Statistics and Probability for 11th Graders - Part 3

Calculating probabilities

Joshua Marie

2025-12-12

Recall: Standard Normal Distribution

What is a standard normal distribution?

Recall: Standard Normal Distribution

What is a standard normal distribution?

A special case of the normal distribution where:

Mean \(\mu = 0\)
Standard deviation \(\sigma = 1\)
Usually a random variable represents as \(Z\)

Recall: Standard Normal Distribution

Standardize \(X\) that scales to standard normal distribution \(Z \sim \mathcal{N}(0, 1)\) through standardization:

\[Z=\frac{X-\mu}{\sigma}\]

You’ll get z-scores

To Calculate Probability from the Normal Distribution

This is how you calculate the cumulative probability:

\[ P(z_1 \leq Z \leq z_2) = F(z) = \int_{z_1}^{z_2} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{z^2}{2} \right) dz \]

The parameters \(\mu\) and \(\sigma\) are already cancelled since \(\mu=0\) and \(\sigma = 1\).

Do not attempt to calculate this. We need tools to find exact probabilities!

You Need a Z-Table

A z-table (also called the standard normal table) gives us the area under the standard normal curve to the LEFT of any z-score.

Access the z-table here: https://slides.joshuamarie.com/resources/materials/z-table/

You Need a Z-Table

Go to this link and download the z-table PDF:

https://slides.joshuamarie.com/resources/materials/z-table/

How to Use the Z-Table

Apparently, I already have set the instruction in the page. Either go look for it, or look after the next slide.

How to Use the Z-Table

Steps to read the z-table:

Find the row corresponding to the first two digits of your z-score (e.g., for \(z = 1.23\), look for 1.2)
Find the column corresponding to the second decimal place (e.g., for \(z = 1.23\), look for 0.03)
The intersection gives you \(P(Z \leq z)\) - the area to the LEFT of that z-score

Example: For \(z = 1.23\), you’d find row “1.2” and column “0.03”

Example: Reading the Z-Table

Find \(P(Z \leq 1.50)\)

Row: 1.5
Column: 0.00
Table value: 0.9332

Interpretation: About 93.32% of the data falls to the LEFT of \(z = 1.50\)

Example: Reading the Z-Table

Find \(P(Z \leq -0.75)\)

Row: -0.7
Column: 0.05
Table value: 0.2266

Interpretation: About 22.66% of the data falls to the LEFT of \(z = -0.75\)

Basics of Calculation

There are three main types of probability calculations:

\(P(Z \leq z)\) - Area to the LEFT (direct from table)
\(P(Z \geq z)\) - Area to the RIGHT (use: 1 - table value)
\(P(z_1 \leq Z \leq z_2)\) - Area BETWEEN two z-scores (subtract table values):

\[P(z_1 \leq Z \leq z_2)=P(Z \leq z_2) - P(Z \leq z_1)\]

Type 1: \(P(Z \leq z)\)

Finding the area to the LEFT of a z-score

This is the easiest case — just look up the z-score directly in the table!

Type 1: \(P(Z \leq z)\)

Finding the area to the LEFT of a z-score

This is the easiest case — just look up the z-score directly in the table!

Example: Find \(P(Z \leq 1.25)\)

Solution:

Look up \(z = 1.25\) in the table
Row: 1.2, Column: 0.05
Table value: 0.8944

Answer: \(P(Z \leq 1.25) = 0.8944\) or 89.44%

Type 2: \(P(Z \geq z)\)

Finding the area to the RIGHT of a z-score

Since the z-table gives LEFT areas, we use:

\[P(Z > z) = 1 - P(Z \leq z)\]

Type 2: \(P(Z \geq z)\)

Finding the area to the RIGHT of a z-score

Since the z-table gives LEFT areas, we use:

\[P(Z > z) = 1 - P(Z \leq z)\]

Example: Find \(P(Z > 0.80)\)

Solution:

Look up \(z = 0.80\) in the table, which gives you 0.7881
Calculate: \(P(Z > 0.80) = 1 - P(Z \leq 0.80) = 1 - 0.7881\)
\(P(Z > 0.80) = \mathbf{0.2119}\)

Answer: \(P(Z > 0.80) = \mathbf{0.2119}\) or 21.19%

Type 3: \(P(z_1 \leq Z \leq z_2)\)

Finding the area BETWEEN two z-scores

We subtract the smaller cumulative area from the larger:

\[P(z_1 \leq Z \leq z_2)=P(Z \leq z_2) - P(Z \leq z_1)\]

Type 3: \(P(z_1 \leq Z \leq z_2)\)

Finding the area BETWEEN two z-scores

We subtract the smaller cumulative area from the larger:

\[P(z_1 \leq Z \leq z_2)=P(Z \leq z_2) - P(Z \leq z_1)\]

Example: Find \(P(-1.00 < Z < 1.50)\)

Solution:

