Let X be a random variable following a Normal (Gaussian) distribution. All Normal distributions have two parameters: mean and standard deviation (or variance). For this X, let μ = 11 and σ = 2.3. An example of where such a distribution may arise is the following:
You have a bag of candy made by Statistics, Inc. The weight of the pieces are not all the same, they are a random variable. This variable follows a Normal distribution with average weight 11 grams and standard deviation 2.3. Define the random variable X as the weight of a randomly selected peice of candy.
For those who like pictures, here is a graphic of the probability density function. It is not a probability, it is a density, a likelihood. It can be used to determine which values are more likely than others. From the graphic, we can tell that weights are more likely around 11 than around 6.4 or 14.45.
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Continuing the candy example, let us calculate the probability density for a weight of 6.3149 grams; that is, calculate f(6.3149).
Your Answer
In the box below, please enter the value of f(6.3149), where X ~ Normal(μ=11; σ=2.3), then click on the “Check your answer!” button. Please round your answer to the ten-thousandths place.
You got the correct answer of f(6.3149) = 0.0218. Congratulations!
Unfortunately, your answer was not correct. Either try again or click on “Show Solution” below to see how to obtain the correct answer.
Thus, the probability density of the next piece of candy weighing 6.3149 grams is approximately 0.0218.
By itself, this value has almost no meaning. However, because f(11) = 0.1735 > 0.0218 = f(6.3149), we know it is more likely that the piece of candy will weigh closer to 11 than to 6.3149.
As a side question, which weight is most likely? That’s right, the modal value, 11. By the way, because the Normal distribution is symmetric and unimodal, the mode is the same as the median and the mean. Look at the graphic above to convince yourself of this.
Show the R Code
Hide the R Code
Copy and paste the following code into your R script window, then run it from there.
dnorm(6.3149, m=11, s=2.3)
In the R output, the number given after running the final line above is the probability density requested, f(6.3149). There are three slots in the function. The first is x; the second, μ; the third, σ.
Show the Excel Formula
Hide the Excel Formula
Copy and paste the following code into an Excel cell, then hit enter.
=NORM.DIST(6.3149,11,2.3,FALSE)
The first slot is the value of x. The second slot is the value of μ. The third slot is the value of σ. The last slot is FALSE for calculating f(x) and TRUE for calculating F(x), P[X ≤ x].
![[Correct!]](images/correct.png "You are correct. Well done!")
You got the correct answer of f(6.3149) = 0.0218. Congratulations!![[Incorrect!]](images/incorrect.png "You are not correct.")
Unfortunately, your answer was not correct. Either try again or click on “Show Solution” below to see how to obtain the correct answer.