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Does Standard Deviation Only Apply to Continuous Variables

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Lesson Explainer: Finding Means and Standard Deviations in Normal Distributions Mathematics

In this explainer, we will learn how to find an unknown mean and/or standard deviation in a normal distribution.

Suppose 𝑋 is a continuous random variable, normally distributed with mean 𝜇 and standard deviation 𝜎 , which we denote by 𝑋 ∼ 𝑁 𝜇 , 𝜎 . Recall that we can code 𝑋 by the linear change of variables 𝑋 ↦ 𝑍 = 𝑋 − 𝜇 𝜎 , where 𝑍 ∼ 𝑁 0 , 1 follows the standard normal distribution and 𝑃 ( 𝑋 < 𝑥 ) = 𝑃 𝑍 < 𝑥 − 𝜇 𝜎 , for all 𝑥 .

We can also use this process to calculate unknown means and standard deviations in normal distributions. Let us look at an example where we need to find the mean.

Example 1: Determining the Mean of a Normal Distribution

Suppose 𝑋 is normally distributed with mean 𝜇 and variance 196. Given that 𝑃 ( 𝑋 ≤ 4 0 ) = 0 . 0 6 6 8 , find the value of 𝜇 .

Answer

In order to find the unknown mean 𝜇 , we code 𝑋 by the change of variables 𝑋 ↦ 𝑍 = 𝑋 − 𝜇 𝜎 , where the standard deviation 𝜎 = √ 1 9 6 = 1 4 . Now 𝑍 ∼ 𝑁 0 , 1 follows the standard normal distribution and 𝑃 ( 𝑋 ≤ 4 0 ) = 𝑃 𝑍 ≤ 4 0 − 𝜇 1 4 = 0 . 0 6 6 8 .

We can now use our calculators or look up 0.0668 in a standard normal distribution table, which tells us that it corresponds to the probability that 𝑍 ≤ − 1 . 5 .

Thus, 4 0 − 𝜇 1 4 = − 1 . 5 − 𝜇 = ( − 1 . 5 ) × 1 4 − 4 0 𝜇 = 6 1 .

We can use exactly the same technique to find unknown standard deviations.

Example 2: Determining the Standard Deviation of a Normal Distribution

Suppose that 𝑋 is a normal random variable whose mean is 𝜇 and standard deviation is 𝜎 . If 𝑃 ( 𝑋 ≤ 3 9 ) = 0 . 0 5 4 8 and 𝜇 = 6 3 , find 𝜎 using the standard normal distribution table.

Answer

In order to find the unknown standard deviation 𝜎 , we code 𝑋 by the change of variables 𝑋 ↦ 𝑍 = 𝑋 − 𝜇 𝜎 , where the mean 𝜇 = 6 3 . Now 𝑍 ∼ 𝑁 0 , 1 follows the standard normal distribution and 𝑃 ( 𝑋 ≤ 3 9 ) = 𝑃 𝑍 ≤ 3 9 − 6 3 𝜎 = 0 . 0 5 4 8 .

We can now look up 0.0548 in a standard normal distribution table, which tells us that it corresponds to the probability that 𝑍 ≤ − 1 . 6 .

Thus, we have 3 9 − 6 3 𝜎 = − 1 . 6 𝜎 = − 2 4 − 1 . 6 = 1 5 .

In the previous examples, we used coding to find an unknown mean or standard deviation when the value of the other parameter was given, along with a probability.

Note that we can find both the mean and the standard deviation simultaneously if two probabilities are given, by solving a pair of simultaneous equations. Here is an example of this type.

Example 3: Determining the Mean and Standard Deviation of a Normal Distribution

Let 𝑋 be a random variable that is normally distributed with mean 𝜇 and standard deviation 𝜎 . Given that 𝑃 ( 𝑋 ≤ 7 2 . 4 4 ) = 0 . 6 4 4 3 and 𝑃 ( 𝑋 ≥ 3 7 . 7 6 ) = 0 . 9 9 4 1 , calculate the values of 𝜇 and 𝜎 .

Answer

In order to find the unknown mean 𝜇 and standard deviation 𝜎 , we code 𝑋 by the change of variables 𝑋 ↦ 𝑍 = 𝑋 − 𝜇 𝜎 . Now 𝑍 ∼ 𝑁 0 , 1 follows the standard normal distribution and 𝑃 ( 𝑋 ≤ 7 2 . 4 4 ) = 𝑃 𝑍 ≤ 7 2 . 4 4 − 𝜇 𝜎 = 0 . 6 4 4 3 and 𝑃 ( 𝑋 ≥ 3 7 . 7 6 ) = 𝑃 𝑍 ≥ 3 7 . 7 6 − 𝜇 𝜎 = 0 . 9 9 4 1 .

