How to Find the Standard Deviation in Continuous Random Variables Given the Mean and Percent

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 ๐‘‹ is a continuous random variable, normally distributed with mean ๐œ‡ and standard deviation ๐œŽ , which we denote by ๐‘‹ ๐‘ ๐œ‡ , ๐œŽ . Recall that we can code ๐‘‹ by the linear change of variables ๐‘‹ ๐‘ = ๐‘‹ ๐œ‡ ๐œŽ , where ๐‘ ๐‘ 0 , 1 follows the standard normal distribution and ๐‘ƒ ( ๐‘‹ < ๐‘ฅ ) = ๐‘ƒ ๐‘ < ๐‘ฅ ๐œ‡ ๐œŽ , for all ๐‘ฅ .

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 ๐‘‹ is normally distributed with mean ๐œ‡ and variance 196. Given that ๐‘ƒ ( ๐‘‹ 4 0 ) = 0 . 0 6 6 8 , find the value of ๐œ‡ .

Answer

In order to find the unknown mean ๐œ‡ , we code ๐‘‹ by the change of variables ๐‘‹ ๐‘ = ๐‘‹ ๐œ‡ ๐œŽ , where the standard deviation ๐œŽ = 1 9 6 = 1 4 . Now ๐‘ ๐‘ 0 , 1 follows the standard normal distribution and ๐‘ƒ ( ๐‘‹ 4 0 ) = ๐‘ƒ ๐‘ 4 0 ๐œ‡ 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 ๐‘ 1 . 5 .

Thus, 4 0 ๐œ‡ 1 4 = 1 . 5 ๐œ‡ = ( 1 . 5 ) × 1 4 4 0 ๐œ‡ = 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 ๐‘‹ is a normal random variable whose mean is ๐œ‡ and standard deviation is ๐œŽ . If ๐‘ƒ ( ๐‘‹ 3 9 ) = 0 . 0 5 4 8 and ๐œ‡ = 6 3 , find ๐œŽ using the standard normal distribution table.

Answer

In order to find the unknown standard deviation ๐œŽ , we code ๐‘‹ by the change of variables ๐‘‹ ๐‘ = ๐‘‹ ๐œ‡ ๐œŽ , where the mean ๐œ‡ = 6 3 . Now ๐‘ ๐‘ 0 , 1 follows the standard normal distribution and ๐‘ƒ ( ๐‘‹ 3 9 ) = ๐‘ƒ ๐‘ 3 9 6 3 ๐œŽ = 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 ๐‘ 1 . 6 .

Thus, we have 3 9 6 3 ๐œŽ = 1 . 6 ๐œŽ = 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 ๐‘‹ be a random variable that is normally distributed with mean ๐œ‡ and standard deviation ๐œŽ . Given that ๐‘ƒ ( ๐‘‹ 7 2 . 4 4 ) = 0 . 6 4 4 3 and ๐‘ƒ ( ๐‘‹ 3 7 . 7 6 ) = 0 . 9 9 4 1 , calculate the values of ๐œ‡ and ๐œŽ .

Answer

In order to find the unknown mean ๐œ‡ and standard deviation ๐œŽ , we code ๐‘‹ by the change of variables ๐‘‹ ๐‘ = ๐‘‹ ๐œ‡ ๐œŽ . Now ๐‘ ๐‘ 0 , 1 follows the standard normal distribution and ๐‘ƒ ( ๐‘‹ 7 2 . 4 4 ) = ๐‘ƒ ๐‘ 7 2 . 4 4 ๐œ‡ ๐œŽ = 0 . 6 4 4 3 and ๐‘ƒ ( ๐‘‹ 3 7 . 7 6 ) = ๐‘ƒ ๐‘ 3 7 . 7 6 ๐œ‡ ๐œŽ = 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 ๐‘ 0 . 3 6 9 9 8 and that ๐‘ 2 . 5 1 8 0 7 .

This yields the pair of simultaneous equations 7 2 . 4 4 ๐œ‡ ๐œŽ = 0 . 3 6 9 9 8 and 3 7 . 7 6 ๐œ‡ ๐œŽ = 2 . 5 1 8 0 7 . We multiply both equations by ๐œŽ : 7 2 . 4 4 ๐œ‡ = 0 . 3 6 9 9 8 ๐œŽ , 3 7 . 7 6 ๐œ‡ = 2 . 5 1 8 0 7 ๐œŽ .

Then, we subtract the second from the first to get 7 2 . 4 4 ๐œ‡ ( 3 7 . 7 6 ๐œ‡ ) = 0 . 3 6 9 9 8 ๐œŽ ( 2 . 5 1 8 0 7 ๐œŽ ) 3 4 . 6 8 = 2 . 8 8 8 0 5 ๐œŽ .

Therefore, we have ๐œŽ = 3 4 . 6 8 2 . 8 8 8 0 5 = 1 2 . 0 0 8 1 .

We can now substitute ๐œŽ = 1 2 . 0 0 8 1 back into the equation 7 2 . 4 4 ๐œ‡ = 0 . 3 6 9 9 8 ๐œŽ , which gives us 7 2 . 4 4 ๐œ‡ = 0 . 3 6 9 9 8 × 1 2 . 0 0 8 1 ๐œ‡ = 6 7 . 9 9 7 2 .

We arrive at values of ๐œ‡ = 6 8 and ๐œŽ = 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 ๐‘‹ ๐‘ 3 . 2 5 , ๐œŽ . Given that ๐‘ƒ ( ๐‘‹ > 2 ๐‘Ž ) = 0 . 1 and ๐‘ƒ ( ๐‘‹ < ๐‘Ž ) = 0 . 3 , find the value of ๐œŽ and the value of ๐‘Ž . Give your answers to one decimal place.