\(P(Z < 1.50)\) from table → 0.9332
\(P(Z < -1.00)\) from table → 0.1587
\(P(-1.00 < Z < 1.50) = 0.9332 - 0.1587 = \mathbf{0.7745}\)

Answer: \(P(-1.00 < Z < 1.50) = \mathbf{0.7745}\) or 77.45%

Practice Problem 1

Problem: Find the following probabilities using the z-table:

\(P(Z \leq 2.00)\)
\(P(Z > -1.50)\)
\(P(0.50 \leq Z \leq 1.80)\)

Solutions:

\(P(Z \leq 2.00) = \mathbf{0.9772}\) (direct from table)
\(P(Z > -1.50) = 1 - P(Z \leq -1.50) = 1 - 0.0668 = \mathbf{0.9332}\)
\(P(0.50 \leq Z \leq 1.80) = P(Z \leq 1.80) - P(Z \leq 0.50) = 0.9641 - 0.6915 = \mathbf{0.2726}\)

Converting to Real-World Problems

Remember the z-score formula:

\[z = \frac{x - \mu}{\sigma}\]

Steps:

Convert the raw score (\(x\)) to a z-score
Use the z-table to find the probability
Interpret in context

Real-World Example 1

Problem: Test scores are normally distributed with \(\mu = 75\) and \(\sigma = 10\). What is the probability that a randomly selected student scored less than 85?

Solution:

Step 1: Convert to z-score \[z = \frac{85 - 75}{10} = \frac{10}{10} = 1.00\]

Step 2: Look up \(z = 1.00\) in table. This will be 0.8413

Interpretation: \(84.13\%\) of students scored less than 85

Real-World Example 2

Problem: Heights of adult men are normally distributed with \(\mu = 175\) cm and \(\sigma = 8\) cm. What percentage of men are taller than 183 cm?

Solution:

Step 1: Convert to z-score \[z = \frac{183 - 175}{8} = \frac{8}{8} = 1.00\]

Step 2: \[P(Z > 1.00) = 1 - P(Z < 1.00) = 1 - 0.8413 = 0.1587\]

Interpretation: \(15.87\%\) of men are taller than 183 cm

Real-World Example 3

Problem: Battery life is normally distributed with \(\mu = 400\) hours and \(\sigma = 50\) hours. What is the probability a battery lasts between 350 and 450 hours?

Solution:

Step 1: Convert both values to z-scores \[z_1 = \frac{350 - 400}{50} = -1.00, \quad z_2 = \frac{450 - 400}{50} = 1.00\]

Step 2: \[P(-1.00 < Z < 1.00) = 0.8413 - 0.1587 = 0.6826\]

Interpretation: \(68.26\%\) of batteries last between 350 and 450 hours

Practice Problem 2

Problem: Your monthly electric bill has \(\mu = 1500\) and \(\sigma = 120\). What is the probability you’ll pay between 1380 and 1740?

Solution:

\[z_1 = \frac{1380 - 1500}{120} = -1.00, \quad z_2 = \frac{1740 - 1500}{120} = 2.00\]

Answer: \(P(-1.00 < Z < 2.00) = 0.9772 - 0.1587 = \mathbf{0.8185}\) or 81.85%

Working Backwards: Finding X from Probability

Sometimes we know the probability and need to find the value (x).

Steps:

Find the z-score that corresponds to the given probability in the z-table
Use the formula: \(x = \mu + z\sigma\)
Interpret the result

Example: Finding X

Problem: Test scores are normally distributed with \(\mu = 70\) and \(\sigma = 12\). What score represents the 90th percentile?

Solution:

Step 1: Find z-score where \(P(Z < z) = 0.90\)

Looking through the table, \(z \approx \mathbf{1.28}\)

Step 2: Calculate x

\[x = 70 + (1.28)(12) = 70 + 15.36 = 85.36\]

Interpretation: A score of approximately 85.36 (around 85 or 86) represents the 90th percentile. Which means you got a better score of 85 than 90% of others, with only 10% scoring higher.

Interactive Z-Score Visualizer

viewof zScore = Inputs.range([-3.5, 3.5], {
    label: "Z-Score",
    value: 1.5,
    step: 0.01
})

viewof calcType = Inputs.radio(
    ["Left tail (P(Z < z))", "Right tail (P(Z > z))", "Between (-z to z)"],
    {label: "Calculate:", value: "Left tail (P(Z < z))"}
)