Using our calculators or looking up 0.6443 and 0.9941 in a standard normal distribution table, we find that these are the probabilities that 𝑍 ≤ 0 . 3 6 9 9 8 and that 𝑍 ≥ − 2 . 5 1 8 0 7 .

This yields the pair of simultaneous equations 7 2 . 4 4 − 𝜇 𝜎 = 0 . 3 6 9 9 8 and 3 7 . 7 6 − 𝜇 𝜎 = − 2 . 5 1 8 0 7 . We multiply both equations by 𝜎 : 7 2 . 4 4 − 𝜇 = 0 . 3 6 9 9 8 𝜎 , 3 7 . 7 6 − 𝜇 = − 2 . 5 1 8 0 7 𝜎 .

Then, we subtract the second from the first to get 7 2 . 4 4 − 𝜇 − ( 3 7 . 7 6 − 𝜇 ) = 0 . 3 6 9 9 8 𝜎 − ( − 2 . 5 1 8 0 7 𝜎 ) 3 4 . 6 8 = 2 . 8 8 8 0 5 𝜎 .

Therefore, we have 𝜎 = 3 4 . 6 8 2 . 8 8 8 0 5 = 1 2 . 0 0 8 1 … .

We can now substitute 𝜎 = 1 2 . 0 0 8 1 … back into the equation 7 2 . 4 4 − 𝜇 = 0 . 3 6 9 9 8 𝜎 , which gives us 7 2 . 4 4 − 𝜇 = 0 . 3 6 9 9 8 × 1 2 . 0 0 8 1 … 𝜇 = 6 7 . 9 9 7 2 … .

We arrive at values of 𝜇 = 6 8 and 𝜎 = 1 2 , to the nearest integer.

We can use this method of simultaneous equations to find other unknown quantities in normal distributions.

Example 4: Finding Unknown Quantities in Normal Distributions

Consider the random variable 𝑋 ∼ 𝑁 3 . 2 5 , 𝜎 . Given that 𝑃 ( 𝑋 > 2 𝑎 ) = 0 . 1 and 𝑃 ( 𝑋 < 𝑎 ) = 0 . 3 , find the value of 𝜎 and the value of 𝑎 . Give your answers to one decimal place.

Answer

In order to find the unknown standard deviation 𝜎 and constant 𝑎 , we code 𝑋 by the change of variables 𝑋 ↦ 𝑍 = 𝑋 − 𝜇 𝜎 , where the mean is 𝜇 = 3 . 2 5 . Now 𝑍 ∼ 𝑁 0 , 1 follows the standard normal distribution and 𝑃 ( 𝑋 > 2 𝑎 ) = 𝑃 𝑍 > 2 𝑎 − 3 . 2 5 𝜎 = 0 . 1 and 𝑃 ( 𝑋 < 𝑎 ) = 𝑃 𝑍 < 𝑎 − 3 . 2 5 𝜎 = 0 . 3 .

Using our calculators or looking up 0.1 and 0.3 in a standard normal distribution table, we find that these are the probabilities that 𝑍 > 1 . 2 8 1 5 5 and that 𝑍 < − 0 . 5 2 4 4 . This yields the pair of simultaneous equations 2 𝑎 − 3 . 2 5 𝜎 = 1 . 2 8 1 5 5 and 𝑎 − 3 . 2 5 𝜎 = − 0 . 5 2 4 4 .

We multiply both equations by 𝜎 : 2 𝑎 − 3 . 2 5 = 1 . 2 8 1 5 5 𝜎 , 𝑎 − 3 . 2 5 = − 0 . 5 2 4 4 𝜎 .

Then, we multiply the second of these by 2: 2 𝑎 − 6 . 5 = − 1 . 0 4 8 8 𝜎 .

We can now eliminate 𝑎 by subtracting the second equation from the first: 2 𝑎 − 3 . 2 5 − ( 2 𝑎 − 6 . 5 ) = 1 . 2 8 1 5 5 𝜎 − ( − 1 . 0 4 8 8 𝜎 ) 3 . 2 5 = 2 . 3 3 0 3 5 𝜎 𝜎 = 1 . 3 9 4 6 4 … .

