Calculate the Normal Distribution Curve with mean and standard deviation. Visualize data spread efficiently.

Normal Distribution Curve

Purpose

The Normal Distribution Curve function helps in visualizing the distribution of data points around a mean value. It is commonly used in statistics to understand the spread of data.

Use Cases

• Visualizing the distribution of test scores in a class
• Analyzing the distribution of heights in a population

How to Use

1. Enter the Mean value (average value around which data points are distributed)
2. Enter the Standard Deviation value (measure of how spread out the data points are)
3. Optionally, adjust the Start and Stop values to focus on a specific range
4. Click on the "Calculate" button to generate the Normal Distribution Curve

Input Values

1. Mean: The average value around which data points are distributed
2. Default unit is inch
3. Standard Deviation: Measure of how spread out the data points are
4. Default unit is inch
5. Start: Starting value for the x-axis
6. Stop: Ending value for the x-axis
7. Variations: Number of points to plot on the curve

Output Values

1. Chart: Visual representation of the Normal Distribution Curve
2. The curve shows how data points are distributed around the mean value

Any other Instruction

• The curve represents how likely it is to find a data point at a specific value
• A higher peak indicates more data points around the mean value
• A wider curve indicates a larger spread of data points

Code Analysis

1. The function calculates the y-values for the Normal Distribution Curve using the given mean and standard deviation
2. It plots the curve based on the specified range and number of variations

Technical Parameters

• mean, standard_devn, start, stop, variations

Return Values

• chart

Example Expressions

You can use the following expressions to directly evaluate in a non-interactive manner using eva():

normal(mean=50.0, standard_devn=10.0, start=0, stop=100, variations=100)
normal(mean=60.0, standard_devn=15.0, start=10, stop=90, variations=50)

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