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A scientist is interested in whether stretching before running 5 kilometers will improve a runner’s time. The scientist decides that an experimental study is the best method to use to answer the question. One group of runners will run a 5-kilometer race without stretching, and a second group of runners will follow a specific stretching routine before running the race. The finishing times will be recorded, along with the group type for each runner.The distribution of times for the group that followed a specific stretching routine before the race is shown. Why is it reasonable to model this distribution with a normal distribution?

A scientist is interested in whether stretching before running 5 kilometers will improve-example-1
User Shinnok
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First, to understand the reason to use a normal distribuition, let's analyse some of it's properties.

The normal distribution is given by the following formula:


f(x)\text{ = }\frac{1}{\sigma\sqrt[]{2\pi}}\text{exp\lbrack}(-1)/(2)((x-\mu)/(\sigma))^2\text{ \rbrack}

The normal distribuition is centered at the mean(given by 'mu' in the equation above) and its growth is controlled by its standard deviation(given by the 'sigma').

It is shaped like a bell(centered, with a gaussian decay, symetric). The data given by the question, have all of those properties. Is centered at the middle with a exponential decay in both 'directions'. Since the data agrees with the gaussian distribution properties, it makes sense to model this distribuition as a normal distribuition.

User Knells
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