To find the value of 𝑎 , we can substitute back into 𝑎 − 3 . 2 5 = − 0 . 5 2 4 4 𝜎 : 𝑎 − 3 . 2 5 = − 0 . 5 2 4 4 × 1 . 3 9 4 6 4 … 𝑎 = 3 . 2 5 − 0 . 7 3 1 3 … = 2 . 5 1 8 6 … .

Thus, rounding to one decimal place, we have 𝜎 = 1 . 4 and 𝑎 = 2 . 5 .

Let us try applying these techniques in a real-life context to find an unknown mean.

Example 5: Determining the Mean of a Normal Distribution in a Real-Life Context

The heights of a sample of flowers are normally distributed with mean 𝜇 and standard deviation 12 cm. Given that 1 0 . 5 6 % of the flowers are shorter than 47 cm, determine 𝜇 .

Answer

We have a normal random variable 𝑋 ∼ 𝑁 𝜇 , 1 2 with unknown mean. To convert the population percentage of 1 0 . 5 6 % into a probability, we divide by 100, so we have 𝑃 ( 𝑋 < 4 7 ) = 0 . 1 0 5 6 .

In order to find the unknown mean 𝜇 , we code 𝑋 by the change of variables 𝑋 ↦ 𝑍 = 𝑋 − 𝜇 𝜎 , where the standard deviation is 𝜎 = 1 2 . Now 𝑍 ∼ 𝑁 0 , 1 follows the standard normal distribution and 𝑃 ( 𝑋 < 4 7 ) = 𝑃 𝑍 < 4 7 − 𝜇 1 2 = 0 . 1 0 5 6 .

We can now use our calculators or look up 0.1056 in a standard normal distribution table, which tells us that it corresponds to the probability that 𝑍 < − 1 . 2 5 0 2 7 . Thus, we have 4 7 − 𝜇 1 2 = − 1 . 2 5 0 2 7 − 𝜇 = ( − 1 . 2 5 0 2 7 ) × 1 2 − 4 7 𝜇 = 6 2 . 0 0 3 2 4 , giving us 𝜇 = 6 2 to the nearest integer.

We can also find unknown standard deviations in real-life contexts.

Example 6: Determining the Standard Deviation of a Normal Distribution in a Real-Life Context

The lengths of a certain type of plant are normally distributed with a mean 𝜇 = 6 3 c m and standard deviation 𝜎 . Given that the lengths of 8 4 . 1 3 % of the plants are less than 75 cm, find the variance.

Answer

We have a normal random variable 𝑋 ∼ 𝑁 6 3 , 𝜎 with unknown variance. To convert the population percentage of 8 4 . 1 3 % into a probability, we divide by 100, so we have 𝑃 ( 𝑋 < 7 5 ) = 0 . 8 4 1 3 .

In order to find the unknown variance 𝜎 , we code 𝑋 by the change of variables 𝑋 ↦ 𝑍 = 𝑋 − 𝜇 𝜎 , where the mean 𝜇 = 6 3 . Now 𝑍 ∼ 𝑁 0 , 1 follows the standard normal distribution and 𝑃 ( 𝑋 < 7 5 ) = 𝑃 𝑍 < 7 5 − 6 3 𝜎 = 0 . 8 4 1 3 .

We can now use our calculators or look up 0.8413 in a standard normal distribution table to find that this is the probability that 𝑍 < 0 . 9 9 9 8 2 . Thus, we have 7 5 − 6 3 𝜎 = 0 . 9 9 9 8 2 𝜎 = 1 2 0 . 9 9 9 8 2 = 1 2 . 0 0 2 1 6 … .

Therefore, our variance is 𝜎 = ( 1 2 . 0 0 2 1 6 … ) = 1 4 4 , to the nearest integer.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Given a normal random variable 𝑋 ∼ 𝑁 𝜇 , 𝜎 and a probability 𝑃 ( 𝑋 < 𝑥 ) = 𝑃 , we can code 𝑋 by the change of variables 𝑋 ↦ 𝑍 = 𝑋 − 𝜇 𝜎 , where 𝑍 ∼ 0 , 1 . Then, we can use the standard normal distribution to find an unknown mean or standard deviation.
  • If we are given two probabilities 𝑃 ( 𝑋 < 𝑥 ) = 𝑃 and 𝑃 ( 𝑋 > 𝑦 ) = 𝑃 , then we can derive a pair of simultaneous equations to find the mean and the standard deviation when both are unknown.
  • We can use these techniques to solve real-world problems involving unknown means and standard deviations in normal distributions.

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Source: https://www.nagwa.com/en/explainers/853196168317/

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