician Carl Friedrich Gauss who used it extensively. However, it was first discovered by Abraham de Moivre as an approximation to the binomial distribution. It shows how even in maths, great ideas are often built on the work of others!
Keep practising and sketching the curves, and you'll become a pro at solving these problems. Good luck!