{
  const width = 800;
  const height = 400;
  const margin = {top: 40, right: 40, bottom: 60, left: 60};
  
  function normalPDF(x) {
    return Math.exp(-0.5 * x * x) / Math.sqrt(2 * Math.PI);
  }
  
  function normalCDF(z) {
    const t = 1 / (1 + 0.2316419 * Math.abs(z));
    const d = 0.3989423 * Math.exp(-z * z / 2);
    const prob = d * t * (0.3193815 + t * (-0.3565638 + t * (1.781478 + t * (-1.821256 + t * 1.330274))));
    return z > 0 ? 1 - prob : prob;
  }
  
  const svg = d3.create("svg")
    .attr("width", width)
    .attr("height", height)
    .attr("viewBox", [0, 0, width, height])
    .attr("style", "max-width: 100%; height: auto; background: white;");
  
  const xScale = d3.scaleLinear()
    .domain([-4, 4])
    .range([margin.left, width - margin.right]);
  
  const yScale = d3.scaleLinear()
    .domain([0, 0.45])
    .range([height - margin.bottom, margin.top]);
  
  const curveData = d3.range(-4, 4.01, 0.05).map(x => ({
    x: x,
    y: normalPDF(x)
  }));
  
  const line = d3.line()
    .x(d => xScale(d.x))
    .y(d => yScale(d.y));
  
  let shadedData;
  if (calcType === "Left tail (P(Z < z))") {
    shadedData = curveData.filter(d => d.x <= zScore);
  } else if (calcType === "Right tail (P(Z > z))") {
    shadedData = curveData.filter(d => d.x >= zScore);
  } else {
    shadedData = curveData.filter(d => d.x >= -Math.abs(zScore) && d.x <= Math.abs(zScore));
  }
  
  const area = d3.area()
    .x(d => xScale(d.x))
    .y0(height - margin.bottom)
    .y1(d => yScale(d.y));
  
  svg.append("path")
    .datum(shadedData)
    .attr("fill", "rgba(76, 175, 80, 0.3)")
    .attr("d", area);
  
  svg.append("path")
    .datum(curveData)
    .attr("fill", "none")
    .attr("stroke", "steelblue")
    .attr("stroke-width", 2)
    .attr("d", line);
  
  if (calcType === "Between (-z to z)") {
    svg.append("line")
      .attr("x1", xScale(-Math.abs(zScore)))
      .attr("x2", xScale(-Math.abs(zScore)))
      .attr("y1", margin.top)
      .attr("y2", height - margin.bottom)
      .attr("stroke", "red")
      .attr("stroke-width", 2)
      .attr("stroke-dasharray", "5,5");
    
    svg.append("line")
      .attr("x1", xScale(Math.abs(zScore)))
      .attr("x2", xScale(Math.abs(zScore)))
      .attr("y1", margin.top)
      .attr("y2", height - margin.bottom)
      .attr("stroke", "red")
      .attr("stroke-width", 2)
      .attr("stroke-dasharray", "5,5");
  } else {
    svg.append("line")
      .attr("x1", xScale(zScore))
      .attr("x2", xScale(zScore))
      .attr("y1", margin.top)
      .attr("y2", height - margin.bottom)
      .attr("stroke", "red")
      .attr("stroke-width", 2)
      .attr("stroke-dasharray", "5,5");
  }
  
  svg.append("g")
    .attr("transform", `translate(0,${height - margin.bottom})`)
    .call(d3.axisBottom(xScale));
  
  svg.append("g")
    .attr("transform", `translate(${margin.left},0)`)
    .call(d3.axisLeft(yScale));
  
  let prob;
  if (calcType === "Left tail (P(Z < z))") {
    prob = normalCDF(zScore);
  } else if (calcType === "Right tail (P(Z > z))") {
    prob = 1 - normalCDF(zScore);
  } else {
    prob = normalCDF(Math.abs(zScore)) - normalCDF(-Math.abs(zScore));
  }
  
  svg.append("text")
    .attr("x", width / 2)
    .attr("y", margin.top - 10)
    .attr("text-anchor", "middle")
    .attr("font-size", "16px")
    .attr("font-weight", "bold")
    .text(`Probability = ${prob.toFixed(4)} (${(prob * 100).toFixed(2)}%)`);
  
  return svg.node();
}

Summary: Using Z-Tables

Tip

Z-tables give cumulative probability to the LEFT
For \(P(Z > z)\), use \(1 - \text{table value}\)
For \(P(z_1 < Z < z_2)\), subtract table values
Always convert real-world problems using \(z = \frac{x-\mu}{\sigma}\)
To find x from probability, look up z first, then use \(x = \mu + z\sigma\)

Assignment

Instructions:

Complete the following problems. Show your complete solution including:

Z-score calculations
Z-table lookups
Final answers with proper units

Problems:

IQ scores are normally distributed with \(\mu = 100\) and \(\sigma = 15\). Find the probability that a randomly selected person has an IQ between 85 and 130.
The lifespan of light bulbs is normally distributed with \(\mu = 800\) hours and \(\sigma = 40\) hours. What percentage of bulbs last more than 900 hours?
If scores on an exam are normally distributed with \(\mu = 78\) and \(\sigma = 8\) what score separates the top 25% of students from the rest?

Additional Resources

For further practice:

Online z-table calculators
Practice worksheets (provided separately)
Interactive normal distribution applets
Review the Empirical Rule (68-95-99.7) as a quick check

Tip

Remember: The z-table is your friend! Practice reading it until it becomes second nature.

Thank You!

Next session: Hypothesis Testing and Confidence Intervals

Questions?