Answer

In order to find the unknown standard deviation ๐œŽ and constant ๐‘Ž , we code ๐‘‹ by the change of variables ๐‘‹ ๐‘ = ๐‘‹ ๐œ‡ ๐œŽ , where the mean is ๐œ‡ = 3 . 2 5 . Now ๐‘ ๐‘ 0 , 1 follows the standard normal distribution and ๐‘ƒ ( ๐‘‹ > 2 ๐‘Ž ) = ๐‘ƒ ๐‘ > 2 ๐‘Ž 3 . 2 5 ๐œŽ = 0 . 1 and ๐‘ƒ ( ๐‘‹ < ๐‘Ž ) = ๐‘ƒ ๐‘ < ๐‘Ž 3 . 2 5 ๐œŽ = 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 ๐‘ > 1 . 2 8 1 5 5 and that ๐‘ < 0 . 5 2 4 4 . This yields the pair of simultaneous equations 2 ๐‘Ž 3 . 2 5 ๐œŽ = 1 . 2 8 1 5 5 and ๐‘Ž 3 . 2 5 ๐œŽ = 0 . 5 2 4 4 .

We multiply both equations by ๐œŽ : 2 ๐‘Ž 3 . 2 5 = 1 . 2 8 1 5 5 ๐œŽ , ๐‘Ž 3 . 2 5 = 0 . 5 2 4 4 ๐œŽ .

Then, we multiply the second of these by 2: 2 ๐‘Ž 6 . 5 = 1 . 0 4 8 8 ๐œŽ .

We can now eliminate ๐‘Ž by subtracting the second equation from the first: 2 ๐‘Ž 3 . 2 5 ( 2 ๐‘Ž 6 . 5 ) = 1 . 2 8 1 5 5 ๐œŽ ( 1 . 0 4 8 8 ๐œŽ ) 3 . 2 5 = 2 . 3 3 0 3 5 ๐œŽ ๐œŽ = 1 . 3 9 4 6 4 .

To find the value of ๐‘Ž , we can substitute back into ๐‘Ž 3 . 2 5 = 0 . 5 2 4 4 ๐œŽ : ๐‘Ž 3 . 2 5 = 0 . 5 2 4 4 × 1 . 3 9 4 6 4 ๐‘Ž = 3 . 2 5 0 . 7 3 1 3 = 2 . 5 1 8 6 .

Thus, rounding to one decimal place, we have ๐œŽ = 1 . 4 and ๐‘Ž = 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 ๐œ‡ and standard deviation 12 cm. Given that 1 0 . 5 6 % of the flowers are shorter than 47 cm, determine ๐œ‡ .

Answer

We have a normal random variable ๐‘‹ ๐‘ ๐œ‡ , 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 ๐‘ƒ ( ๐‘‹ < 4 7 ) = 0 . 1 0 5 6 .

In order to find the unknown mean ๐œ‡ , we code ๐‘‹ by the change of variables ๐‘‹ ๐‘ = ๐‘‹ ๐œ‡ ๐œŽ , where the standard deviation is ๐œŽ = 1 2 . Now ๐‘ ๐‘ 0 , 1 follows the standard normal distribution and ๐‘ƒ ( ๐‘‹ < 4 7 ) = ๐‘ƒ ๐‘ < 4 7 ๐œ‡ 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 ๐‘ < 1 . 2 5 0 2 7 . Thus, we have 4 7 ๐œ‡ 1 2 = 1 . 2 5 0 2 7 ๐œ‡ = ( 1 . 2 5 0 2 7 ) × 1 2 4 7 ๐œ‡ = 6 2 . 0 0 3 2 4 , giving us ๐œ‡ = 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 ๐œ‡ = 6 3 c m and standard deviation ๐œŽ . 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 ๐‘‹ ๐‘ 6 3 , ๐œŽ with unknown variance. To convert the population percentage of 8 4 . 1 3 % into a probability, we divide by 100, so we have ๐‘ƒ ( ๐‘‹ < 7 5 ) = 0 . 8 4 1 3 .

In order to find the unknown variance ๐œŽ , we code ๐‘‹ by the change of variables ๐‘‹ ๐‘ = ๐‘‹ ๐œ‡ ๐œŽ , where the mean ๐œ‡ = 6 3 . Now ๐‘ ๐‘ 0 , 1 follows the standard normal distribution and ๐‘ƒ ( ๐‘‹ < 7 5 ) = ๐‘ƒ ๐‘ < 7 5 6 3 ๐œŽ = 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 ๐‘ < 0 . 9 9 9 8 2 . Thus, we have 7 5 6 3 ๐œŽ = 0 . 9 9 9 8 2 ๐œŽ = 1 2 0 . 9 9 9 8 2 = 1 2 . 0 0 2 1 6 .

Therefore, our variance is ๐œŽ = ( 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 ๐‘‹ ๐‘ ๐œ‡ , ๐œŽ and a probability ๐‘ƒ ( ๐‘‹ < ๐‘ฅ ) = ๐‘ƒ , we can code ๐‘‹ by the change of variables ๐‘‹ ๐‘ = ๐‘‹ ๐œ‡ ๐œŽ , where ๐‘ 0 , 1 . Then, we can use the standard normal distribution to find an unknown mean or standard deviation.
  • If we are given two probabilities ๐‘ƒ ( ๐‘‹ < ๐‘ฅ ) = ๐‘ƒ and ๐‘ƒ ( ๐‘‹ > ๐‘ฆ ) = ๐‘ƒ , 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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