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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 𝑋𝑁𝜇,𝜎. Recall that we can code 𝑋 by the linear change of variables 𝑋𝑍=𝑋𝜇𝜎, where 𝑍𝑁0,1 follows the standard normal distribution and 𝑃(𝑋<𝑥)=𝑃𝑍<𝑥𝜇𝜎, 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(𝑋40)=0.0668, find the value of 𝜇.

Answer

In order to find the unknown mean 𝜇, we code 𝑋 by the change of variables 𝑋𝑍=𝑋𝜇𝜎, where the standard deviation 𝜎=196=14. Now 𝑍𝑁0,1 follows the standard normal distribution and 𝑃(𝑋40)=𝑃𝑍40𝜇14=0.0668.

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 𝑍1.5.

Thus, 40𝜇14=1.5𝜇=(1.5)×1440𝜇=61.

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(𝑋39)=0.0548 and 𝜇=63, 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 𝑋𝑍=𝑋𝜇𝜎, where the mean 𝜇=63. Now 𝑍𝑁0,1 follows the standard normal distribution and 𝑃(𝑋39)=𝑃𝑍3963𝜎=0.0548.

We can now look up 0.0548 in a standard normal distribution table, which tells us that it corresponds to the probability that 𝑍1.6.

Thus, we have 3963𝜎=1.6𝜎=241.6=15.

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(𝑋72.44)=0.6443 and 𝑃(𝑋37.76)=0.9941, 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 𝑋𝑍=𝑋𝜇𝜎. Now 𝑍𝑁0,1 follows the standard normal distribution and 𝑃(𝑋72.44)=𝑃𝑍72.44𝜇𝜎=0.6443 and 𝑃(𝑋37.76)=𝑃𝑍37.76𝜇𝜎=0.9941.

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 𝑍0.36998 and that 𝑍2.51807.

This yields the pair of simultaneous equations 72.44𝜇𝜎=0.36998 and 37.76𝜇𝜎=2.51807. We multiply both equations by 𝜎: 72.44𝜇=0.36998𝜎,37.76𝜇=2.51807𝜎.

Then, we subtract the second from the first to get 72.44𝜇(37.76𝜇)=0.36998𝜎(2.51807𝜎)34.68=2.88805𝜎.

Therefore, we have 𝜎=34.682.88805=12.0081.

We can now substitute 𝜎=12.0081 back into the equation 72.44𝜇=0.36998𝜎, which gives us 72.44𝜇=0.36998×12.0081𝜇=67.9972.

We arrive at values of 𝜇=68 and 𝜎=12, 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 𝑋𝑁3.25,𝜎. Given that 𝑃(𝑋>2𝑎)=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 𝑋𝑍=𝑋𝜇𝜎, where the mean is 𝜇=3.25. Now 𝑍𝑁0,1 follows the standard normal distribution and 𝑃(𝑋>2𝑎)=𝑃𝑍>2𝑎3.25𝜎=0.1 and 𝑃(𝑋<𝑎)=𝑃𝑍<𝑎3.25𝜎=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>1.28155 and that 𝑍<0.5244. This yields the pair of simultaneous equations 2𝑎3.25𝜎=1.28155 and 𝑎3.25𝜎=0.5244.

We multiply both equations by 𝜎: 2𝑎3.25=1.28155𝜎,𝑎3.25=0.5244𝜎.

Then, we multiply the second of these by 2: 2𝑎6.5=1.0488𝜎.

We can now eliminate 𝑎 by subtracting the second equation from the first: 2𝑎3.25(2𝑎6.5)=1.28155𝜎(1.0488𝜎)3.25=2.33035𝜎𝜎=1.39464.

To find the value of 𝑎, we can substitute back into 𝑎3.25=0.5244𝜎: 𝑎3.25=0.5244×1.39464𝑎=3.250.7313=2.5186.

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 10.56% of the flowers are shorter than 47 cm, determine 𝜇.

Answer

We have a normal random variable 𝑋𝑁𝜇,12 with unknown mean. To convert the population percentage of 10.56% into a probability, we divide by 100, so we have 𝑃(𝑋<47)=0.1056.

In order to find the unknown mean 𝜇, we code 𝑋 by the change of variables 𝑋𝑍=𝑋𝜇𝜎, where the standard deviation is 𝜎=12. Now 𝑍𝑁0,1 follows the standard normal distribution and 𝑃(𝑋<47)=𝑃𝑍<47𝜇12=0.1056.

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<1.25027. Thus, we have 47𝜇12=1.25027𝜇=(1.25027)×1247𝜇=62.00324, giving us 𝜇=62 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=63cm and standard deviation 𝜎. Given that the lengths of 84.13% of the plants are less than 75 cm, find the variance.

Answer

We have a normal random variable 𝑋𝑁63,𝜎 with unknown variance. To convert the population percentage of 84.13% into a probability, we divide by 100, so we have 𝑃(𝑋<75)=0.8413.

In order to find the unknown variance 𝜎, we code 𝑋 by the change of variables 𝑋𝑍=𝑋𝜇𝜎, where the mean 𝜇=63. Now 𝑍𝑁0,1 follows the standard normal distribution and 𝑃(𝑋<75)=𝑃𝑍<7563𝜎=0.8413.

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<0.99982. Thus, we have 7563𝜎=0.99982𝜎=120.99982=12.00216.

Therefore, our variance is 𝜎=(12.00216)=144, to the nearest integer.

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

Key Points

  • Given a normal random variable 𝑋𝑁𝜇,𝜎 and a probability 𝑃(𝑋<𝑥)=𝑃, we can code 𝑋 by the change of variables 𝑋𝑍=𝑋𝜇𝜎, where 𝑍0,1. 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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